sensitivities in asset liability management

Sensitivities in Asset Liability Management

Author:

Hao Sun

ANR: 433182

MSc. Quantitative Finance and Actuarial Science, Tilburg University 2016

Tilburg School of Economics and Management

Tilburg University

Thesis supervisor:

Prof. dr. Hans Schumacher

Second reader:

Prof. dr. B.J.M. Werker

1

Abstract

The typical approach for computing quantities of interest in ALM is Monte Carlo. Either a real-

world or a risk-neutral scenario set is used, depending on whether the quantity of interest

relates to risk management or represents a market value. In order to compute sensitivities, the

standard method is to generate a new scenario set in which the input parameter is slightly

changed, and to compare the new value of the quantity of interest with the one obtained from

the original scenario set. This is the so called “bump-and-revalue” method. Over the past

decades, several methods have been developed that allow the estimation of sensitivities on the

basis of a single scenario set. An example of such a method is the pathwise method. I apply

both bump-and-revalue method and pathwise method in the ALM model of a life insurance

company. It is found that the bump and revalue method is easy to implement but estimators

have bias and large standard errors. In contrast, although pathwise method requires additional

theoretical differentiation work before simulation, the thesis shows that it has clear advantages

and gives the best estimators.

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Table of Contents 1. Introduction ................................................................................................................................ 3

2. Estimation methods .................................................................................................................... 5

Bump and revalue ....................................................................................................................... 6

Symmetric method ................................................................................................................. 7

Confidence Interval ..................................................................................................................... 8

Pathwise method ........................................................................................................................ 8

3. Application - Life Insurance Company ...................................................................................... 11

Model definition ....................................................................................................................... 11

Investment Policy ...................................................................................................................... 13

Actuarial Reserve ...................................................................................................................... 15

Financial model ......................................................................................................................... 18

4: Target variable and parameter variable ................................................................................... 20

Summary of formulas................................................................................................................ 26

Equity expression (target variable) ........................................................................................... 27

5. Differentiation Process ............................................................................................................. 28

Simulation and results .............................................................................................................. 30

6. Sensitivity of Standard Deviation of Equity w.r.t. Asset Mix Parameter .................................. 36

7. Vasicek Model ........................................................................................................................... 42

8. Conclusion ................................................................................................................................. 50

Appendix ....................................................................................................................................... 52

A.1 Mortality rate table ............................................................................................................ 52

A.2 Matlab code ........................................................................................................................ 53

A.2.1 Code for bump and revalue method in Part5.............................................................. 53

A.2.2 Code for pathwise method in Part5 ............................................................................ 54





Bibliography .................................................................................................................................. 66

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1. Introduction

Traditional analysis of asset management used to focus only on assets allocation without taking

liabilities into consideration, which leads to poor performance in risk management. In 1970s,

Asset Liability Management (ALM) is firstly pioneered by Anglo-Saxon financial institutions

because of increasingly fluctuating interest rates. At the beginning, ALM is just an idea of

maturity matching of assets and liabilities to mitigate the risk of failing to reach the planning

targets, such as meeting current and future cash flow obligations. In current years, ALM has

become a complete framework which consists of comprehensive terms. The success of ALM

could result in larger profits and better risk management structure. Furthermore, especially in

the financial field, ALM has been widely applied. It can be seen as a systematic mechanism to

manage the risk that results from mismatches between assets and liabilities.

The purpose of asset liability management (ALM) is to obtain an insight in the development

tendency and interdependency of assets and liabilities of entities. Taking a life insurance

company as example, available capital needs to be invested profitably (such as stock, bonds) in

terms of capital part while payments for policyholders need to be made according to specific

insurance policies in the contract in terms of liability part.

Sensitivity analysis is the study of how a different value of a parameter variable influences the

value of a related target variable. In reality, when a situation (parameter variable) becomes

different, sensitivity analysis is a method to predict the corresponding outcome (target

variable). This plays an important role in establishing the validity of any proposed financial

conclusion through simulation.

In ALM, the reasons (factors) causing the mismatch between assets and liabilities could be

fluctuating interest rates or liquidity of assets. Therefore, it is important to estimate the

sensitivity with respect to these factors in order to reach a better performance. In this thesis, I

investigate the applicability of some modern sensitivity estimation methods into ALM models

and combine sensitivity analysis and ALM to reach a better understanding and recognition.

The typical approach for computing quantities of interest in ALM is Monte Carlo. Generally,

there are two categories of approaches in terms of sensitivity analysis, which are finite

perturbation and infinitesimal perturbation. In this thesis, we investigate the applicability of the

pathwise method and the bump-and-revalue method in ALM models. In the following figure,

the relationship of several possible methods is shown:

4

The bump-and-revalue method in the category of finite perturbation while pathwise method is

in the category of infinitesimal perturbation. In particular, infinitesimal perturbation sensitivity

estimation method could be regarded as the limit of finite perturbation sensitivity estimation

method. The specific derivation and example of pathwise method would be shown in the part 2

- “Estimation methods”.

In a word, the purpose of this thesis is to investigate the application of the pathwise method

and the bump and revalue method in ALM models.

The remaining parts of this thesis are structured as follows. In part 3, a life insurance company

is introduced. Part 4 describes target variable and parameter variable in the fixed interest

model. I take the partial derivative with respect to the fixed interest rate and run the simulation

of sensitivities estimation in part 5. In part 6, I change the target variable and parameter

variable into standard deviation of equity and proportion of capital invested in risky assets

respectively. In part 7, another uncertainty is introduced and Vasicek model is developed.

Lastly, Part 8 concludes this thesis.

Monte Carlo Sensitivity Approaches

Finite perturbation

independent samples correlated samples

Infinitesimal perturbation

Likelihood ratiopathwise

differentiation

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2. Estimation methods

In financial mathematics, calculation of the value of derivatives is a central problem. In

mathematics, a derivative represents a rate of change. In finance, a derivative could be defined

as a contract between parties which specifies the particular conditions under which payments

are to be made between these two parties. In the field of financial derivatives, taking options as

an example, we are generally able to observe the price of an option any time in a given market.

However, the derivatives of the option price, “Greeks”, cannot be observed directly in the

market, which makes it more important and meaningful to calculate the sensitivities accurately

and directly. In the process of estimation, there are both theoretical and operational

challenges. The Monte Carlo method is often used to determine the prices of options. This

results partially from Monte Carlo’s efficiency when calculating multi-dimension financial

questions such as the ones involving interest rates or multiple assets. Besides, the reason could

also be the simple computational application, because Monte Carlo can be parallelized across

many calculating clusters.

As known, these sensitivities are called “Greeks”. To be more specific, the term "Greek" is used

in finance to refer to derivatives (in the mathematical sense of the word). In particular,

derivatives of option prices with respect to various parameters or variables are used. The term

"Greeks" refers to both first-order and second-order (higher-order) derivatives. The following

are the main examples:

Delta: S

C

; Vega:

C

Gamma: 2

2

S

C

Volga:

2

2

C

Theta: t

C

Vanna:

S

C2

Where C stands for the option price, S stands for the price of the underlying asset, stands for

the volatility, and t stands for time.

There are two directions to solve sensitivity estimation problem at the moment, which are

estimation with simulation of at least two values of the parameter of differentiation and

estimation without, respectively. In other words or specifically, the first method, bump and

revalue method, is to generate a new scenario set in which the input parameter is slightly

changed, and to compare the new value of the quantity of interest with the one obtained from

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the original scenario set. The second method, pathwise method, is just based on only the

original scenario set. As mentioned in the introduction part, the second method applies

infinitesimal perturbation sensitivity estimation method, which uses differentiation of sample

paths to estimate sensitivities.

Bump and revalue

(Glasserman, 2003) In the bump and revalue (also called Finite-Difference) method, consider a

model which depends on a parameter 𝜃, which is ranging over some interval of the real time.

𝑌(𝜃) means the output at the given parameter 𝜃.

𝛼(𝜃) = 𝐸[𝑌(𝜃)]

Glasserman (2003) states that in the example of option pricing, 𝑌(𝜃) refers to the discounted

payoff. Therefore, 𝛼(𝜃) is the price of this option because the expectation under risk-neutral

measure of the discounted payoff of the option is equal to the price. The derivatives of option

price, “Greeks”, would be different when 𝜃 refers to different components.

When 𝜃 refers to initial price of the underlying asset, the first order derivative of 𝛼(𝜃) is delta

(𝜕𝛼(𝜃)

𝜕𝜃) and the second order derivative is gamma (

𝜕2𝛼(𝜃)

𝜕𝜃2).

The different discounted payoffs obtained from 𝑛 independent simulations are denoted

by 𝑌1(𝜃),… , 𝑌𝑛(𝜃). In addition, 𝑌1(𝜃 + ℎ),… , 𝑌𝑛(𝜃 + ℎ) are another n replications of the paths

with bump size ℎ. From this, we obtain the forward-difference estimator (or one-sided

estimator):

∆̂𝐹= ∆̂𝐹(𝑛, ℎ) =�̅�𝑛(𝜃 + ℎ) − �̅�𝑛(𝜃)

ℎ

In addition, �̅�𝑛(𝜃 + ℎ) and �̅�𝑛(𝜃) represent the expectation approximated through a Monte

Carlo simulation.

�̅�𝑛(𝜃) =1

𝑛∑ 𝑌𝑖(𝜃)

𝑛

𝑖=1

The expectation of this forward-difference estimator is following:

𝐸[∆̂𝐹] = ℎ−1[𝛼(𝜃 + ℎ) − 𝛼(𝜃)]

Supposing 𝛼(𝜃) is at least twice differentiable at 𝜃, then according to Taylor’s theorem,

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𝛼(𝜃 + ℎ) = 𝛼(𝜃) + 𝛼′(𝜃)ℎ +1

2𝛼′′(𝜃)ℎ2 + 𝑜(ℎ2)

Therefore, we can have the bias in this forward-difference when )( is exactly twice

differentiable at :

𝐵𝑖𝑎𝑠[∆̂𝐹] = 𝐸[∆̂𝐹 − 𝛼′(𝜃)] =

1

2𝛼′′(𝜃)ℎ + 𝑜(ℎ)

By taking smaller h, we tend to reach a more accurate estimator. However, the bias still

remains. As a result, we can conclude that the bias in bump and revalue method is inevitable.

Symmetric method

The estimator of the “bump and revalue” method is the following:

∆̂𝐹= ∆̂𝐹(𝑛, ℎ) =�̅�𝑛(𝜃 + ℎ) − �̅�𝑛(𝜃)

ℎ

Here the bump is one-sided.

An alternative is the central-difference (a.k.a. symmetric method) estimator, which is the

following:

∆̂𝐶= ∆̂𝐶(𝑛, ℎ) =�̅�𝑛(𝜃 + ℎ) − �̅�𝑛(𝜃 − ℎ)

ℎ

Here the bump occurs both upwards and downwards.

The symmetric method is more costly because simulation of bump-and-revalue method (one

sided bump) only requires one point 𝜃 + ℎ while the symmetric method requires two points.

Supposing 𝛼(𝜃) is at least twice differentiable at 𝜃, then according to Taylor’s theorem,

𝛼(𝜃 + ℎ) = 𝛼(𝜃) + 𝛼′(𝜃)ℎ +1

2𝛼′′(𝜃)ℎ2 + 𝑜(ℎ2)

𝛼(𝜃 − ℎ) = 𝛼(𝜃) − 𝛼′(𝜃)ℎ +1

2𝛼′′(𝜃)ℎ2 + 𝑜(ℎ2)

Therefore,

𝑏𝑖𝑎𝑠(∆̂𝐶) =𝛼(𝜃 + ℎ) − 𝛼(𝜃 − ℎ)

2ℎ− 𝛼′(𝜃) = 𝑜(ℎ)

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The bias from the symmetric method is apparently less than the bias from the forward-

difference method. Both the forward-difference estimator (one-sided estimator) and the

symmetric estimator (two-sided estimator) incorporate a bias which is due to the finite-

difference approximation (bump size h). However, in symmetric method, there would be an

improvement in terms of bias correction.

It is noted that in the bump and revalue method, paths 𝑌(𝜃) and 𝑌(𝜃 + ℎ) use common

random numbers, which means that two paths are generated form the same sequence. In

simulations, it can be realized by simulating 𝜃 and 𝜃 + ℎ with the same seed for the random

number generator.

Confidence Interval

An interval estimate of a parameter is an interval or a range of values used to estimate the

parameter 𝜇. This estimate may or may not contain the value of the parameter being

estimated. In the case of a 95% confidence interval, such as [a, b], we are 95% confident that

the true parameter is in the interval [a, b]. In other words, there is 5% possibility that the true

parameter is not in the interval [a, b].

The specific calculation method is as follows.

The central limit theorem states that when the sample size (n) is large, approximately 95% of

the sample means (�̅�) will fall within ± 1.96 standard errors (𝜎) of the population mean, that is,

𝑃 (�̅� − 1.96𝜎

√𝑛< 𝜇 < �̅� + 1.96

𝜎

√𝑛) = 0.95

Therefore, [�̅� − 1.96𝜎

√𝑛, �̅� + 1.96

𝜎

√𝑛] is the confidence interval for the parameter 𝜇.

Pathwise method

Pathwise method is described by Glasserman (2003). The method tries to take advantage of

additional information about parameter dependence and the dynamics in the process of

simulation. To be more specific, additional information is about the differentiation of target

variable with respect to the parameter variable. In this approach, it is required to differentiate

on a sample-by-sample (path-by-path) basis, and compute expectation only after that.

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Mathematically, we have the following.

As noted above, while taking smaller h, we can tend to reach a more accurate estimator.

Therefore, this time we take the limit as h tends to zero and we obtain the first order derivative

of the equation:

𝛼(𝜃) = 𝐸[𝑌(𝜃)]

𝑌′(𝜃) = limℎ→0

𝑌(𝜃 + ℎ) − 𝑌(𝜃)

ℎ

Then we take the expectation of this estimator, 𝐸[𝑌′(𝜃)]. It is required that following

transformation of differentiation and equality of expectation is justified:

𝐸 [𝑑

𝑑𝜃𝑌(𝜃)] =

𝑑

𝑑𝜃𝐸[𝑌(𝜃)]

Then term 1

𝑛∑ 𝑌𝑖′(𝜃)𝑛𝑖=1 is an unbiased estimator of the sensitivity 𝛼′(𝜃).

For instance, we could consider the delta of a call option under the Black-Scholes model using

the pathwise method, which is the sensitivity of option price (C) w.r.t. 𝑆0 (initial value of

underlying asset). It is assumed that the value of underlying asset follows a Geometric

Brownian Motion (GBM).

Firstly, the equation of discounted payoff of this call option, 𝑌, is the following:

𝑌 = 𝑒−𝑟𝑇max (𝑆𝑇 − 𝐾, 0)

With 𝑆𝑇 = 𝑆0𝑒(𝑟−

1

2𝜎2)𝑇+𝜎√𝑇𝑍, 𝑍~𝑁(0,1).

In this equation, terms 𝑟, 𝜎, 𝐾, 𝑇 and 𝑆𝑇 represent the risk-free rate, volatility of underlying

asset, strike price in this call option, time to maturity and value of underlying asset at maturity

respectively.

After applying the chain rule for differentiation, we obtain:

𝑑𝑌

𝑑𝑆0=𝑑𝑌

𝑑𝑆𝑇

𝑑𝑆𝑇𝑑𝑆0

𝑑𝑌

𝑑𝑆𝑇=

𝑑

𝑑𝑆𝑇𝑒−𝑟𝑇max (𝑆𝑇 − 𝐾, 0)

It is noticed that the first term 𝑑𝑌

𝑑𝑆𝑇 fails to exist when 𝑆𝑇 = 𝐾, but this event has zero

probability. As a result, we have the equation:

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𝑑𝑌

𝑑𝑆𝑇= 𝑒−𝑟𝑇𝟏 {𝑆𝑇 > 𝐾}

Where the term 𝟏 {𝑆𝑇 > 𝐾} is the indicator function of event {𝑆𝑇 > 𝐾}.

With respect to the second term 𝑑𝑆𝑇

𝑑𝑆0, it can be noticed that 𝑆𝑇 is linear in 𝑆0. Therefore, the

following equation is obtained:

𝑑𝑆𝑇𝑑𝑆0

=𝑆𝑇𝑆0

In this way, by introducing different simulation paths 𝑆𝑇1 , … , 𝑆𝑇𝑛 which start at the same initial

value of underlying asset, we obtain the pathwise estimator for the delta:

∆̂𝑝=1

𝑛∑ 𝑒−𝑟𝑇

𝑆𝑇𝑛𝑆0𝟏 {𝑆𝑇𝑛 > 𝐾}

𝑛

𝑖=1

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3. Application - Life Insurance Company

In this part, I turn to a specific example. A life insurance company is simulated to describe an

ALM model. To be more specific, in this life insurance company, we consider a portfolio of life

insurance contracts, in which there are 1000 participants in each of the 4 following groups:

healthy male, healthy female, disabled male and disabled female. It is assumed that all

participants start at 18 years old. In addition, the insurance is paid for by an annual constant

premium which is the same for all policyholders. The benefits paid by the insurance contracts

are as follows:

(i) An annual benefit which is paid as long as the participant is alive

(ii) A one-time benefit upon death of the participant, which is indexed by the interest

rate.

The life benefits are increasing in time according to a fixed schedule. The death benefits are

indexed by the interest rate, but the base amount is lower when the time of death comes later,

again according to a fixed schedule.

It is assumed that there are no new policyholders in this case. In other words, we regard

participants of this insurance contract as a closed market. More details are given in the

following content.

Model definition

In general, the balance sheet of the ALM model in this thesis is simply the following:

Assets Liabilities

Capital 𝐂𝐤

Actuarial reserve 𝐃𝐤

Equity 𝐐𝐤

In assets part, there is only the capital part Ck. In liabilities part, it consists of two parts:

actuarial reserve part Dk and equity part Qk.

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In this thesis, actuarial reserve is defined as the following: the current value of benefits to be

paid, minus the current value of premiums to be received. The purpose of actuarial reserve is to

accomplish the promise of the insurance company to the insurance policyholders.

Equity part is paid to the shareholders of this insurance company, which is the dividend for the

individual investors. In other words, the equity part is the surplus between the capital part and

the actuarial reserve part, for which the equation is 𝑄𝑘 = 𝐶𝑘 − 𝐷𝑘. The surplus would be

distributed to shareholders.

The horizon of this life insurance product is T (years), which can be divided into k periods. The

rebalancing time interval between two periods is defined as T/k=∆t. Normally, ∆t stands for one

year, as it is in this case.

In addition, for each group 𝑖 = 1,2,3,4 at each period 𝑘 = 1, . . , 𝑇, we need the information

about death benefits 𝑇𝑘𝑖 and survival benefits 𝐸𝑘

𝑖 , premium income 𝑃𝑘𝑖 for all 𝑖 and k.

On top of that, the data of mortality rate 𝑞𝑘𝑖 should be given for all 𝑖 and k.

About the initial data, we make the following assumptions.

There are four groups of policyholders (m=4): healthy male and disabled male, healthy female

and disabled female. The initial data of each group is given as follows.

The initial survival benefit 𝐸1𝑖 is 5, 5.1, 4.8 and 4.9, which means the disabled have more survival

benefit than the healthy and the male have more than the female. In addition, survival benefits

increase 0.2 after each period for all 4 groups to compensate policyholders for the decreasing

survival probabilities.

The initial death benefit 𝑇1𝑖 is 4200, 4160, 4240 and 4200, which implies that the healthy have

more death benefit than the disabled and the female have more than the male. It is noted that

death benefits decrease 100 after each period for all 4 groups to compensate insurance

company for the increasing mortality rate.

The premium 𝑃𝑘𝑖 income is set to 17, which is constant for every policyholder at each period.

The mortality rate 𝑞𝑘𝑖 is given in the appendix.

This life insurance product is valid from age 18 to 37, which means the horizon is 20 (T=20).

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Investment Policy

As a whole, the income of this life insurance company is the premium paid by all policyholders,

which is denoted by Pk (total premium income at the beginning of period k). Investment policy

here could be thought of as the capital allocation, which is the investment distribution of the

insurance company income. In this part, I am going to determine the allocation of capital part

of this insurance company. To be more specific, it is assumed that this insurance company

invests capital both in stock and bonds.

It is assumed here that the bond investment is exclusively in zero coupon bonds with fixed

maturity (𝜏 = 15). Also it is assumed that the bonds cannot be sold until they reach the

maturity. This assumption is aimed to simplify the model because without fixed maturity

periods additional variables would be introduced. Therefore, we can focus more on the asset

liability part.

As mentioned above, the horizon of this life insurance product is divided into K periods.

Therefore, I assume that the insurance company updates its investment policy at the beginning

of all K periods. However, the investment policy is not updated arbitrarily and certain rules are

set here. To be more specific, investment managers of this insurance company distribute a fixed

percentage of the capital into the stock market with an exception as specified below.

Details about financial modeling of stocks and bonds are discussed later in the model building

part.

The following is about the allocation of capital assets to profit from investment. As stated

before, the capital is invested into stocks and zero coupon bonds.

The company holds a fixed proportion 𝑤𝑠 ∈ [0,1] of capital in stocks as an investment strategy

at each period, to avoid too much risk. The remaining capital is invested into bonds with a given

maturity of 𝜏 periods.

It is assumed that it is impossible to short bonds, to avoid too complex problems.

First of all, we need to define the variables one by one in this case:

𝑃𝑘: Premium income when kth period begins.

𝐶𝑘−1: Sum of value of assets at the end of the last period (k-1).

𝐷𝑘: Sum of value of actuarial reserve at the end of kth period

𝐸𝑘: Sum of value of survival benefits at the end of kth period

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𝑇𝑘: Sum of value of death benefits at the end of kth period, which are indexed.

𝑁𝑘: Available capital for new investment at the beginning of kth period.

𝑛𝑘−𝑖: Number of zero-coupon bonds bought at the beginning of period (k- 𝑖).

𝐴𝑘: Capital invested in stocks at the beginning of kth period.

𝑝𝑘: Return of investment portfolio.

𝑏(𝑘, 𝜏 − 𝑖): Price of the zero coupon bond at the end of kth period with maturity 𝜏 − 𝑖.

The available capital for new investment at the kth period is the capital that is free to distribute.

However, it is assumed that the zero coupon bond cannot be sold until reaching the maturity.

Therefore, the value of holding bonds is excluded from the available capital. Mathematically, it

is required to subtract the value of zero coupon bonds which are still not reaching their

maturity.

As mentioned above, the premium payment is received at the beginning of every period and

the payments are executed at the end of every period. Therefore, the payment is excluded in

the capital in the last period.

To sum up, we write following equation which defines the available capital for new investment

at the kth period:

𝑁𝑘 = 𝐶𝑘−1 + 𝑃𝑘 − ∑ 𝑛𝑘−𝑖𝑏(𝑘 − 1, 𝜏 − 𝑖)𝜏−1𝑖=1 .

The term ∑ 𝑛𝑘−𝑖𝑏(𝑘 − 1, 𝜏 − 𝑖)𝜏−1𝑖=1 is the sum of the values of zero-coupon bond held at the

end of period k-1.

If it happens that the available capital for new investment 𝑁𝑘 is less than the planned capital

invested into the stock market, then the capital that is invested in stock market is just the

available capital for new investment 𝑁𝑘, because of the assumption that bonds cannot be sold

before the maturity.

The capital invested in stocks is computed as follows:

𝐴𝑘 = min{𝑁𝑘, 𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘)}

Then, the remaining capital 𝑁𝑘 − 𝐴𝑘 at the beginning of kth period is invested in the bond

market. The number of zero-coupon bond with maturity 𝜏 at the beginning of the kth period is 𝑁𝑘−𝐴𝑘

𝑏(𝑘,𝜏).

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Actuarial Reserve

The liability of this life insurance company consists of the actuarial reserve for the policyholders

and the equity part for the shareholders. Since the equity part is the target variable in this part,

we study the composition of actuarial reserve part.

To be efficient, we assume the contracts of insurance products can be described with 4 groups

(m=4), which are healthy male and disabled male, healthy female and disabled female. These 4

groups are able to describe the general characteristics of the participants in this insurance

products.

Firstly, the following variables are defined.

𝛿0𝑖 = 1000: We assumed that the initial number of participants in all 4 groups is 1000.

Number of contracts in group 𝑖 at the end of kth period: 𝛿𝑘𝑖 = (1 − 𝑞𝑘

𝑖 )𝛿𝑘−1𝑖 , 𝑖 = 1,2,3,4.

Where 𝑞𝑘𝑖 represents the mortality rate of policyholders of the group 𝑖 during period k.

Term 𝐷𝑘𝑖 is the value of actuarial reserve for group 𝑖 at the end of the kth period.

Term 𝑃𝑘𝑖 is the premium paid by policyholders in group 𝑖 at the beginning of the kth period.

Term 𝐸𝑘𝑖 is the payment to policyholders in group 𝑖 when policyholders still survive at the end of

the kth period.

Term 𝑇𝑘𝑖 is the payment to policyholders in group 𝑖 when policyholders die at the end of the kth

period, which is interest related.

We now consider the special case in which the interest rate is assumed to be fixed. Term 𝜃 is

the fixed interest rate, which is 1% in the following application and simulation.

The formula of death benefit is as follows:

𝑇𝑘+1 =∑𝑞𝑘+1𝑖 𝛿𝑘

𝑖𝑇𝑘+1𝑖

𝑚

𝑖=1

(1 + 𝜃)k+1

According to the definition at the beginning of application part, we acknowledge that actuarial

reserve is defined as follows: the current value of benefits to be paid, minus the current value

of premiums to be received. As a result, it is essential to calculate the current value of benefits

to be paid and current value of premiums to be received separately. The general formula for

actuarial reserve is given as follows:

16

𝐷0 = (𝑑0,1(𝐸1 + 𝑇1) + 𝑑0,2(𝐸2 + 𝑇2) + ⋯+ 𝑑0,20(𝐸20 + 𝑇20)) − (𝑃1 + 𝑑0,1𝑃2 +⋯+ 𝑑0,19𝑃20)

𝐷1 = (𝑑1,1(𝐸2 + 𝑇2) + ⋯+ 𝑑1,19(𝐸20 + 𝑇20)) − (𝑃2 + 𝑑1,1𝑃3…+ 𝑑1,18𝑃20)

…

𝐷19 = 𝑑19,1(𝐸20 + 𝑇20) − 𝑃20

Therefore, we have 𝐷𝑡 = ∑ 𝑑𝑡,𝑖−𝑡(𝐸𝑖 + 𝑇𝑖)20𝑖=𝑡+1 − ∑ 𝑑𝑡,𝑖−𝑡𝑃𝑖+1

19𝑖=𝑡 .

Where 𝑑𝑖,𝑗 (𝑖 = 0,1,2… .19; 𝑗 = 1,2… .20) is the discount factor for each period.

With respect to value of actuarial reserve, the following formula (Fackler Formula) holds:

𝐷𝑘𝑖𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 ) = (1 + 𝜃)(𝐷𝑘−1𝑖 + 𝑃𝑘

𝑖)𝛿𝑘−1𝑖 − 𝛿𝑘−1

𝑖 𝑞𝑘𝑖 𝑇𝑘

𝑖(1 + 𝜃)𝑘 − 𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 )𝐸𝑘𝑖

𝐷𝑘𝑖 =

((1 + 𝜃)(𝐷𝑘−1𝑖 + 𝑃𝑘

𝑖) − 𝑞𝑘𝑖 𝑇𝑘

𝑖(1 + 𝜃)𝑘)

1 − 𝑞𝑘𝑖

− 𝐸𝑘𝑖

Then the actuarial reserve can be calculated through the recursion formula. According to the

general formula of actuarial reserve, the discount factor 𝑑𝑖,𝑗 equals 1

(1+𝜃)𝑗 in the case of fixed

interest rate. Therefore, the initial actuarial reserve is as follows:

𝐷0 = (𝐸1 + 𝑇11 + 𝜃

+𝐸2 + 𝑇2(1 + 𝜃)2

+⋯+𝐸20 + 𝑇20(1 + 𝜃)20

) − (𝑃1 +𝑃21 + 𝜃

+⋯+𝑃20

(1 + 𝜃)19)

Based on the initial actuarial reserve and recursive formula, value of actuarial reserve can be

obtained. In addition, actuarial reserve can also be deduced backwards through the general

formula:

𝐷1 = (𝐸2 + 𝑇2(1 + 𝜃)

+ ⋯+𝐸20 + 𝑇20(1 + 𝜃)19

) − (𝑃2 +𝑃31 + 𝜃

+⋯+𝑃20

(1 + 𝜃)18)

…

𝐷19 =𝐸20 + 𝑇20(1 + 𝜃)

− 𝑃20

In this formula, the idea is studying how the actuarial reserve 𝐷𝑘𝑖 at the end of kth period

relates to the actuarial reserve 𝐷𝑘−1𝑖 at the end of (k-1) period. Therefore, value of actuarial

reserve at the kth period could be obtained based on the actuarial reserve at the (k-1) period.

So basically, I will determine the actuarial reserve at the kth period based on the actuarial

reserve at the (k-1) period only in the case of fixed interest rates.

17

The relationship can be illustrated using the following figure:

Through this figure, the relationship between term 𝐷𝑘𝑖 and term 𝐷𝑘−1

𝑖 is revealed. After taking

premium payment, death payment and survival payment into account, an equation between 𝐷𝑘𝑖

and 𝐷𝑘−1𝑖 can be found, which can reflect this relationship. In this process, the premium

payment is the cash inflow. In other part, the death payment and survival payment are the cash

outflow.

Actuarial reserve Dk𝑖 multiplied by contracts number at kth period δ𝑘

i (which is equal to

term 𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 )) is equal to the total capital of actuarial reserve 𝐷𝑘 account at kth period.

Notice that premium is paid at the beginning of each period. Therefore, total payment at kth

period is the premium multiplied by the number of contracts at the end of last period (k-1).

As noted above, we have:

𝐷𝑘𝑖 =

((1 + 𝜃)(𝐷𝑘−1𝑖 + 𝑃𝑘

𝑖) − 𝑞𝑘𝑖 𝑇𝑘

𝑖(1 + 𝜃)𝑘)

1 − 𝑞𝑘𝑖

− 𝐸𝑘𝑖

Now multiply both sides by 𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 ), then the following is obtained:

𝐷𝑘𝑖𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 ) = (1 + 𝜃)(𝐷𝑘−1𝑖 + 𝑃𝑘

𝑖)𝛿𝑘−1𝑖 − 𝛿𝑘−1


𝑖(1 + 𝜃)𝑘 − 𝛿𝑘−1𝑖 (1 − 𝑞𝑘

𝑖 )𝐸𝑘𝑖

After substituting δ𝑘i into 𝛿𝑘−1

𝑖 (1 − 𝑞𝑘𝑖 ) and changing the format, we have:

(1 + 𝜃)(𝐷𝑘−1𝑖 + 𝑃𝑘

𝑖)𝛿𝑘−1𝑖 = 𝐷𝑘

𝑖 δ𝑘i + δ𝑘

i 𝐸𝑘𝑖 + 𝛿𝑘−1


𝑖(1 + 𝜃)𝑘

Therefore,

(1 + 𝜃)(𝐷𝑘−1 + 𝑃𝑘) = 𝐷𝑘 + 𝐸𝑘 + 𝑇𝑘

𝐴𝑐𝑡𝑢𝑎𝑟𝑖𝑎𝑙 𝑅𝑒𝑠𝑒𝑟𝑣𝑒 𝐴𝑐𝑐𝑜𝑢𝑛𝑡

𝑎𝑡 (𝑘 − 1) 𝑝𝑒𝑟𝑖𝑜𝑑: 𝐴𝑐𝑐𝑜𝑢𝑛𝑡 𝐷𝑘−1𝑖

𝐴𝑐𝑡𝑢𝑎𝑟𝑖𝑎𝑙 𝑅𝑒𝑠𝑒𝑟𝑣𝑒 𝐴𝑐𝑐𝑜𝑢𝑛𝑡

𝑎𝑡 𝑘𝑡ℎ 𝑝𝑒𝑟𝑖𝑜𝑑: Account 𝐷𝑘𝑖

Premium Payment 𝑃𝑘𝑖 Death & Survival Payment: 𝐸𝑘

𝑖 & 𝑇𝑘𝑖

18

The above formula illustrates how the actuarial reserve at the kth period is computed based on

the data at the last period. From this formula, we are able to deduce the following actuarial

reserve from above recursive formula.

Financial model

The investment tools are the zero coupon bonds and stocks. I will illustrate these two

investment tools in the following.

Firstly, we discuss the zero coupon bond for the case in which the interest rate is fixed.

Supposing that the face value of this zero-coupon bond is FV and the fixed interest rate is 𝜃, the

value of the zero coupon bond at kth period still with maturity term 𝑖 can be calculated in this

way:

bk(𝑖) =FV

(1 + 𝜃)𝑖

The period 𝜏 (tau) is referred to as the fixed maturity, which also represents full maturity here.

In the formula, we can notice that the value of this zero coupon bond increases as you get

closer to maturity.

For instance, at the very beginning of this zero coupon bond, with full maturity, its value

reaches the minimum, which is:

bk(𝜏) =FV

(1 + 𝜃)𝜏

The second investment tool is stocks. In this thesis, Black – Scholes model is used to describe

the financial market. It means that the innovations on the market are driven by a Brownian

motion, namely:

𝑑𝑆𝑡 = 𝜇𝑆𝑡𝑑𝑡 + 𝜎𝑠𝑆𝑡𝑑𝑊𝑠𝑡

which describes the dynamics of the stock market. The stock follows the geometric Brownian

motion.

In this Black – Scholes model, we have the following rules about the value of stock:

𝑆𝑡 = 𝑆0exp ((𝜇 −𝜎𝑠2

2)𝑡 + 𝜎𝑠𝑊𝑠𝑡)

19

𝐸(𝑆𝑡) = 𝑆0e𝜇𝑡

𝑆𝑡𝑆𝑡−1

= exp (𝜇 −𝜎𝑠2

2+ 𝜎𝑠(𝑊𝑠𝑡 −𝑊𝑠(𝑡−1)))

To make above the formulas simpler, the following terms are defined:

𝑍𝑡 = 𝑊𝑠𝑡 −𝑊𝑠(𝑡−1)

𝜶(𝒁𝒔,𝒕) = exp (𝜇 −𝜎𝑠2

2+ 𝜎𝑠(𝑊𝑠𝑡 −𝑊𝑠(𝑡−1)))

Therefore,

𝑆𝑡𝑆𝑡−1

= 𝜶(𝒁𝒔,𝒕)

Lastly, the values of the parameters in the above financial model are defined in the following

table. These values are also used in the matlab simulation.

parameter value

𝜇 7,50%

𝜎𝑠 20%

𝜃 1%

20

4: Target variable and parameter variable

In this part, the interest rate is assumed to be fixed. In sensitivity analysis, it is essential to

distinguish target variable and parameter variable. To be more specific, the parameter variable

is the independent variable. The target variable is the variable that is the output in this study.

We take the expectation of the equity Qk, 𝐸[Qk], as the target variable.

The parameter variable is the fixed interest rate: 𝜃.

To reach the equation of the equity part in the asset liability management model of the life

insurance company, it is required to write down the recursive equations of the ALM model

based on the content above concerning the balance sheet model, which includes random

variables that drive the stock price process (Brownian motion).

The balance sheet with the formulas of the assets part and the liabilities part is shown as

follows:

Assets part Liabilities part

Capital 𝐂𝐤:

Ck = (1 + 𝑝𝑘)(𝐶𝑘−1 + 𝑃𝑘) − 𝐸𝑘 − 𝑇𝑘

Actuarial reserve 𝐃𝐤:

𝐷𝑘 =∑𝛿𝑘𝑖

𝑚

𝑖=1

𝐷𝑘𝑖

Equity 𝐐𝐤:

Qk = Ck − 𝐷𝑘

Before writing a complete set of ALM equations, it is necessary to clarify the prerequisite of

pathwise method. In the process of pathwise differentiation, we can do the differentiation

while treating random variables as fixed parameters only if these random variables are

exogenous.

Therefore, in the complete set of ALM equations, it will be necessary to keep track of the

detailed composition of the bond portfolio (either in terms of the number of units of bonds of

different maturities, or in terms of the values of the different parts of the bond portfolio; both

21

are possible). The reason is that, at time t, the bonds that were bought time 𝜏 (tau) ago are

converted to cash, and this part of the portfolio becomes available for new investment. In the

simulation, to represent the detailed composition of the bond portfolio in the Matlab

implementation, a vector of fixed length 𝜏 (tau) is introduced. At each step, it is noted that the

last element of the vector represents the part that is converted to cash and becomes available

for investment; other elements are shifted one position, and a new element comes in at the

beginning. This is the technique that is used in the implementation of the matlab simulation.

Period k Period (k+1)

At the beginning of the kth period, after benefits have been paid, contributions have been

received, and the new composition of the investment portfolio has been determined, the

company has assets consisting of stocks with value 𝐴𝑘 and a bond portfolio consisting of 𝑛𝑘,𝑗

units of bonds with time to maturity 𝜏 − 𝑗 (j=0,…,tau-1). For instance, 𝑛𝑘,0 means the units of

zero coupon bonds at k-th period with time to maturity 𝜏 (full maturity) and 𝑛𝑘,𝜏−1 means the

units of zero coupon bonds at kth period with time to maturity 1 (maturing after one period).

There are 𝛿𝑘𝑖 participants in group 𝑖 at the end of kth period.

Therefore, we have:

𝑃𝑘+1 =∑𝛿𝑘𝑖𝑃𝑘+1𝑖

𝑚

𝑖=1

𝐸𝑘+1 =∑𝛿𝑘+1𝑖 𝐸𝑘+1

𝑖

𝑚

𝑖=1


𝑖𝑇𝑘+1𝑖

𝑚

𝑖=1

(1 + 𝜃)k+1

Therefore, the state variable Xk, which contains the value of stock and number of zero coupon

bonds at each period within 𝜏 (tau) periods, can be represented like this:

(𝐴𝑘 𝑛𝑘,0 ⋯ 𝑛𝑘,𝜏−1).

The length of this vector is 𝜏 + 1.

Here, 𝐴𝑘 means the capital invested into stocks at the kth period and 𝑛𝑘 means the number of

newly bought zero coupon bonds at the kth period, as mentioned in the investment policy.

22

The vector (

𝑛𝑘,0⋮

𝑛𝑘,𝜏−1) contains the numbers of zero coupon bonds held, of maturities from 1 to

full maturity at kth period.

The value of the bond portfolio is as follows:

𝑣𝑎𝑙𝑢𝑒 (

𝑛𝑘,0⋮

𝑛𝑘,𝜏−1) =∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖)

𝜏−1

𝑖=0

To make the terms easier to see, I use the following expressions rather than writing out the

components all of the time:

𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑 (

𝑛𝑘,0⋮

𝑛𝑘,𝜏−1) = 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘)

The following graph illustrates the composition of the capital part at the end of each period:

Capital at the end of each period

Holding Bonds Benefits A_k New Bonds

23

At the end of the k-th period, the new value of the stocks is given by:

𝐴𝑘𝑆𝑘+1𝑆𝑘

= 𝐴𝑘𝛼(𝑍𝑠,𝑘+1)

The new value of the bond portfolio is as follows:

𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = (1 + 𝜃)∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖)

𝜏−1

𝑖=0

=∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

It is noted that this new value is the value of the zero coupon bond portfolio after one period,

which includes the maturing zero coupon bonds.

The surviving number of participants in the group 𝑖 is given by:

𝛿𝑘+1𝑖 = 𝛿𝑘

𝑖 (1 − 𝑞𝑘𝑖 )

This number is also the number of participants at the end of next period (k+1).

The capital at the end of the kth period is: value of stocks + value of bonds - death benefits - life

benefits, which is as follows:

𝐶𝑘 = 𝐴𝑘𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) − 𝐸𝑘 − 𝑇𝑘

It is noted here that in the capital part, the term contribution 𝑃𝑘 is not included. The reason is

that the contribution has been already transferred into the available capital part, then

transferred into the stocks and bonds portfolio, which means that the term 𝐴𝑘𝛼(𝑍𝑠,𝑘+1) +

𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) includes the contribution.

In this process, the number of zero coupon bonds reaching maturity is equal to 𝑛𝑘,𝜏−1 (last row

of the bond portfolio vector). It is mentioned above that 𝑛𝑘,𝜏−1 means the units of zero coupon

bonds at k-th period with time to maturity 1 (mature after one period).

The part of the capital that is locked up in bonds at the end of kth period is as follows:

𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑 (

𝑛𝑘,0⋮

𝑛𝑘,𝜏−2) = (1 + 𝜃)∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖)

𝜏−2

𝑖=0

Also to make the terms easier to see, I use following expressions:

24

𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑 (

𝑛𝑘,0⋮

𝑛𝑘,𝜏−2) = 𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘)

The capital that is available for investment at the beginning of (k+1)th period is therefore as follows:

𝑁𝑘+1 = 𝐶𝑘 + 𝑃𝑘+1 − 𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘)

According to the investment policy, the available capital is invested into bonds of time to

maturity 𝜏 and assets according to the following strategy.

As mentioned before, we assume that the company rebalances its assets at the beginning of

each period. Consequently, investors hold a fixed proportion ws ∈ [0,1] of total capital

(including the capital that is locked up in bonds) in stocks as an investment strategy to avoid too

much risk.

Therefore, the following relation holds:

𝐴𝑘+1 = min{𝑁𝑘+1, 𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1)}

The number of newly bought zero coupon bonds is given by:

𝑛𝑘+1,0 =𝑁𝑘+1 − 𝐴𝑘+1𝑏(𝑘 + 1, 𝜏)

Therefore at the beginning of the (k+1)th period, the company has assets consisting of stocks

with value 𝐴𝑘+1 and a bond portfolio consisting of 𝑛𝑘+1,𝑖 units of bonds with time to maturity

𝜏 − 𝑖 (𝑖 =0,…, 𝜏 − 𝑖). The vector of this state variable at the beginning of the (k+1)th period is

follows:

(𝐴𝑘+1 𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1)

It is noted that there is the following relationship of numbers of zero coupon bonds between

the consecutive periods.

𝑛𝑘+1,1 = 𝑛𝑘,0

⋮

𝑛𝑘+1,𝜏−1 = 𝑛𝑘,𝜏−2

To illustrate this relationship, we use the following graph:

25

(

𝑛𝑘+1,0𝑛𝑘+1,1⋮

𝑛𝑘+1,,𝜏−2𝑛𝑘+1,𝜏−1)

(

𝑛𝑘,0𝑛𝑘,1⋮

𝑛𝑘,,𝜏−2𝑛𝑘,𝜏−1)

In general, for 𝑗 = 1,2, … , 𝜏 − 1, we have:

𝑛𝑘+2,𝑗+1 = 𝑛𝑘+1,𝑗

Therefore, we obtain the relationship between state variable (𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1) and state

variable (𝑛𝑘,0 ⋯ 𝑛𝑘,𝜏−1), which is as follows:

(𝑛𝑘+1,0 𝑛𝑘+1,1 ⋯ 𝑛𝑘+1,𝜏−2 𝑛𝑘+1,𝜏−1) = (𝑁𝑘+1 − 𝐴𝑘+1𝑏(𝑘 + 1, 𝜏)

𝑛𝑘,0 ⋯ 𝑛𝑘,𝜏−3 𝑛𝑘,𝜏−2)

From this point, the cycle for capital repeats.

To write the complete and systematic ALM equation, we are supposed to make it of this form in

the following:

Xk+1 = 𝑓(Xk, 𝑌𝑘, 𝑍𝑘)

Where, X, Y and Z are all vectors. To be more specific, the vector Xk represents the variables

that are updated in each step at time period k (the state variables), the vector 𝑍𝑘 represents

exogenous random variables (in this case, it is referred as the increment of the Brownian

motion 𝑊𝑠𝑡 that appears in the Black-Scholes equation), and the vector 𝑌𝑘 represents

deterministic input variables (in this case, these would be the survival and death benefits per

policy holder).

After analyzing the state variable, we need to write the complete ALM equation. As known, the

state variable is in this form:

(𝐴𝑘+1 𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1)

To sum up, we will investigate it separately: 𝐴𝑘+1 and (𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1).

According to the equations above:

𝐴𝑘+1 = min{𝑁𝑘+1, 𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1)}

𝐶𝑘 = 𝐴𝑘𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) − 𝐸𝑘 − 𝑇𝑘


26

After obtaining 𝐴𝑘+1, we look at the bond portfolio (𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1).

For the bond portfolio (𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1), we have:

𝑛𝑘+1,0 =𝑁𝑘+1 − 𝐴𝑘+1𝑏(𝑘 + 1, 𝜏)

=max{0, 𝑁𝑘+1 −𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1)}

𝑏(𝑘 + 1, 𝜏)

In the present case of fixed interest rate, we have:

𝑛𝑘+1,0 =max{0, 𝑁𝑘+1 − 𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1)} (1 + 𝜃)

𝜏

𝐹𝑉

Summary of formulas

The process is as follows from kth period to (k+1)th period:

𝑁𝑘+1 >> 𝐴𝑘+1 >> 𝑛𝑘+1 >> 𝐶𝑘+1 >> 𝑁𝑘+2

Given that all information at k-th period and mortality rate at all of time are known:

1. Determine 𝑁𝑘+1: Available capital for new investment at the beginning of (k+1) period.

𝑃𝑘+1 =∑𝛿𝑘𝑖𝑃𝑘+1𝑖

𝑚

𝑖=1


2. Determine 𝐴𝑘+1: Capital invested in stocks at the beginning of (k+1) period, which is equal to: 𝐴𝑘+1 = min{𝑁𝑘+1, 𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1)}

3. Determine the number of newly bought zero coupon bonds 𝑛𝑘+1,0 at the beginning of this

period

𝑛𝑘+1,0 =𝑁𝑘+1 − 𝐴𝑘+1𝑏(𝑘 + 1, 𝜏)

𝑏(𝑘 + 1, 𝜏) =𝐹𝑉

(1 + 𝜃)𝜏

Therefore, 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1) and 𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘+1) are known.

4. Determine the capital 𝐶𝑘+1 at the end of this period

𝛿𝑘+1𝑖 = 𝛿𝑘

𝑖 (1 − 𝑞𝑘𝑖 )

𝐸𝑘+1 =∑𝛿𝑘+1𝑖 𝐸𝑘+1

𝑖

𝑚

𝑖=1

27


𝑖𝑇𝑘+1𝑖

𝑚

𝑖=1

(1 + 𝜃)k+1

𝐶𝑘+1 = 𝐴𝑘+1 ∗ 𝛼(𝑍𝑠,𝑘+2) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1) − 𝐸𝑘+1 − 𝑇𝑘+1

𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = (1 + 𝜃)∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖)

𝜏−1

𝑖=0

=∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

Then it is the cycle of this development:

𝑁𝑘+2 = 𝐶𝑘+1 + 𝑃𝑘+2 − 𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘+1)

With respect to the actuarial reserve, we have:

𝐷𝑘+1 = (1 + 𝜃)(𝐷𝑘 + 𝑃𝑘+1) − 𝐸𝑘 − 𝑇𝑘

Equity expression (target variable)

In the above, we have obtained:

𝐶𝑘+1 = 𝐴𝑘+1 ∗ 𝛼(𝑍𝑠,𝑘+2) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1) − 𝐸𝑘+1 − 𝑇𝑘+1

In the investment policy, we also have found the expression of the actuarial reserve:

𝐷𝑘+1 = (1 + 𝜃)(𝐷𝑘 + 𝑃𝑘+1) − 𝐸𝑘+1 − 𝑇𝑘+1

According to the definition of the equity part, the formula of the equity account is given by:

𝑄𝑘+1 = 𝐶𝑘+1 − 𝐷𝑘+1

28

5. Differentiation Process

In the differentiation process, all is about partial derivative with respect to the interest rate 𝜃.

To avoid writing long expressions, we use the following expression:

DQk+1 =𝜕𝑄𝑘+1𝜕𝜃

Where D refers to the partial derivative with respect to interest rate 𝜃.

Similarly,

DAk+1 =𝜕𝐴𝑘+1𝜕𝜃

Therefore, the differentiation expression of equity is as follows:

𝐷𝑄𝑘+1 = 𝐷𝐶𝑘+1 − 𝐷𝐷𝑘+1

We will differentiate each part step by step in the following.

Firstly, we need to write down the expression of 𝐷𝐶𝑘+1:

𝐷𝐶𝑘+1 = 𝛼(𝑍𝑠,𝑘+2)𝐷𝐴𝑘+1 + 𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1) − 𝐷𝑇𝑘+1

To solve above differentiation, we need to solve 𝐷𝐴𝑘+1, 𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1) and 𝐷𝑇𝑘+1:

𝐷𝐴𝑘+1 = 𝐼(𝑁𝑘+1 ≤ 𝑤𝑠(𝐶𝑘 + 𝑃𝑘+1 ))𝐷𝑁𝑘+1 + 𝐼(𝑁𝑘 > 𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘 ))𝑤𝑠𝐷(𝐶𝑘 + 𝑃𝑘+1 )

Then we need to write the expressions for 𝐷𝑁𝑘+1.

According to the formulas in the summary part:


𝐷𝑁𝑘+1 = 𝐷𝐶𝑘 − 𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘)

Then we need to write the expression for 𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘)

𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘) = D∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−2

𝑖=0

Similarly, we have:

29

𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = D∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

As observed above: from the relation between the state variable (𝑛𝑘+1,0 ⋯ 𝑛𝑘+1,𝜏−1) and

the state variable (𝑛𝑘,0 ⋯ 𝑛𝑘,𝜏−1), we know the following.

For 𝑘 ≥ 𝜏, there is following relationship holds:

𝑛𝑘,𝜏−1 = 𝑛𝑘−1,𝜏−2 = ⋯ = 𝑛𝑘−𝜏+2,1 = 𝑛𝑘−𝜏+1,0

To illustrate this relationship, we use the following graph:

(

𝑛𝑘,0𝑛𝑘,1⋮

𝑛𝑘,,𝜏−2𝑛𝑘,𝜏−1)

(

𝑛𝑘−1,0𝑛𝑘−1,1⋮

𝑛𝑘−1,,𝜏−2𝑛𝑘−1,𝜏−1)

…….

(

𝑛𝑘−𝜏+2,0𝑛𝑘−𝜏+2,1

⋮𝑛𝑘−𝜏+2,𝜏−2𝑛𝑘−𝜏+2,𝜏−1)

(

𝑛𝑘−𝜏+1,0𝑛𝑘−𝜏+1,1

⋮𝑛𝑘−𝜏+1,𝜏−2𝑛𝑘−𝜏+1,𝜏−1)

Therefore, 𝐷𝑛𝑘−𝜏+1,0 = 𝐷𝑛𝑘−𝜏+2,1 = ⋯ = 𝐷𝑛𝑘−1,𝜏−2 = 𝐷𝑛𝑘,𝜏−1

𝐷𝑛𝑘,𝜏−1 = 𝐷𝑛𝑘−𝜏+1,0 = 𝐷(𝑁𝑘−𝜏+1 − 𝐴𝑘−𝜏+1)(1 + 𝜃)

𝜏

𝐹𝑉

Similarly, we have:

𝐷𝑛𝑘,𝑖 = 𝐷𝑛𝑘−𝑖,0 = 𝐷(𝑁𝑘−𝑖 − 𝐴𝑘−𝑖)(1 + 𝜃)

𝜏

𝐹𝑉

Then based on the bonds portfolio, we find the result of term 𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘+1). In the same

way, we can also obtain the result of term 𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘).

𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = D∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

=∑𝐷(𝑁𝑘−𝑖 − 𝐴𝑘−𝑖)(1 + 𝜃)

𝜏

𝐹𝑉∗ 𝑏(𝑘, 𝜏 − 𝑖 − 1) + 𝑛𝑘−𝑖,0 ∗ 𝐷𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘) = D∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−2

𝑖=0

=∑𝐷(𝑁𝑘−𝑖 − 𝐴𝑘−𝑖)(1 + 𝜃)

𝜏

𝐹𝑉∗ 𝑏(𝑘, 𝜏 − 𝑖 − 1) + 𝑛𝑘−𝑖,0 ∗ 𝐷𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−2

𝑖=0

30

Concerning the term 𝐷𝑇𝑘+1, we have:

𝐷𝑇𝑘+1 =∑(𝑘 + 1) ∗ 𝑞𝑘+1𝑖 𝛿𝑘

𝑖𝑇𝑘+1𝑖

𝑚

𝑖=1

(1 + 𝜃)k

Secondly, we need to write down the expression of 𝐷𝑘+1:

𝐷𝐷𝑘+1 = 𝐷𝑘 + 𝑃𝑘+1 + (1 + 𝜃)𝐷𝐷𝑘 − 𝐷𝑇𝑘+1

Simulation and results

In the matlab, we use 50,000 scenarios (nscenarios=50,000) to run the simulation. After running

the simulation of this model, we have the following results.

For the bump and revalue method (where bump size h=0.005), we have the following figure

showing the sensitivity of expected equity account with respect to interest rate (figure 1):

𝐹𝑖𝑔𝑢𝑟𝑒 1: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑢𝑚𝑝 𝑎𝑛𝑑 𝑟𝑒𝑣𝑎𝑙𝑢𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

0 2 4 6 8 10 12 14 16 18 20-10.5

-10

-9.5

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5x 10

6

31

In this figure, the x-axis stands for the age from 18 to 37 and the y-axis stands for sensitivity of

the expected equity part with respect to fixed interest rate. As shown, the value of the

sensitivity is negative and decreases as time goes by.

To investigate why the value of sensitivity is negative, we can check the change of capital part

and actuarial reserve part before and after the change of interest rate. It can be noted that an

increase of the interest rate after a bump results in a decrease of the capital part and an

increase of the actuarial reserve part.

Mathematically, we have:

𝑄𝑘 = 𝐶𝑘 − 𝐷𝑘

Consequently, the equity part decreases after the increase of interest rate, which results in the

negative sensitivity.

𝐹𝑖𝑔𝑢𝑟𝑒 1.1: 𝐸𝑥𝑝𝑒𝑐𝑡𝑒𝑑 𝑒𝑞𝑢𝑖𝑡𝑦 𝑝𝑎𝑟𝑡 (𝐸(𝑄𝑘)) 𝑏𝑒𝑓𝑜𝑟𝑒 𝑎𝑛𝑑 𝑎𝑓𝑡𝑒𝑟 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑒𝑠𝑡 𝑟𝑎𝑡𝑒

The changes of the capital part and the actuarial reserve part are shown as follows:

Terms/periods 9 10 11

𝐶𝑘 before bump 67457 74188 81062

0 5 10 15 20-2

0

2

4

6

8

10

12x 10

4

EQ before bump

EQ after bump

32

𝐶𝑘 after bump 59149 63753 68225

∆𝐶𝑘 -8308 -10435 -12837

𝐷𝑘 before bump 23420 25647 27752

𝐷𝑘 after bump 48751 49844 50648

∆𝐷𝑘 25331 24197 22896

With respect to the expected value of equity, it can be concluded that raising the interest rate

by 50 basis points reduces the expected value of equity by 71% on a 10-year horizon, and by

46% on a 20-year horizon.

We can obtain the confidence interval of the sensitivity of expected equity value for different

periods with 95% confidence level.

Figure 1: Confidence interval of 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘))

0 2 4 6 8 10 12 14 16 18 20-10.5

-10

-9.5

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5x 10

6

DEQ

Lower Bound

Upper Bound

33

In the above figure, we have 3 lines, which are DEQ line (estimated sensitivity of 𝐸(𝑄𝑘)), lower

bound (lower bound of confidence interval, �̅� − 1.96𝜎

√𝑛) and upper bound line (upper bound of

confidence interval, �̅� + 1.96𝜎

√𝑛).

In the next steps, the confidence interval will be applied in all cases.

In the bump and revalue method, we investigate the influence of the increment size h on the

sensitivity value next. To be more specific, we decrease the increment h from 0.005 to 0.001

and run the simulation again.

Then we have the following figure 2:

Figure 2: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑢𝑚𝑝 𝑎𝑛𝑑 𝑟𝑒𝑣𝑎𝑙𝑢𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

For the pathwise method, we have the following figure 3 concerning the sensitivity of the

expected equity account with respect to the interest rate:

0 2 4 6 8 10 12 14 16 18 20-10

-9.5

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5x 10

6

DEQ

Lower Bound

Upper Bound

34

Figure 3: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ𝑤𝑖𝑠𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

Other than the bump and revalue method (with one-sided bump) and the pathwise method, we

also try a symmetric method, in which both positive and negative bumps are applied, rather

than only a one-sided bump.

Theoretically, the symmetric method should lead to a smaller bias.

The figure of the simulation results of symmetric method is the following:

Figure 3.1: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑒𝑡ℎ𝑜𝑑

0 2 4 6 8 10 12 14 16 18 20-10

-9.5

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5x 10

6

DEQ

Lower Bound

Upper Bound

0 5 10 15 20-10

-9.5

-9

-8.5

-8

-7.5

-7

-6.5

-6

-5.5x 10

6

DEQ

Lower Bound

Upper Bound

35

Confidence intervals of sensitivities from period 18 to period 20 in the above methods are

summarized in the following table:

Method/

periods

18 19 20

Pathwise −8.796 × 106 ± 1.5 × 104 −9.137 × 106 ± 1.7 × 104 −9.52 × 106 ± 1.99 × 104

Bump-and-revalue(one sided)

ℎ = 0.001 −8.88 × 106 ± 1.6 × 104 −9.23 × 106 ± 1.8 × 104 −9.63 × 106 ± 2.05 × 104

ℎ = 0.005 −9.26 × 106 ± 1.7 × 104 −9.65 × 106 ± 1.9 × 104 −1.01 × 107 ± 2.2 × 104

symmetric method

ℎ = 0.001 −8.803 × 106 ± 1.53 × 104 −9.147 × 106 ± 1.75 × 104 −9.525 × 106 ± 2.0 × 104

With respect to the bias, from the above figures and table, it can be noted that the curve of

bump and revalue method converges to the curve of pathwise method as bump size decreases.

However, even when the bump size is small, the bias still remains. As a result, a decrease of the

bump size enhances the accuracy of estimation in the bump and revalue method but does not

eliminate the bias.

In addition, the Taylor series approximation suggests that the bias of the one-sided

approximation should be approximately linear in the size of the bump, which means that a five

times larger bump should produce a five times larger bias. This can also be seen in the above

table.

Lastly, it can also be noticed that there is no significant difference between the symmetric

method and the pathwise method. This means that the symmetric method is more accurate

than the one-sided bump and revalue method as indicated by Taylor series expansion

characteristics.

36

6. Sensitivity of Standard Deviation of Equity w.r.t. Asset Mix Parameter

In this part, we introduce another target variable: the standard deviation of the equity part,

which is denoted as follows:

𝜎(𝑄𝑘): Standard deviation of equity part at the end of kth period.

The parameter variable 𝑤𝑠 is the proportion of capital invested in risky assets (asset mix

parameter). The parameter 𝑤𝑠 will be referred to as the investment proportion.

In this part the interest rate is still assumed fixed. Then as noted above, the actuarial reserve is

deterministic. Therefore it has no volatility, which means 𝜎(𝐷𝑘) = 0.

Consequently, we have the following equation:

𝜎(𝑄𝑘) = 𝜎(𝐶𝑘 − 𝐷𝑘) = 𝜎(𝐶𝑘)

Firstly, to investigate the standard deviation of equity part, we need to obtain the variance of

the equity account: 𝑉𝑎𝑟(𝐶𝑘) because the standard deviation is the square root of variance.

𝑉𝑎𝑟(𝐶𝑘) = 𝐸[(𝐶𝑘 − 𝐸(𝐶𝑘))2]

This can also be written as:

𝑉𝑎𝑟(𝐶𝑘) = 𝐸(𝐶𝑘2) − 𝐸2(𝐶𝑘)

In the differentiation process, all is about partial derivative with respect to investment

proportion. To avoid writing long expressions, we use the following differentiation expressions:

D𝜎(𝑄𝑘) =𝜕𝜎(𝑄𝑘)

𝜕𝑤𝑠

Where D refers to the partial derivative with respect to investment proportion.

Similarly,

D𝜎(𝐶𝑘) =𝜕𝜎(𝐶𝑘)

𝜕𝑤𝑠

Therefore,

D𝜎(𝐶𝑘) = 𝐷√𝑉𝑎𝑟(𝐶𝑘) =1

2√𝑉𝑎𝑟(𝐶𝑘)𝐷𝑉𝑎𝑟(𝐶𝑘) =

1

2𝜎(𝐶𝑘)𝐷𝑉𝑎𝑟(𝐶𝑘)

37

According to the formulas of the capital part:

𝐶𝑘 = 𝐴𝑘 ∗ 𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) − 𝐸𝑘 − 𝑇𝑘

𝐴𝑘 = min{𝑁𝑘, 𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘)}

𝑁𝑘 = 𝐶𝑘−1 + 𝑃𝑘 − 𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘−1)

Therefore,

𝑉𝑎𝑟(𝐶𝑘) = 𝑉𝑎𝑟{𝐴𝑘 ∗ 𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) − 𝐸𝑘 − 𝑇𝑘}

= 𝑉𝑎𝑟{𝐴𝑘 ∗ 𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘)}

Let

𝑋𝑘 = 𝐴𝑘 ∗ 𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘)

𝑉𝑎𝑟(𝐶𝑘) = 𝑉𝑎𝑟(𝑋𝑘)

We have

𝑉𝑎𝑟(𝑋𝑘) = 𝐸[(𝑋𝑘 − 𝐸(𝑋𝑘))2]

𝑉𝑎𝑟(𝑋𝑘) = 𝐸(𝑋𝑘2) − 𝐸2(𝑋𝑘)

Then

𝐷𝑉𝑎𝑟(𝐶𝑘) = 𝐷𝑉𝑎𝑟(𝑋𝑘) = 𝐸(2𝑋𝑘𝐷𝑋𝑘) − 2𝐸{𝑋𝑘}𝐸{𝐷𝑋𝑘}

So in the next step, we need to differentiate the term X:

𝐷𝐶𝑘 = 𝐷𝑋𝑘 = 𝐷{𝐴𝑘 ∗ 𝛼(𝑍𝑠,𝑘+1) + 𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘)}

To investigate above differentiation, we need to differentiate the terms one by one.

𝐷{𝐴𝑘} = 𝐼(𝑁𝑘 ≤ 𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘 ))𝐷𝑁𝑘 + 𝐼(𝑁𝑘 > 𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘 ))𝐷(𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘 ))

𝐷𝑁𝑘 = 𝐷𝐶𝑘−1 − 𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘−1)

𝐷(𝑤𝑠(𝐶𝑘−1 + 𝑃𝑘 )) = 𝐶𝑘−1 + 𝑃𝑘 + 𝑤𝑠 ∗ 𝐷𝐶𝑘−1

Then we need to write the expression for 𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘)

38

𝐷𝑣𝑎𝑙𝑢𝑒𝑙𝑜𝑐𝑘𝑒𝑑(𝑛𝑘) = 𝐷∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−2

𝑖=0

Similarly, we have:

𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = 𝐷∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

As noted above, for 𝑘 ≥ 𝜏, the following relationship holds:

𝑛𝑘,𝜏−1 = 𝑛𝑘−1,𝜏−2 = ⋯ = 𝑛𝑘−𝜏+2,1 = 𝑛𝑘−𝜏+1,0

𝐷𝑛𝑘,𝜏−1 = 𝐷𝑛𝑘−𝜏+1,0 = 𝐷(𝑁𝑘−𝜏+1 − 𝐴𝑘−𝜏+1)(1 + 𝜃)

𝜏

𝐹𝑉

Therefore,

𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) = 𝐷∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

= ∑𝐷(𝑁𝑘−𝑖 − 𝐴𝑘−𝑖)(1 + 𝜃)

𝜏

𝐹𝑉∗ 𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

After differentiation of the above equations, we run the simulation of this model. In matlab, we

use 50,000 scenarios (nscenarios=50,000) to run the simulation. After running the simulation of

this model, we have the following results:

For the bump and revalue method (where the bump size h=0.1), we have the following figure

concerning the standard deviation of the equity account with respect to the investment

proportion 𝑤𝑠.

39

Figure 4: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑡𝑦 𝑝𝑎𝑟𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑢𝑚𝑝_𝑟𝑒𝑣𝑎𝑙𝑢𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

In this figure, the x-axis stands for the age from 18 to 37 and the y-axis stands for the sensitivity

of the standard deviation of equity part with respect to investment proportion 𝑤𝑠. As shown,

the value of the sensitivity is positive and increases as time goes by.

The reason is that the increase of the investment proportion into the stock increases the

volatility of the equity part.

In this bump and revalue method, we investigate the influence of the increment size h on the

sensitivity value next. To be specific, the increment h is decreased from 0.1 to 0.01. We run the

simulation again.

Then we have the following figure 5 (on the next page):

0 5 10 15 200

0.5

1

1.5

2

2.5

3

3.5

4x 10

5

DstdQ

Lower Bound

Upper Bound

40

Figure 5: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑡𝑦 𝑝𝑎𝑟𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑏𝑢𝑚𝑝_𝑟𝑒𝑣𝑎𝑙𝑢𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

For the pathwise method, we have the following figure 6 concerning the sensitivity of the

standard deviation of the equity account with respect to the investment proportion 𝑤𝑠.

Figure 6: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦 𝑜𝑓 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑞𝑢𝑖𝑡𝑦 𝑝𝑎𝑟𝑡 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ𝑤𝑖𝑠𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

0 2 4 6 8 10 12 14 16 18 200

0.5

1

1.5

2

2.5

3

3.5

4x 10

5

DstdQ

Lower Bound

Upper Bound

0 5 10 15 200

0.5

1

1.5

2

2.5

3

3.5x 10

5

DstdQ

Lower Bound

Upper bound

41

The confidence intervals of sensitivities from period 18 to period 20 in the above methods are

summarized in the following table:

Method/periods 18 19 20

Pathwise 2.81 × 105 ± 6.6 × 102 3.09 × 105 ± 8.3 × 102 3.39 × 105 ± 8.8 × 102

Bump-and-revalue

ℎ = 0.01 2.88 × 105 ± 2.5 × 103 3.15 × 105 ± 2.8 × 103 3.46 × 105 ± 3.1 × 103

ℎ = 0.1 3.24 × 105 ± 2.9 × 103 3.57 × 105 ± 3.1 × 103 3.93 × 105 ± 3.5 × 103

It can be noted that the result of the bump-and-revalue method is closer to the result of

pathwise method as the bump size decreases. However, even with small bump size, the bias

still remains. As a result, a decrease of the bump size enhances the accuracy of estimation in

bump and revalue method but does not eliminate the bias. In addition, the standard deviation

of estimation in pathwise method is smaller as well, as suggested by the comparison of

confidence intervals in the above table. As the bias increases owing to a larger bump size, the

confidence interval in the bump and revalue method becomes wider.

Therefore, the pathwise method has unbiased estimation result and possibly less standard

deviation, which means that the pathwise method is an advanced method in our application of

investigating the sensitivities of the standard deviation of the equity part with respect to the

investment proportion in the asset liability management model of life insurance products that

is used here.

42

7. Vasicek Model

Compared to the fixed interest rate, a new uncertainty is introduced into the interest rate to

make it volatile. In practice, interest rates are not constant.

We use a one factor short rate model, namely the Vasicek model, to model the interest rate

environment (other models, such as Hull-White model or Cox-Ingersoll–Ross model can be

applied as well).

𝑑𝑟𝑡 = −𝛼(𝑟(𝑡) − 𝜃)𝑑𝑡 + 𝜎𝑟𝑑𝑊𝑟𝑡

Terms 𝛼, 𝜃, 𝜎𝑟 mean the speed of reversion, long-term expected value of the short rate and

volatility of the short interest rate respectively.

This is called “mean reverting process”. To be specific, when 𝑟(𝑡) > 𝜃, we have a negative drift,

while when 𝑟(𝑡) < 𝜃, we have a positive drift. As a result, the drift is always directed to the

long-term expected value 𝜃 of the short rate.

When 0<=s<=t, the short rate satisfies:

𝑟(𝑡) = 𝑟(𝑠)𝑒−𝛼(𝑡−𝑠) + 𝜃(1 − 𝑒−𝛼(𝑡−𝑠)) + 𝜎𝑟∫ 𝑒−𝛼(𝑡−𝑢)𝑑𝑊(𝑢)𝑡

𝑠

When s=0 and the initial interest rate 𝑟(0) is known, we have:

𝑟(𝑡) = 𝑟(0)𝑒−𝛼𝑡 + 𝜃(1 − 𝑒−𝛼𝑡) + 𝜎𝑟∫ 𝑒−𝛼(𝑡−𝑢)𝑑𝑊(𝑢)𝑡

0

It can be concluded that:

(1) The short rate 𝑟𝑡 has a normal distribution;

(2) The expected value of 𝑟𝑡 is equal to 𝑟(0)𝑒−𝛼𝑡 + 𝜃(1 − 𝑒−𝛼𝑡) because the mean of the

stochastic integral is 0.

In this model, the price of a zero coupon bond at time 𝑡 ∈ [0,20] that pays one unit of currency

at maturity 𝜏 is expressed as follows (Mamon, 2004):

𝑏(𝑡, 𝜏) = 𝐴(𝜏)𝑒−𝐵(𝜏)𝑟(𝑡)

This is an exponential affine function of the prevailing short interest rate 𝑟(𝑡) with

𝐴(𝜏) = exp {(𝜃 −𝜎𝑟2

2𝛼2) (𝐵(𝜏) − 𝜏) −

𝜎𝑟2

4𝛼𝐵2(𝜏)};

43

𝐵(𝜏) =1−e−𝛼𝜏

𝛼.

Values of the parameters in the above financial model are defined in the following table, which

are also used in the matlab simulation.

parameter value

𝛼 15%

𝜎𝑠 1.5%

𝜃 1%

In addition, we assume that the initial interest rate 𝑟0 is 0.5%. Then we can simulate the

interest rate at different periods in the Vasicek model.

As mentioned with respect to the “actuarial reserve” in part 3, according to the definition at the

beginning of application part, we acknowledge that actuarial reserve is defined as follows: the

current value of benefits to be paid, minus the current value of premiums to be received. We

have the following equations:

𝐷0 = (𝑑0,1(𝐸1 + 𝑇1) + 𝑑0,2(𝐸2 + 𝑇2) + ⋯+ 𝑑0,20(𝐸20 + 𝑇20)) − (𝑃1 + 𝑑0,1𝑃2 +⋯+ 𝑑0,19𝑃20)

𝐷1 = (𝑑1,1(𝐸2 + 𝑇2) + ⋯+ 𝑑1,19(𝐸20 + 𝑇20)) − (𝑃2 + 𝑑1,1𝑃3…+ 𝑑1,18𝑃20)

…

𝐷19 = 𝑑19,1(𝐸20 + 𝑇20) − 𝑃20

Therefore, we have 𝐷𝑡 = ∑ 𝑑𝑡,𝑖−𝑡(𝐸𝑖 + 𝑇𝑖)20𝑖=𝑡+1 − ∑ 𝑑𝑡,𝑖−𝑡𝑃𝑖+1

19𝑖=𝑡

Where 𝑑𝑖,𝑗 (𝑖 = 0,1,2… .19; 𝑗 = 1,2… .20) is the discount factor for each period and the

equation for the discount factor is 𝑑𝑖,𝑗 = 𝑏(𝑖, 𝑗).

Concerning death benefits 𝑇𝑘+1 when 𝑡 = 𝑘 + 1, it is defined that the equation is as follows:


𝑖𝑇𝑘+1𝑖

𝑚

𝑖=1

𝐵𝐿𝑘+1

44

𝐵𝐿𝑘+1 =1

𝑑𝑘,1𝐵𝐿𝑘

Where term 𝐵𝐿𝑘 is the benefit level. In this case, death benefit is indexed by benefit level.

Then we need to differentiate this asset liability model in the Vasicek model. For simplicity, we

use term D to stand for the differentiation with respect to the long-term mean of the short

rate 𝜃.

Compared to the differentiation process above in the fixed interest rate case, the difference in

this part is in the following terms:

𝐷𝑏(𝑡, 𝜏) =𝜕𝑏(𝑡, 𝜏)

𝜕𝜃= 𝐷𝐴(𝜏) ∗ 𝑒−𝐵(𝜏)𝑟(𝑡) + 𝐴(𝜏) ∗ 𝐷𝑒−𝐵(𝜏)𝑟(𝑡)

𝐷𝑟(𝑡) = 1 − 𝑒−𝛼𝑡

𝐷𝐴(𝜏) = (𝐵(𝜏) − 𝜏)𝐴(𝜏)

𝐷𝑒−𝐵(𝜏)𝑟(𝑡) = −𝐵(𝜏)𝑒−𝐵(𝜏)𝑟(𝑡)𝐷𝑟(𝑡) = −𝐵(𝜏)𝑒−𝐵(𝜏)𝑟(𝑡)(1 − 𝑒−𝛼𝑡)

Therefore, we will have:

𝐷𝑏(𝑡, 𝜏) = (𝐵(𝜏) − 𝜏)𝐴(𝜏) ∗ 𝑒−𝐵(𝜏)𝑟(𝑡) − 𝐴(𝜏) ∗ 𝐵(𝜏)(1 − 𝑒−𝛼𝑡)𝑒−𝐵(𝜏)𝑟(𝑡)

= 𝑒−𝐵(𝜏)𝑟(𝑡)[𝐴(𝜏)𝐵(𝜏)𝑒−𝛼𝑡 − 𝐴(𝜏) ∗ 𝜏]

Therefore, we obtain the partial derivative with respect to 𝜃 of the number of bonds at

different periods:

𝐷𝑛𝑘,0 = 𝐷𝑁𝑘 − 𝐴𝑘𝑏(𝑘, 𝜏)

=𝐷(𝑁𝑘 − 𝐴𝑘) ∗ 𝑏(𝑘, 𝜏) − 𝐷𝑏(𝑘, 𝜏) ∗ (𝑁𝑘 − 𝐴𝑘)

𝑏(𝑘, 𝜏)2

𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) =∑𝑛𝑘,𝑖𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

𝐷𝑣𝑎𝑙𝑢𝑒𝑒𝑛𝑑(𝑛𝑘) =∑𝐷𝑛𝑘,𝑖 ∗ 𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

+∑𝑛𝑘,𝑖 ∗ 𝐷𝑏(𝑘, 𝜏 − 𝑖 − 1)

𝜏−1

𝑖=0

𝐷𝑑𝑖,𝑗 = 𝑒−𝐵(𝑗)𝑟(𝑖)[𝐴(𝑗)𝐵(𝑗)𝑒−𝛼𝑖 − 𝐴(𝑗) ∗ 𝑗]

𝐷𝐵𝐿𝑘 = 𝐷(1

𝑑𝑘−1,1𝐵𝐿𝑘−1) = 𝐵𝐿𝑘−1𝐷

1

𝑑𝑘−1,1+

1

𝑑𝑘−1,1𝐷𝐵𝐿𝑘−1

45

𝐷𝑇𝑘 =∑𝑞𝑘𝑖 𝛿𝑘−1𝑖 𝑇𝑘

𝑖

𝑚

𝑖=1

𝐷𝐵𝐿𝑘

As noted above: 𝐷𝑡 = ∑ 𝑑𝑡,𝑖−𝑡(𝐸𝑖 + 𝑇𝑖)20𝑖=𝑡+1 − ∑ 𝑑𝑡,𝑖−𝑡𝑃𝑖+1

19𝑖=𝑡 , when 𝑡 = 𝑘

𝐷𝐷𝑘 = ∑ 𝐷𝑑𝑡,𝑖−𝑡 ∗ (𝐸𝑖 + 𝑇𝑖)

20

𝑖=𝑡+1

+ ∑ 𝑑𝑡,𝑖−𝑡 ∗ 𝐷(𝐸𝑖 + 𝑇𝑖)

20

𝑖=𝑡+1

−∑𝐷𝑑𝑡,𝑖−𝑡 ∗ 𝑃𝑖+1

19

𝑖=𝑡

After differentiation of the above equations, then we run the simulation of this model. In

matlab, we use 50,000 scenarios (nscenarios=50,000) to run the simulation. After running the

simulation of this model, we have the following results.

For the bump and revalue method (where the bump size h=0.01), we have the following figure

concerning the sensitivity of the equity account with respect to the long-term expected value of

the short interest rate (figure 7):


In this figure, the x-axis stands for the age from 18 to 37 and the y-axis stands for the sensitivity

of the expected equity with respect to the long-term expected value of the short interest rate.

As shown, the value of sensitivity is negative and decreases as time goes by.

0 2 4 6 8 10 12 14 16 18 20-6

-5.5

-5

-4.5

-4

-3.5

-3x 10

6

DEQ

Lower Bound

Upper Bound

46

To investigate why the value of sensitivity is negative, we can check the change of the capital

part and the actuarial reserve part before and after the change of the long-term expected value

of the short interest rate. It can be noted that an increase of the long-term expected value of

the short interest rate after a bump results in a decrease of the capital part and an increase of

the actuarial reserve part. Consequently, the equity part decreases after the increase of the

long-term expected value of the short interest rate, which results in the negative sensitivity.

Mathematically, we have:

𝑄𝑘 = 𝐶𝑘 − 𝐷𝑘

The changes of the capital part and the actuarial reserve part are shown as follows:

Terms/periods 9 10 11

𝐶𝑘 before bump 70293 77258 84389

𝐶𝑘 after bump 65113 70373 75398

∆𝐶𝑘 -5180 -6885 -8991

𝐷𝑘 before bump 18461 21535 24582

𝐷𝑘 after bump 47906 50187 52231

∆𝐷𝑘 29445 28652 27649

In the context of the bump and revalue method, we investigate the influence of the increment

size h on the sensitivity value in the next step. To be specific, the increment h is decreased from

0.01 to 0.005. We run the simulation again. Then we have the following figure 8:

47


For the symmetric method, the following figure 8.1 is obtained:

Figure 8.1: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑠𝑦𝑚𝑚𝑒𝑡𝑟𝑖𝑐 𝑚𝑒𝑡ℎ𝑜𝑑

0 5 10 15 20-5.5

-5

-4.5

-4

-3.5

-3

-2.5x 10

6

DEQ

Lower Bound

Upper Bound

0 2 4 6 8 10 12 14 16 18 20-5.5

-5

-4.5

-4

-3.5

-3x 10

6

DEQ

Lower Bound

Upper Bound

48

For the pathwise method, we have the following figure 9 and table concerning the sensitivity of

expected equity account with respect to the long-term expected value of the short interest

rate:

Figure 9: 𝑆𝑒𝑛𝑠𝑖𝑡𝑖𝑣𝑖𝑡𝑦(𝐷𝐸(𝑄𝑘)) 𝑖𝑛 𝑡ℎ𝑒 𝑝𝑎𝑡ℎ𝑤𝑖𝑠𝑒 𝑚𝑒𝑡ℎ𝑜𝑑

Confidence intervals of sensitivities from period 18 to period 20 in above methods are

summarized in following table:

Method/periods 18 19 20

Pathwise −4.29 × 106 ± 1.65 × 104 −4.47 × 106 ± 1.87 × 104 −4.67 × 106 ± 2.15 × 104

Bump-and-revalue

ℎ = 0.005 −4.69 × 106 ± 3.8 × 105 −4.89 × 106 ± 4.06 × 105 −5.09 × 106 ± 4.3 × 105

ℎ = 0.01 −4.92 × 106 ± 1.97 × 105 −5.14 × 106 ± 2.1 × 105 −5.41 × 106 ± 2.2 × 105

Symmetric method

ℎ = 0.005 −4.51 × 106 ± 1.8 × 105 −4.70 × 106 ± 1.9 × 105 −4.89 × 106 ± 2 × 105

0 5 10 15 20-4.8

-4.6

-4.4

-4.2

-4

-3.8

-3.6

-3.4

-3.2

-3x 10

6

DEQ

Lower Bound

Upper Bound

49

It can also be noted that the curve of the bump and revalue method converges to the curve of

pathwise method as the bump size decreases. However, even when the bump size is small, the

bias still remains. As a result, a decrease of the bump size enhances the accuracy of estimation

in the bump and revalue method but does not eliminate the bias, which suggests that pathwise

method is a more advanced estimation method in our application of investigating the

sensitivities of expected equity part with respect to the long-term expected value of the short

interest rate in Vasicek model.

The standard deviation of the estimation in the pathwise method is smaller as well, as seen

from the comparison of confidence intervals in the above table. Therefore, the pathwise is a

more advanced method with unbiased estimation and lower standard deviation. While the bias

increases owing to larger bump size, the confidence interval in the bump and revalue method

appears to become narrower. In addition, the Taylor series approximation suggests that the

bias of the one-sided approximation should be approximately linear in the size of the bump,

which means twice larger bump should produce twice larger bias. This could also be shown in

the above table.

Lastly, it can also be noticed that the symmetric method obtains a more accurate result and

lower standard deviation than the one-sided bump and revalue method. The result is closer to

the result in the pathwise method but still shows a significant difference. Owing to the

additional stochastic element in the Vasicek model, the standard deviation of the symmetric

method is significantly larger than the standard deviation in the fixed interest rate model, which

results in a wider confidence interval in the Vasicek model.

In this case, the standard deviation of the pathwise method is substantially lower, which is not

found in the other cases. This implies that advantages of the pathwise method are more

remarkable when another uncertainty is introduced.

50

8. Conclusion

In this thesis, we have investigated the application of the pathwise method and the bump and

revalue method in estimating sensitivities in an asset liability model of a life insurance

company.

To illustrate and compare the two methods, we use three cases to run the simulation

separately, which are the following:

Cases/variables Target variable Parameter variable

Case 1 Expectation of equity Fixed interest rate

Case 2 Standard deviation of equity Investment proportion

Case 3 Expectation of equity Long-term mean of short rate

To sum up, strength and weakness of two methods are listed as follows:

Strength Weakness

Bump and revalue method Easily to implement Biased results

Pathwise method Unbiased results with less

variance

Additional work of

differentiation

Through three cases in this thesis, It can be concluded that unbiased estimates of sensitivities

with lower standard errors could be obtained using the pathwise method because of the step

by step differentiation, while the bump and revalue method gives biased estimates. With

smaller bump size, the bias can be decreased accordingly. In the figures of each case, the curve

of sensitivity in the bump and revalue method is closer to the curve of sensitivity in the

pathwise method as the bump size decreases.

In addition, when the bump occurs both upwards and downwards, the bump and revalue

method becomes the symmetric method. In the above three cases, it has been found that the

symmetric method obtains a more accurate result and a lower standard deviation than in the

one-sided bump and revalue method.

51

Computational speeds of the pathwise method and the bump and revalue method are

expected to be comparable. To be specific, if only one input parameter variable is introduced,

then the bump and revalue method has one additional simulation, while the pathwise method

has additional differentiation steps, which are likely to lead to almost the same computational

speed. As the number of parameter variables increases, the pathwise method and the one-

sided method should still have approximately the same computational speed. However, the

symmetric method is expected to become relative slower when more input parameters are

used.

To summarize briefly, when estimating sensitivities, the bump and revalue method is easy to

implement but has estimators with bias and larger standard errors. In contrast, although the

pathwise method requires additional theoretical differentiation work before simulation, this

thesis has shown that it has clear advantages and gives the best estimations.

52

Appendix

A.1 Mortality rate table

Mortality rate table applied in this thesis:

Source: RP-2014 Mortality Tables released by the Society of Actuaries' (SOA’s) Retirement Plans

Experience Committee (RPEC).

AGE Healthy Male Disabled Male Healthy Female Disabled Female

18 0,000449 0,005744 0,000206 0,002162

19 0,000444 0,006462 0,000218 0,002231

20 0,000446 0,007110 0,000231 0,002231

21 0,000452 0,007863 0,000244 0,002231

22 0,000463 0,008546 0,000258 0,002231

23 0,000477 0,008914 0,000272 0,002286

24 0,000492 0,009036 0,000286 0,002328

25 0,000508 0,008476 0,000300 0,002383

26 0,000523 0,008090 0,000318 0,002465

27 0,000536 0,007863 0,000339 0,002576

28 0,000551 0,007775 0,000365 0,002700

29 0,000570 0,007810 0,000396 0,002837

30 0,000595 0,007915 0,000433 0,003003

31 0,000628 0,008108 0,000477 0,003182

32 0,000671 0,008353 0,000529 0,003361

33 0,000725 0,008616 0,000589 0,003553

34 0,000793 0,008896 0,000657 0,003746

35 0,000876 0,009159 0,000733 0,003939

36 0,000973 0,009386 0,000816 0,004132

37 0,001087 0,009649 0,000906 0,004380

53

A.2 Matlab code

A.2.1 Code for bump and revalue method in Part5 % ALM analysis of life insurance product % bump and revalue method in fixed interest rate case clear load E load q load T

% given parameters nscenarios=50000; sigma=0.2; mu=0.075; tau=15; FV=10; theta_bm=0.01;% benchmark interest h=0.005;% bump size K=20; %length of years: 18 to 37 years old w_S=0.5; % investment proportion P=17; % premium

BEQ=zeros(2,K); Z=randn(nscenarios,K); thetas=[theta_bm-h theta_bm+h]; for j=1:2 theta=thetas(j); delta=[1000 1000 1000 1000]; Benefit=0; Premium=0; for i=1:20 Premium=Premium+P*sum(delta)* (1+theta)^(-i)*(1+theta); Benefit=Benefit+theta_matrix(i)*(E(i,:)*((1-

q(i,:)).*delta)'+(T(i,:)*(1+theta)^i)*(q(i,:).*delta)'); delta=(1-q(i,:)).*delta;% new delta end D0=Benefit-Premium;%initial actuarial reserve

EC=zeros(1,K); D=zeros(1,K); EQ=zeros(1,K); % the cycle would be N >> A >> n >> C >> N delta=[1000 1000 1000 1000]; n=zeros(nscenarios,tau); %n(:,1) indicates number of bonds with maturity

tau N=sum(delta)*P*ones(nscenarios,1); A=w_S*N; n(:,1)=(N-A)/((1+theta)^(-tau)*FV); % compute quantities at the end of first period V_end=(1+theta)*sum(n*diag((1+theta).^(-tau:-1)),2)*FV; C=A.*exp(mu-sigma^2/2+sigma*Z(:,1))+V_end-E(1,:)*((1-q(1,:)).*delta)'-

(1+theta)*T(1,:)*(q(1,:).*delta)';% EC(1)=mean(C);

54

D(1)=(1+theta)*(D0+sum(delta)*P)-E(1,:)*((1-q(1,:)).*delta)'-

(1+theta)*T(1,:)*(q(1,:).*delta)'; EQ(1)=EC(1)-D(1); delta=(1-q(1,:)).*delta; %survivors at the end of period 1 % loop for t=2:K V_locked=sum(n(:,1:tau-1)*diag((1+theta).^(-tau+1:-1)),2)*FV; N=C+P*sum(delta)-V_locked; A=min(N,w_S*(C+P*sum(delta))); n(:,2:tau)=n(:,1:tau-1); n(:,1)=(N-A)/((1+theta)^(-tau)*FV); V_end=(1+theta)*sum(n*diag((1+theta).^(-tau:-1)),2)*FV; C=A.*exp(mu-sigma^2/2+sigma*Z(:,t))+V_end-E(t,:)*((1-

q(t,:)).*delta)'-(T(t,:)*(1+theta)^t)*(q(t,:).*delta)'; EC(t)=mean(C); D(t)=(1+theta)*(D(t-1)+P*sum(delta))-E(t,:)*((1-q(t,:)).*delta)'-

(T(t,:)*(1+theta)^t)*(q(t,:).*delta)'; delta=(1-q(t,:)).*delta; % new delta EQ(t)=EC(t)-D(t); end BEQ(j,:)=EQ; end DEQ=(BEQ(2,:)-BEQ(1,:))/(thetas(2)-thetas(1)); plot(DEQ);

A.2.2 Code for pathwise method in Part5 % ALM analysis of life product % pathwise method in fixed interest rate case clear load E load q load T

%Given parameters nscenarios=50000; sigma=0.2; mu=0.075; tau=15; FV=10; theta=0.01;% benchmark interest K=20; % length of years: 18 to 37 years old w_S=0.5; P=17; % premium Z=randn(nscenarios,K);

%Discount future benefits and premium to determine current actuarial reserve Be=zeros(1,K);%future benefits DBe=zeros(1,K); Pr=zeros(1,K);%future premium

DC=zeros(nscenarios,K); DD=zeros(1,K); EC=zeros(1,K); DN=zeros(nscenarios,K);

55

DA=zeros(nscenarios,K);

%value of bond with different maturities b=zeros(1,16); %b(1)=b(tau=15);b(16)=b(tau=0) for i=1:16 b(i)=FV/(1+theta)^(16-i); end %derivative value of bond with different maturities Db=zeros(1,16); for i=1:16 Db(i)=(i-16)*FV/(1+theta)^(17-i); end

% the cycle would be N >> A >> n >> C >> N delta=[1000 1000 1000 1000]; n=zeros(nscenarios,tau); Dn=zeros(nscenarios,tau); Pr(1)=sum(delta)*P; N=Pr(1)*ones(nscenarios,1); A=w_S*N; n(:,1)=(N-A)/b(1); Dn(:,1)=-Db(1)*(N-A)/b(1)^2; V_end=n(:,1)*b(2); DV_end=Dn(:,1)*b(2)+n(:,1)*Db(2); Be(1)=E(1,:)*((1-q(1,:)).*delta)'+(1+theta)*T(1,:)*(q(1,:).*delta)'; DBe(1)=T(1,:)*(q(1,:).*delta)'; C=A.*exp(mu-sigma^2/2+sigma*Z(:,1))+V_end-Be(1); EC(1)=mean(C); DC(:,1)=DV_end-DBe(1); delta=(1-q(1,:)).*delta; %survivors at the end of period 1 % loop

for t=2:K

V_locked=sum(n(:,1:tau-1)*diag(b(1,2:15)),2); DV_locked=sum(Dn(:,1:tau-1)*diag(b(1,2:15)),2)+sum(n(:,1:tau-

1)*diag(Db(1,2:15)),2); Pr(t)=P*sum(delta); N=C+Pr(t)-V_locked; DN(:,t)=DC(:,t-1)-DV_locked; A=min(N,w_S*(C+P*sum(delta)));

DA(:,t)=(N<=(w_S*(C+P*sum(delta)))).*DN(:,t)+(N>(w_S*(C+P*sum(delta)))).*w_S.

*DC(:,t-1); n(:,2:tau)=n(:,1:tau-1); Dn(:,2:tau)=Dn(:,1:tau-1); n(:,1)=(N-A)/b(1); Dn(:,1)=(DN(:,t)-DA(:,t))/b(1)-Db(1)*(N-A)/b(1)^2; V_end=sum(n*diag(b(1,2:16)),2); DV_end=sum(Dn*diag(b(1,2:16)),2)+sum(n*diag(Db(:,2:16)),2);

Be(t)=E(t,:)*((1-

q(t,:)).*delta)'+(1+theta)^t*T(t,:)*(q(t,:).*delta)'; DBe(t)=t*(1+theta)^(t-1)*T(t,:)*(q(t,:).*delta)'; C=A.*exp(mu-sigma^2/2+sigma*Z(:,t))+V_end-Be(t); DC(:,t)=DA(:,t).*exp(mu-sigma^2/2+sigma*Z(:,t))+DV_end-DBe(t);

56

EC(t)=mean(C); delta=(1-q(t,:)).*delta; % new delta end

%actuarial reserve part %discount factor DF=zeros(1,K); DDF=zeros(1,K); for i=1:20 DF(i)=1/(1+theta)^(i-1); end for i=1:20 DDF(i)=(1-i)/(1+theta)^i; end for k=1:19 X=0; for i=1:(20-k) X=X+DDF(i+1)*Be(i+k)+DF(i+1)*DBe(i+k)-DDF(i)*Pr(i+k); end DD(k)=X; end

sigma_Q=std(DC); Pa_Sensitivity_Q=mean(DC)-DD; plot(Pa_Sensitivity_Q); hold on plot(Pa_Sensitivity_Q-1.96/sqrt(nscenarios)*sigma_Q); hold on plot(Pa_Sensitivity_Q+1.96/sqrt(nscenarios)*sigma_Q); hold off

A.2.3 Code for bump and revalue method in Part6 % ALM analysis of life insurance product % bump and revalue method in Part 6 clear load E load q load T

% Given parameters nscenarios=50000; sigma=0.2; mu=0.075; tau=15; FV=10; theta=0.01; h=0.01;% bump size K=20; % length of years: 18 to 37 years old w_S_bm=0.5; P=17; % premium Z=randn(nscenarios,K); D0=-10615;%inital actuarial reserve from fixed interest rate case w_SV=[w_S_bm-h w_S_bm+h]; % standard deviation of Q

57

stdV=zeros(2,K);

for j=1:2 w_S=w_SV(j); AC=zeros(nscenarios,K);%for the calculation of standard deviation

% the cycle would be N >> A >> n >> C >> N delta=[1000 1000 1000 1000]; n=zeros(nscenarios,tau); %n(:,1) indicates number of bonds with maturity

tau N=sum(delta)*P*ones(nscenarios,1); A=w_S*N; n(:,1)=(N-A)/((1+theta)^(-tau)*FV); % compute quantities at the end of first period V_end=(1+theta)*sum(n*diag((1+theta).^(-tau:-1)),2)*FV; C=A.*exp(mu-sigma^2/2+sigma*Z(:,1))+V_end-E(1,:)*((1-q(1,:)).*delta)'-

(1+theta)*T(1,:)*(q(1,:).*delta)';% AC(:,1)=C; stdV(j,1)=std(C); delta=(1-q(1,:)).*delta; %survivors at the end of period 1 % loop

for t=2:K

V_locked=sum(n(:,1:tau-1)*diag((1+theta).^(-tau+1:-1)),2)*FV; N=C+P*sum(delta)-V_locked; A=min(N,w_S*(C+P*sum(delta))); n(:,2:tau)=n(:,1:tau-1); n(:,1)=(N-A)/((1+theta)^(-tau)*FV); V_end=(1+theta)*sum(n*diag((1+theta).^(-tau:-1)),2)*FV; C=A.*exp(mu-sigma^2/2+sigma*Z(:,t))+V_end-E(t,:)*((1-

q(t,:)).*delta)'-(T(t,:)*(1+theta)^t)*(q(t,:).*delta)'; AC(:,t)=C; stdV(j,t)=std(C); delta=(1-q(t,:)).*delta; % new delta end end DEQ=(stdV(2,:)-stdV(1,:))/(w_SV(2)-w_SV(1)); plot(DEQ);

A.2.4 Code for pathwise method in Part6 % compute sensitivity of standard deviation of equity part using pathwise

method % this is pathwise method in the following, which is not in vectors

calculation % but solve each scenario from 1 to 50,000 clear load E load q load T %% Given parameters nscenarios=50000; % number of scenarios m=4; % Model points: normal and disabled male; normal and disabled female sigma = 0.20; %volatility of stock market

58

mu = 0.075; %drift term in the stock market tau = 15; % the maturity length of the bond FV = 10; % the face value of this zero coupon bond theta =0.01; %the fixed yearly interest rate 1% K = 20; % length of years: 18 to 37 years old w_S = 0.5; % proportion of capital invested into stocks P = 17; % Constant Premium income per policyholder delta = zeros(K,m); % number of policyholders with T*m, which is the same

as Benefits matrix Z = randn(nscenarios,K); %ranmdom matrix in the standard normal distribution %% Initialization of variables n = zeros(nscenarios,K); % initialization of the number of zero coupon

bonds A = zeros(nscenarios,K); % initialization of the value of stocks C = zeros(nscenarios,K); % initialization of the capital part N = zeros(nscenarios,K); % initialization of the new available capital part V_end=zeros(nscenarios,K); DA = zeros(nscenarios,K); % partial derivative of A with respect to theta DD = zeros(1,K); % partial derivative of D with respect to theta Dn = zeros(nscenarios,K); % partial derivative of n with respect to theta DC = zeros(nscenarios,K); DN = zeros(nscenarios,K); DT = zeros(1,K); Dvalue_end= zeros(nscenarios,K); delta_0=10^3; %initial number of policyholders in each age group DQ=zeros(nscenarios,K); D0=-10615; % This method calculate 50,000 scenario for nscenario = 1:nscenarios N(nscenario,1)=m*delta_0*P; A(nscenario,1)=w_S*N(nscenario,1); DA(nscenario,1)=N(nscenario,1); n(nscenario,1)=(N(nscenario,1)-A(nscenario,1))*(1+theta)^tau/FV; V_end(nscenario,1)=(1+theta)*n(nscenario,1)*FV/(1+theta)^tau; Dn(nscenario,1)=(DN(nscenario,1)-

DA(nscenario,1))*(1+theta)^tau/FV;%differntiate (1+theta)^tau/FV Dvalue_end(nscenario,1)=Dn(nscenario,1)*FV/(1+theta)^(tau-1); delta(1,:)=(1-q(1,:))*delta_0; C(nscenario,1)=A(nscenario,1)*exp(mu-

sigma^2/2+sigma*Z(nscenario,1))+V_end(nscenario,1)-E(1,:)*delta(1,:)'-

(T(1,:)*(1+theta))*(delta_0-delta(1,:))';%at the 1st period DC(nscenario,1)=DA(nscenario,1)*exp(mu-

sigma^2/2+sigma*Z(nscenario,1))+Dvalue_end(nscenario,1); %%to compare N and w_S*(C_k-1+P_k), we need to compute both of them for t=2:K %! in this stage, there is no maturing bond. As a result, if t<=tau delta(t,:)=(1-q(t,:)).*delta(t-1,:); N(nscenario,t)=C(nscenario,t-1)+P*sum(delta(t-1,:))-

V_end(nscenario,t-1); A(nscenario,t)=min(N(nscenario,t),w_S*(C(nscenario,t-

1)+P*sum(delta(t-1,:)))); n(nscenario,t)=(N(nscenario,t)-A(nscenario,t))*(1+theta)^tau/FV; V_end(nscenario,t)=(1+theta)*V_end(nscenario,t-

1)+(1+theta)*(N(nscenario,t)-A(nscenario,t)); C(nscenario,t)=A(nscenario,t)*exp(mu-

sigma^2/2+sigma*Z(nscenario,t))+V_end(nscenario,t)-E(t,:)*delta(t,:)'-

(T(t,:)*(1+theta)^t)*(delta(t-1,:)-delta(t,:))';

59

DN(nscenario,t)=DC(nscenario,t-1)-Dvalue_end(nscenario,t-1); DA(nscenario,t)=(N(nscenario,t)<=(w_S*(C(nscenario,t-

1)+P*sum(delta(t-

1,:)))))*DN(nscenario,t)+(N(nscenario,t)>(w_S*(C(nscenario,t-

1)+P*sum(delta(t-1,:)))))*(C(nscenario,t-1)+P*sum(delta(t-

1,:))+w_S*DC(nscenario,t-1)); Dn(nscenario,t)=(1+theta)^tau/FV*(DN(nscenario,t)-

DA(nscenario,t)); Dvalue_end(nscenario,t)=(1+theta)*Dvalue_end(nscenario,t-

1)+(1+theta)*(DN(nscenario,t)-DA(nscenario,t));%accoring the equation of

V_end DC(nscenario,t)=DA(nscenario,t)*exp(mu-

sigma^2/2+sigma*Z(nscenario,t))+Dvalue_end(nscenario,t); else delta(t,:)=(1-q(t,:)).*delta(t-1,:); %dot: this is to multiply

the element in the two matrices V_locked=0; for i =2:15 %"cycle" to calculate the value of locked bonds

portfolio V_locked=V_locked+(1+theta)*(n(nscenario,i+t-tau-

1)*FV/(1+theta)^i); %for i=t-1, n(nscenario,t-1)and 1; end N(nscenario,t)=C(nscenario,t-1)+P*sum(delta(t-1,:))-V_locked; A(nscenario,t)=min(N(nscenario,t),w_S*(C(nscenario,t-

1)+P*sum(delta(t-1,:)))); n(nscenario,t)=(N(nscenario,t)-A(nscenario,t))*(1+theta)^tau/FV; V_end(nscenario,t)=(1+theta)*V_locked+(1+theta)*(N(nscenario,t)-

A(nscenario,t)); C(nscenario,t)=A(nscenario,t)*exp(mu-

sigma^2/2+sigma*Z(nscenario,t))+V_end(nscenario,t)-E(t,:)*delta(t,:)'-

(T(t,:)*(1+theta)^t)*(delta(t-1,:)-delta(t,:))';%this step is to finish the

loop for "N >> A >> n >> C >> N" D2=0;%Dvalue_locked(t-1)=D2 for i=2:15 D2=D2+FV/(1+theta)^(i-1)*Dn(nscenario,i+t-tau);%Note that

the number for (1+theta)^ end DN(nscenario,t)=DC(nscenario,t-1)-D2;%Dvalue_locked(t-1)=D2 DA(nscenario,t)=(N(nscenario,t)<=(w_S*(C(nscenario,t-

1)+P*sum(delta(t-

1,:)))))*DN(nscenario,t)+(N(nscenario,t)>(w_S*(C(nscenario,t-

1)+P*sum(delta(t-1,:)))))*(C(nscenario,t-1)+P*sum(delta(t-

1,:))+w_S*DC(nscenario,t-1)); Dn(nscenario,t)=(1+theta)^tau/FV*(DN(nscenario,t)-

DA(nscenario,t)); Dvalue_end(nscenario,t)=(1+theta)*D2+(1+theta)*(DN(nscenario,t)-

DA(nscenario,t)); DC(nscenario,t)=DA(nscenario,t)*exp(mu-

sigma^2/2+sigma*Z(nscenario,t))+Dvalue_end(nscenario,t); end end end DVC=mean(2*C.*DC)-2*mean(C).*mean(DC);%according to the formula between

expectation and variance DstdQ=DVC./(2*std(C)); plot(DstdQ);

60

A.2.5 Code for bump and revalue method in Part7 % ALM analysis of life product % bump and revalue method in Vasicek model clear load E load q load T %given parameters nscenarios=50000; sigma=0.2; mu=0.075; tau=15; FV=10; theta_bm=0.01;% benchmark interest h=0.01;% bump size K=20; % length of years: 18 to 37 years old w_S=0.5; P=17; % premium BC=[zeros(nscenarios,K) zeros(nscenarios,K)]; BD=[zeros(nscenarios,K) zeros(nscenarios,K)]; BQ=[zeros(nscenarios,K) zeros(nscenarios,K)]; BEQ=zeros(2,K); Z=randn(nscenarios,K); C_vec=zeros(2,K); D_vec=zeros(2,K); thetas=[theta_bm theta_bm+h]; % Vasicek parameters r0=0.005; a=0.15;% speed of mean reversion sigma_r=0.015;%volatility of interest rate % Incorporate stochastic interest rates into the ALM model for j=1:2 Be=zeros(nscenarios,K);%future benefits Pr=zeros(1,K);%future premiums theta=thetas(j); AC=zeros(nscenarios,K);%for the calculation of standard deviation AD=zeros(nscenarios,K);%for the calculation of standard deviation EC=zeros(1,K); ED=zeros(1,K); BL=zeros(nscenarios,K);%benefit level %% vasicek part %B(1)=(1-exp(-a*(time to maturity)))/a; B=zeros(1,20); % vector from left to right: maturity from 19 to zero;

B(1,1)=B(t=1,tau=19); for i=1:20 B(1,i)=(1-exp(-a*(20-i)))/a; end %A_r(tau)=exp((theta-sigma_r^2/(2*a^2))*(B(tau)-tau)-

sigma_r^2/(4*a)*B(tau)^2); A_r=zeros(1,20); for i=1:20 A_r(1,i)=exp((theta-sigma_r^2/(2*a^2))*(B(1,i)-(20-i))-

sigma_r^2/(4*a)*B(1,i)^2); end

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%bond price formula: b(t,tau)=A_r(tau)*exp(-B(tau)*r_t); b_0=FV*A_r(1,5)*exp(-B(1,5)*r0);

%% bump and revalue method r=zeros(nscenarios,K); Z_r=randn(nscenarios,K);%stochastic movements in the vasicek model b=zeros(nscenarios,16); % the cycle would be N >> A >> n >> C >> N delta=[1000 1000 1000 1000]; n=zeros(nscenarios,tau); %n(:,1) indicates number of bonds with maturity

tau Pr(1)=sum(delta)*P; N=Pr(1)*ones(nscenarios,1); A=w_S*N; n(:,1)=(N-A)/b_0; r(:,1)=r0-a*(r0-theta)+sigma_r*Z_r(:,1); for i=1:16 b(:,i)=FV*A_r(1,i+4)*exp(-B(1,i+4)*r(:,1)); end % compute quantities at the end of first period V_end=n(:,1).*b(:,2); % let the death benefits be indexed on a floating-rate basis. % this means that the benefit level at time t+1 is equal to the benefit % level at time t, multiplied by the inverse of the value at time t of a % bond with face value 1 that matures at time t+1. BL(:,1)=1./(A_r(1,19)*exp(-B(1,19)*r0));%benefit level = inverse of the

value %at time t of a bond with maturity 1 Be(:,1)=E(1,:)*((1-q(1,:)).*delta)'+BL(:,1)*(T(1,:)*(q(1,:).*delta)'); C=A.*exp(mu-sigma^2/2+sigma*Z(:,1))+V_end-Be(:,1);% AC(:,1)=C; EC(1)=mean(C); delta=(1-q(1,:)).*delta; %survivors at the end of period 1 % loop for t=2:K V_locked=sum(n(:,1:tau-1).*b(:,2:15),2); Pr(t)=P*sum(delta); N=C+Pr(t)-V_locked; A=min(N,w_S*(C+Pr(t))); n(:,2:tau)=n(:,1:tau-1); n(:,1)=(N-A)./b(:,1); r(:,t)=r(:,t-1)-a*(r(:,t-1)-theta)+sigma_r*Z_r(:,t);%development of

Vasicek interest b=zeros(nscenarios,16); for i=1:16 b(:,i)=FV*A_r(1,i+4)*exp(-B(1,i+4)*r(:,t)); end V_end=sum(n.*b(:,2:16),2); BL(:,t)=BL(:,t-1).*(1./(A_r(1,19)*exp(-B(1,19)*r(:,t-1))));%benefit

level Be(:,t)=E(t,:)*((1-

q(t,:)).*delta)'+BL(:,t)*(T(t,:)*(q(t,:).*delta)'); C=A.*exp(mu-sigma^2/2+sigma*Z(:,t))+V_end-Be(:,t); AC(:,t)=C; EC(t)=mean(C); %D(t)=(1+r_t(t))*(D(t-1)+P*sum(delta))-E(t,:)*((1-q(t,:)).*delta)'-

(T(t,:)*(1+r_t(t))^t)*(q(t,:).*delta)';

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delta=(1-q(t,:)).*delta; % new delta end BC(:,(K+1)^(j-1):K*j)=AC; %actuarial reserve %discount factor DF=zeros(nscenarios,K); for k=1:19 for i=1:20 DF(:,i)=A_r(1,21-i)*exp(-B(1,21-i)*r(:,k)); end X=0; for i=1:(20-k) X=X+DF(:,i+1).*Be(:,k+i)-DF(:,i)*Pr(k+i); end AD(:,k)=X; ED(k)=mean(X); end BD(:,(K+1)^(j-1):K*j)=AD; BQ(:,(K+1)^(j-1):K*j)=BC(:,(K+1)^(j-1):K*j)-BD(:,(K+1)^(j-1):K*j); EQ=EC-ED; BEQ(j,:)=EQ; C_vec(j,:)=EC; D_vec(j,:)=ED; end DEQ=(BEQ(2,:)-BEQ(1,:))/(thetas(2)-thetas(1)); sigma_Q=std((BQ(:,K+1:2*K)-BQ(:,1:K))/(thetas(2)-thetas(1))); plot(DEQ); hold on plot(DEQ-1.96/sqrt(nscenarios)*sigma_Q); hold on plot(DEQ+1.96/sqrt(nscenarios)*sigma_Q); hold off

A.2.6 Code for pathwise method in Part7 % ALM analysis of life product % pathwise method clear load E load q load T %Given parameters nscenarios=50000; sigma=0.2; mu=0.075; tau=15; FV=10; theta=0.01;% benchmark interest K=20; % length of years: 18 to 37 years old w_S=0.5; P=17; % premium BEQ=zeros(2,K); Z=randn(nscenarios,K); % Vasicek parameters r0=0.005;

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a=0.15;% speed of mean reversion sigma_r=0.015;%volatility of interest rate Be=zeros(nscenarios,K);%future benefits DBe=zeros(nscenarios,K);%differentiation of Be Pr=zeros(1,K);%future premiums DC=zeros(nscenarios,K); DD=zeros(nscenarios,K); EC=zeros(1,K); DN=zeros(nscenarios,K); DA=zeros(nscenarios,K); BL=zeros(nscenarios,K);%benefit level DBL=zeros(nscenarios,K); %% vasicek part r=zeros(nscenarios,K);%nscenarios*K Z_r= randn(nscenarios,K);%stochastic movements in the vasicek model r(:,1)=r0-a*(r0-theta)+sigma_r*Z_r(:,1); for i=2:K dr=-a*(r(:,i-1)-theta)+sigma_r*Z_r(:,i); r(:,i)=r(:,i-1)+dr; end % Calculate Bonds value at different periods with maturities from 0 to

15. %B(1)=(1-exp(-a*(time to maturity)))/a; B=zeros(1,20); % vector from left to right: maturity from 19 to zero;

B(1,1)with matuirity of 19; for i=1:20 B(1,i)=(1-exp(-a*(20-i)))/a; end %A_r(tau)=exp((theta-sigma_r^2/(2*a^2))*(B(tau)-tau)-

sigma_r^2/(4*a)*B(tau)^2); A_r=zeros(1,20); for i=1:20 A_r(1,i)=exp((theta-sigma_r^2/(2*a^2))*(B(1,i)-(20-i))-

sigma_r^2/(4*a)*B(1,i)^2); end %bond price formula: b(t,tau)=A_r(tau)*exp(-B(tau)*r_t); b_0=FV*A_r(1,5)*exp(-B(1,5)*r0); %%Incorporate stochastic interest rates into the ALM model % Pathwise method % the cycle would be N >> A >> n >> C >> N delta=[1000 1000 1000 1000]; n=zeros(nscenarios,tau); %n(:,1) indicates number of bonds with maturity

tau Dn=zeros(nscenarios,tau); Pr(1)=sum(delta)*P; N=Pr(1)*ones(nscenarios,1); A=w_S*N; n(:,1)=(N-A)/b_0; Dn(:,1)=-FV*(B(1,5)-15)*A_r(1,5)*exp(-B(1,5)*r0)*(N-A)/b_0^2; % compute quantities at the end of first period b=zeros(nscenarios,16); for i=1:16 b(:,i)=FV*A_r(1,i+4)*exp(-B(1,i+4)*r(:,1)); end Db=zeros(nscenarios,16); for i=1:16 Db(:,i)=FV*(A_r(1,i+4)*B(1,i+4)*exp(-a)-A_r(1,i+4)*(16-i))*exp(-

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B(1,i+4)*r(:,1)); end V_end=n(:,1).*b(:,2); DV_end=Dn(:,1).*b(:,2)+Db(:,2).*n(:,1); BL(:,1)=1./(A_r(1,19)*exp(-B(1,19)*r0));%benefit level DBL(:,1)=(-1./(A_r(1,19)*exp(-B(1,19)*r0))^2)*(B(1,19)-1)*A_r(1,19)*exp(-

B(1,19)*r0); Be(:,1)=E(1,:)*((1-q(1,:)).*delta)'+BL(:,1)*T(1,:)*(q(1,:).*delta)'; DBe(:,1)=DBL(:,1)*T(1,:)*(q(1,:).*delta)'; C=A.*exp(mu-sigma^2/2+sigma*Z(:,1))+V_end-Be(:,1); EC(1)=mean(C); DC(:,1)=DV_end-DBe(:,1); delta=(1-q(1,:)).*delta; %survivors at the end of period 1 %DEQ(1)=mean(DC(:,1))-DD(1); % loop for t=2:K V_locked=sum(n(:,1:tau-1).*b(:,2:15),2); DV_locked=sum(Dn(:,1:tau-1).*b(:,2:15),2)+sum(n(:,1:tau-

1).*Db(:,2:15),2); Pr(t)=P*sum(delta); N=C+Pr(t)-V_locked; DN(:,t)=DC(:,t-1)-DV_locked; A=min(N,w_S*(C+P*sum(delta)));

DA(:,t)=(N<=(w_S*(C+P*sum(delta)))).*DN(:,t)+(N>(w_S*(C+P*sum(delta)))).*w_S.

*DC(:,t-1); n(:,2:tau)=n(:,1:tau-1); Dn(:,2:tau)=Dn(:,1:tau-1); n(:,1)=(N-A)./b(:,1); Dn(:,1)=(DN(:,t)-DA(:,t))./b(:,1)-Db(:,1).*(N-A)./(b(:,1).*b(:,1)); b=zeros(nscenarios,16); for i=1:16 b(:,i)=FV*A_r(1,i+4)*exp(-B(1,i+4)*r(:,t)); end Db=zeros(nscenarios,16); for i=1:16 Db(:,i)=FV*(A_r(1,i+4)*B(1,i+4)*exp(-a*t)-A_r(1,i+4)*(16-

i))*exp(-B(1,i+4)*r(:,t)); end V_end=sum(n.*b(:,2:16),2); DV_end=sum(Dn.*b(:,2:16),2)+sum(n.*Db(:,2:16),2); BL(:,t)=BL(:,t-1).*(1./(A_r(1,19)*exp(-B(1,19)*r(:,t-1))));%benefit

level DBL(:,t)=DBL(:,t-1).*(1./(A_r(1,19)*exp(-B(1,19)*r(:,t-1))))+BL(:,t-

1).*(-1./(A_r(1,19)*exp(-B(1,19)*r(:,t-1))).^2).*(A_r(1,19)*B(1,19)*exp(-

a*(t-1))-A_r(1,19)).*exp(-B(1,19)*r(:,t-1)); Be(:,t)=E(t,:)*((1-q(t,:)).*delta)'+BL(:,t)*T(t,:)*(q(t,:).*delta)'; DBe(:,t)=DBL(:,t)*T(t,:)*(q(t,:).*delta)'; C=A.*exp(mu-sigma^2/2+sigma*Z(:,t))+V_end-Be(:,t); DC(:,t)=DA(:,t).*exp(mu-sigma^2/2+sigma*Z(:,t))+DV_end-DBe(:,t); EC(t)=mean(C); delta=(1-q(t,:)).*delta; % new delta end %actuarial reserve %discount factor DF=zeros(nscenarios,K); DDF=zeros(nscenarios,K);

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for k=1:19 for i=1:20 DF(:,i)=A_r(1,21-i)*exp(-B(1,21-i)*r(:,k)); end for i=1:20 DDF(:,i)=(A_r(1,21-i)*B(1,21-i)*exp(-a*k)-A_r(1,21-i)*(i-

1))*exp(-B(1,21-i)*r(:,k)); X=0; end for i=1:(20-k) X=X+DDF(:,i+1).*Be(:,i+k)+DF(:,i+1).*DBe(:,i+k)-DDF(:,i)*Pr(i+k); end DD(:,k)=X; end DQ=DC-DD; sigma_Q=std(DC-DD); Pa_Sensitivity_Q=mean(DQ); plot(Pa_Sensitivity_Q); hold on plot(Pa_Sensitivity_Q-1.96/sqrt(nscenarios)*sigma_Q); hold on plot(Pa_Sensitivity_Q+1.96/sqrt(nscenarios)*sigma_Q); hold off

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