P1.T2. Quantitative Analysis Bionic Turtle FRM Practice Questions Reading 13 Michael Miller, Mathematics and Statistics for Financial Risk Management, 2nd Edition Probabilities (Chapter 2) Basic Statistics (Chapter 3) Distributions (Chapter 4) Bayesian Analysis (Chapter 6) Hypothesis Testing and Confidence Intervals

(Chapter 7) This is a super-collection of quantitative practice questions. It represents several years of cumulative history mapped to the current Reading 10 (Miller’s Mathematics and Statistics for Financial Risk Management) By David Harper, CFA FRM CIPM www.bionicturtle.com

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This question set contains not only recently written questions, but also an accumulation of prior questions. The reason we did not delete many of the prior questions is simple: although the FRM’s econometrics readings have churned in recent years (specifically, for Probabilities and Statistics, from Gujarati to Stock and Watson to Miller), the learning outcomes (AIMs) have remain essentially unchanged. Further, the testable concepts are generally quite durable over time. For this reason, for example, the Gujarati Q&A is highly relevant. Nevertheless, within an assigned readings, our practice questions are sequenced in reverse chronological order (the most recent questions appear first). For example, in regard to assigned Miller Chapter 3 (Statistics), you will notice there are fully three (3) sets of questions: Miller Chapter 3 (T2.303 to .308) Stock & Watson (T2.208 to .213): written in 2012. Optional, but relevant Gujarati (T2.57 to .82): written prior to 2012. Optional, but relevant Therefore, do not feel obligated to review all of the questions in this document! Rather, consider the additional questions as merely a supplemental, optional resource for those who will to spend additional time with the concepts. A fine strategy is to merely review the most recent questions (“Miller”) within each Chapter. The major sections are: Probabilities (assigned is Miller Chapter 2)

o Most recent BT questions, Miller Chapter 2 (T2.300 to .302) o Previous BT questions, Stock & Watson Chapter 2 (T2.201 to .207)

Statistics (assigned is Miller Chapter 3) o Most recent BT questions, Miller Chapter 3 (T2.303 to .308) o Previous BT questions, Stock & Watson Chapter 3 (T2.208 to .213) o Previous BT questions, Gujarati (T2.57 to .82)

Distributions (assigned is Miller Chapter 4) o Most recent BT questions, Miller Chapter 4 (T2.309 to .312) o Previous BT questions, Rachev Chapters 2 & 3 (T2.110 to .126)

Bayesian Analysis (assigned is Miller Chapter 6)

o Most recent BT questions, Miller Chapter 6 (T2.500 to T2.501) o Previous BT questions, Miller Chapter 2 (T2.302)

Hypothesis Testing & Confidence Intervals (assigned is Miller Chapter 7) o Most recent BT questions, Miller Chapter 5 (T2.313 – .315)

Appendix o Annotated Gujarati (encompassing, highly relevant)

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PROBABILITIES - KEY IDEAS ................................................................................................. 4

PROBABILITIES (MILLER CHAPTER 2) .................................................................................. 6 P1.T2.300. PROBABILITY FUNCTIONS (MILLER) ......................................................................... 6 P1.T2.301. MILLER'S PROBABILITY MATRIX ................................................................................ 9

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Probabilities - Key Ideas

Risk measurement is largely the quantification of uncertainty. We quantify uncertainty by characterizing outcomes with random variables. Random variables have distributions which are either discrete or continuous.

In general, we observe samples; and use them to make inferences about a population (in practice, we tend to assume the population exists but it not available to us)

We are concerned with the first four moments of a distribution:

o Mean

o Variance, the square of the standard deviation. Annualized standard deviation is called volatility; e.g., 12% volatility per annum.

o Skew (a function of the third moment about the mean): a symmetrical distribution has zero skew or skewness

o Kurtosis (a function of the fourth moment about the mean).

The normal distribution has kurtosis = 3.0

As excess kurtosis = 3 – kurtosis, a normal distribution has zero excess kurtosis

Kurtosis > 3.0 refers to a heavy-tailed distribution (a.k.a., leptokurtosis), which will also tend to have a higher peaked.

The concepts of joint, conditional and marginal probability are important.

To test a hypothesis about a sample mean (i.e., is the true population mean different than some value), we use a student t or normal distribution

o Student t if the population variance is unknown (it usually is unknown)

o If the sample is large, the student t remains applicable, but as it approximates the normal, for large samples the normal is used since the difference is not material

To test a hypothesis about a sample variance, we use the chi-squared

To test a joint hypothesis about regression coefficients, we use the F distribution

In regard to the normal distribution:

o N(mu, σ^2) indicates the only two parameters required. For example, N(3,10) connotes a normal distribution with mean of 3 and variance of 10 and, therefore, standard deviation of SQRT(10)

o The standard normal distribution is N(0,1) and therefore requires no parameter specification: by definition it has mean of zero and variance of 1.0.

o Please memorize, with respect to the standard normal distribution:

For N(0,1) Pr(Z < -2.33) ~= 1.0% (CDF is one-tailed)

For N(0,1) Pr (Z< -1.645)~ = 5.0% (CDF is one-tailed)

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The definition of a random sample is technical: the draws (or trials) are independent and identically distributed (i.i.d.)

o Identical: same distribution

o Independence: no correlation (in a time series, no autocorrelation)

The assumption of i.i.d. is a precondition for:

o Law of large numbers

o Central limit theorem (CLT)

o Square root rule (SRR) for scaling volatility; e.g., we typically scales a daily volatility of (V) to an annual volatility with V*SQRT(250). Please note that i.i.d. returns is the unrealistic precondition.

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Probabilities (Miller Chapter 2) P1.T2.300. Probability functions (Miller) P1.T2.301. Miller's probability matrix

P1.T2.300. Probability functions (Miller)

AIMs: Describe the concept of probability. Describe and distinguish between continuous and discrete random variables. Define and distinguish between the probability density function, the cumulative distribution function and the inverse cumulative distribution function, and calculate probabilities based on each of these functions.

300.1. Assume the probability density function (pdf) of a zero-coupon bond with a notional value of $10.00 is given by f(x) = x/8 - 0.75 on the domain [6,10] where x is the price of the bond:

What is the probability that the price of the bond is between $8.00 and $9.00?

a) 25.750% b) 28.300% c) 31.250% d) 44.667%

300.2. Assume the probability density function (pdf) of a zero-coupon bond with a notional value of $5.00 is given by f(x) = (3/125)*x^2 on the domain [0,5] where x is the price of the bond:

Although the mean of this distribution is $3.75, assume the expected final payoff is a return of the full par of $5.00. If we apply the inverse cumulative distribution function and find the price of the bond (i.e., the value of x) such that 5.0% of the distribution is less than or equal to (x), let this price be represented by q(0.05); in other words, a 5% quantile function. If the 95.0% VaR is given by -[q(0.05) - 5] or [5 - q(0.05)], which is nearest to this 95.0% VaR?

a) $1.379 b) $2.842 c) $2.704 d) $3.158

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300.3. Assume a loss severity given by (x) can be characterized by a probability density function (pdf) on the domain [1, e^5]. For example, the minimum loss severity = $1 and the maximum possible loss severity = exp(5) ~= $148.41. The pdf is given by f(x) = c/x as follows:

What is the 95.0% value at risk (VaR); i.e., given that losses are expressed in positive values, at what loss severity value (x) is only 5.0% of the distribution greater than (x)?

a) $54.42 b) $97.26 c) $115.58 d) $139.04

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Answers: 300.1. C. 31.250% The anti-derivative is F(X) = x^2/16 - 0.75*x + c. We can confirm it is a probability by evaluating it on the domain [x = 6, x = 10] = 10^2/16 - 0.75*10 - 6^2/16 - 0.75*6 = -1.25 - (-2.25) = 1.0. Probability [8 <= x <= 9] = [9^2/16 - 0.75*9] - [8^2/16 - 0.75*8] = -1.68750 - (-2.000) = 31.250% 300.2. D. $3.158 As f(x) = 3/125*x^2, F(x) = 3/125*(1/3)*x^3 = p, such that: p = F(x) = (3/125)*(1/3)*x^3 = x^3/125, solving for x: x = (125*p)^(1/3) = 5*p^(1/3). For p = 5%, x = 5*5%^(1/3) = $1.8420. As q(0.05) = $1.8420, 95% VaR = $5.00 - $1.8420 = $3.1580 300.3. C. $115.58 We need d/dx [ln(x)] = 1/x; see http://en.wikipedia.org/wiki/Natural_logarithm#The_natural_logarithm_in_integration If f(x) = c/x, then anti-derivative F'(x) = c*ln(x) + a; It must be the case that, under a probability function, F'(e^5) = 1.0 such that 1.0 = c*ln(e^5) = c*5, and therefore c = 1/5. Put another way, under the definition of a probability:

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1 ln( )1.0 ( ) 1.0

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ee ec xdx F x dxx x

As F'(x) = p = ln(x)/5, now solving for x: p = ln(x)/5, 5p = ln(x), and taking exp() of both sides: exp(5p) = x, such that for the 95% quantile function: exp(5*0.95%) = $115.58 Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-300-probability-functions-miller.6728/

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P1.T2.301. Miller's probability matrix

AIMs: Calculate the probability of an event given a discrete probability function. Distinguish between independent and mutually exclusive events. Define joint probability, describe a probability matrix and calculate joint probabilities using probability matrices.

301.1. A random variable is given by the discrete probability function f(x) = P[X = x(i)] = a*X^3 such that x(i) is a member of {1, 2, 3} and (a) is a constant. That is, X has only three discrete outcomes. What is the probability that X will be greater than its mean? (bonus: what is the distribution's variance?)

a) 45.8% b) 50.0% c) 62.3% d) 75.0%

301.2. A credit asset has a principal value of $6.0 with probability of default (PD) of 3.0% and a loss given default (LGD) characterized by the following continuous probability density function (pdf): f(x) = x/18 such that 0 ≤ x ≤ $6. Let expected loss (EL) = E[PD*LGD]. If PD and LGD are independent, what is the asset's expected loss? (note: why does independence matter?)

a) $0.120 b) $0.282 c) $0.606 d) $1.125

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301.3. In analyzing a company, Analyst Sam prepared a probability matrix which is a joint (aka, bivariate) probability mass function that characterizes two discrete variables, equity performance versus a benchmark (over or under) and bond rating change. The company's equity performance will result in one of three mutually exclusive outcomes: under-perform, track the benchmark, or over-perform. The company's bond will either be upgraded, downgraded, or remain unchanged. Unfortunately, before Sam could share his probability matrix, he spilled coffee on it, and unfortunately some cells are not visible.

Two questions: what is the joint Prob [equity over-performs, bond has no change]; and are the two discrete variables independent?

a) 7.0%, yes b) 12.0%, yes c) 19.0%, no d) 22.0%, no

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Answers: 301.1. D. 75.0% Because it is a probability function, a*1^3 + a*2^3 + a*3^3 = 1.0; i.e., 1a + 8a + 27a = 1.0, such that a = 1/36. Mean = 1*(1/36) + 2*(8/36) + 3*(27/36) = 2.722. The P [X > 2.2722] = P[X = 3] = (1/36)*3^3 = 27/36 = 75.0% Bonus: Variance = (1 -2.722)^2*(1/36) + (2 -2.722)^2*(8/36) + (3 -2.722)^2*(27/36) = 0.2562, with standard deviation = SQRT(0.2562) = 0.506135 301.2. A. $0.120 If PD and LGD are not independent, then E[PD*LGD] <> E(PD) * E(LGD); for example, if they are positively correlated, then E[PD*LGD] > E(PD) * E(LGD). For the E[LGD], we integrate the pdf: if f(x) = x/18 s.t. 0 < x < $6, then F'(x) = (1/18)*(1/2)*x^2 = x^2/36 (note this satisfied the definition of a probability over the domain (0,6) as 6^2/36 = 1.0). The mean of f(x) integrates xf(x) where xf(x) = x*x/18 = x^2/18, which integrates to 1/18*(x^3/3) = x^3/54, so E[LGD] = 6^3/54 = $4.0.

Therefore, the expected loss = E[PD * LGD] = 3.0%*$4.0 = $0.120.

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301.3. C. 19.0%, no Joint Prob[under-perform, upgrade] = 4%, such that marginal (aka, unconditional) Prob[upgrade] = 4% + 8% + 11% = 23%. The marginal (unconditional) Prob[no change] = 100% - 23% - 13% = 64%, and therefore: Joint Prob[over-perform, no change] = 64% - 15% - 30% = 19.0%. The variables are independent if and only if (iif) the joint probability is equal to the product of marginal pmfs (pdfs); In this case, joint Prob[over-perform, no change] = 19.0% but the product of marginals = 32%*64% = 20.48%; i.e., 19% <> Prob[over-perform]*Prob[no change] Discuss in forum here: http://www.bionicturtle.com/forum/threads/p1-t2-301-millers-probability-matrix.6757/