jump

Chapter 14

Stochastic Calculus for Jump Processes

The modelling of risky asset by stochastic processes with continuous paths,based on Brownian motions, suffers from several defects. First, the path con-tinuity assumption does not seem reasonable in view of the possibility ofsudden price variations (jumps) resulting of market crashes. Secondly, themodeling of risky asset prices by Brownian motion relies on the use of theGaussian distribution which tends to underestimate the probabilities of ex-treme events.

A solution is to use stochastic processes with jumps, that will account forsudden variations of the asset prices. On the other hand, such jump modelsare generally based on the Poisson distribution which has a slower tail decaythan the Gaussian distribution. This allows one to assign higher probabilitiesto extreme events, resulting in a more realistic modeling of asset prices.

14.1 The Poisson Process

The most elementary and useful jump process is the standard Poisson processwhich is a stochastic process (Nt)tR+ with jumps of size +1 only, and whosepaths are constant in between two jumps, i.e. at time t, the value Nt of theprocess is given by

Nt =k=1

1[Tk,)(t), t R+, (14.1)

where

The notation Nt is not to be confused with the same notation used for numeraireprocesses in Chapter 10.

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1[Tk,)(t) =

1 if t Tk,0 if 0 t < Tk,

k 1, and (Tk)k1 is the increasing family of jump times of (Nt)tR+ suchthat

limk

Tk = +.

In addition, (Nt)tR+ satisfies the following conditions:

1. Independence of increments: for all 0 t0 < t1 < < tn and n 1 therandom variables

Nt1 Nt0 , . . . , Ntn Ntn1 ,

are independent.

2. Stationarity of increments: Nt+h Ns+h has the same distribution asNt Ns for all h > 0 and 0 s t.

The meaning of the above stationarity condition is that for all fixed k Nwe have

P(Nt+h Ns+h = k) = P(Nt Ns = k),

for all h > 0, i.e. the value of the probability

P(Nt+h Ns+h = k)

does not depend on h > 0, for all fixed 0 s t and k N.

The next figure represents a sample path of a Poisson process.

0

1

2

3

4

5

6

7

0 2 4 6 8 10

Nt

t

Fig. 14.1: Sample path of a Poisson process (Nt)tR+ .

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Notes on Stochastic Finance

Based on the above assumption, given a time value T > 0 a natural questionarises:

what is the probability distribution of the random variable NT ?

We already know that Nt takes values in N and therefore it has a discretedistribution for all t R+.

It is a remarkable fact that the distribution of the increments of (Nt)tR+ ,can be completely determined from the above conditions, as shown in thefollowing theorem.

As seen in the next result, cf. [6], Nt Ns has the Poisson distributionwith parameter (t s).

Theorem 14.1. Assume that the counting process (Nt)tR+ satisfies theabove Conditions 1 and 2. Then for all fixed 0 s t we have

P (Nt Ns = k) = e(ts)((t s))k

k!, k N, (14.2)

for some constant > 0.

The parameter > 0 is called the intensity of the Poisson process (Nt)tR+and it is given by

:= limh0

1

hP(Nh = 1). (14.3)

The proof of the above Theorem 14.1 is technical and not included here,cf. e.g. [6] for details, and we could in fact take this distribution property(14.2) as one of the hypotheses that define the Poisson process.

Precisely, we could restate the definition of the standard Poisson process(Nt)tR+ with intensity > 0 as being a process defined by (14.1), which isassumed to have independent increments distributed according to the Poissondistribution, in the sense that for all 0 t0 t1 < < tn,

(Nt1 Nt0 , . . . , Ntn Ntn1)

is a vector of independent Poisson random variables with respective param-eters

((t1 t0), . . . , (tn tn1)).

In particular, Nt has the Poisson distribution with parameter t, i.e.

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P(Nt = k) =(t)k

k!et, t > 0.

The expected value E[Nt] of Nt can be computed as

IE[Nt] = t, (14.4)

cf. Exercise 16.1.

Short Time Behaviour

From (14.3) above we deduce the short time asymptotics

P(Nh = 1) = heh ' h, h 0,

andP(Nh = 0) = eh ' 1 h, h 0.

By stationarity of the Poisson process we find more generally that

P(Nt+h Nt = 1) = heh ' h, h 0,

andP(Nt+h Nt = 0) = eh ' 1 h, h 0,

for all t > 0.

This means that within a short interval [t, t + h] of length h, the in-crement Nt+hNt behaves like a Bernoulli random variable with parameterh. This fact can be used for the random simulation of Poisson process paths.

We also find that

P(Nt+h Nt = 2) ' h22

2, h 0, t > 0,

and more generally

P(Nt+h Nt = k) ' hkk

k!, h 0, t > 0.

The intensity of the Poisson process can in fact be made time-dependent (e.g.by a time change), in which case we have

We use the notation f(h) ' hk to mean that limh0 f(h)/hk = 1.

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P(Nt Ns = k) = exp(

w t

s(u)du

) (r ts(u)du

)kk!

, k 0.

In particular,

P(Nt+dt Nt = k) =

e(t)dt ' 1 (t)dt, k = 0,

(t)e(t)dtdt ' (t)dt, k = 1,

o(dt), k 2,

and P(Nt+dt Nt = 0), P(Nt+dt Nt = 1) coincide respectively with (13.2)and (13.3) above. The intensity process ((t))tR+ can also be made randomin the case of Cox processes.

Poisson Process Jump Times

In order to prove the next proposition we note that we have the equivalence

{T1 > t} {Nt = 0},

and more generally{Tn > t} {Nt n 1},

for all n 1.

In the next proposition we compute the distribution of Tn with its density.It coincides with the gamma distribution with integer parameter n 1, alsoknown as the Erlang distribution in queueing theory.

Proposition 14.1. For all n 1 the probability distribution of Tn has thedensity function

t 7 net tn1

(n 1)!on R+, i.e. for all t > 0 the probability P(Tn t) is given by

P(Tn t) = nw

tes

sn1

(n 1)!ds.

Proof. We have

P(T1 > t) = P(Nt = 0) = et, t R+,

and by induction, assuming that

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P(Tn1 > t) = w

tes

(s)n2

(n 2)!ds, n 2,

we obtain

P(Tn > t) = P(Tn > t Tn1) + P(Tn1 > t)= P(Nt = n 1) + P(Tn1 > t)

= et(t)n1

(n 1)!+

w

tes

(s)n2

(n 2)!ds

= w

tes

(s)n1

(n 1)!ds, t R+,

where we applied an integration by parts to derive the last line.

In particular, for all n Z and t R+, we have

P(Nt = n) = pn(t) = et(t)n

n!,

i.e. pn1 : R+ R+, n 1, is the density function of Tn.

Similarly we could show that the time

k := Tk+1 Tk

spent in state k N, with T0 = 0, forms a sequence of independent identi-cally distributed random variables having the exponential distribution withparameter > 0, i.e.

P(0 > t0, . . . , n > tn) = e(t0+t1++tn), t0, . . . , tn R+.

Since the expectation of the exponentially distributed random variable kwith parameter > 0 is given by

IE[k] =1

,

we can check that the higher the intensity (i.e. the higher the probabilityof having a jump within a small interval), the smaller is the time spent ineach state k N on average.

In addition, given that {NT = n}, the n jump times on [0, T ] of the Poissonprocess (Nt)tR+ are independent uniformly distributed random variables on[0, T ]n.

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Compensated Poisson Martingale

From (14.4) above we deduce that

IE[Nt t] = 0, (14.5)

i.e. the compensated Poisson process (Ntt)tR+ has centered increments.

Since in addition (Nt t)tR+ also has independent increments we getthe following proposition.

Proposition 14.2. The compensated Poisson process

(Nt t)tR+

is a martingale with respect to its own filtration (Ft)tR+ .

Extensions of the Poisson process include Poisson processes with time-dependent intensity, and with random time-dependent intensity (Cox pro-cesses). Renewal processes are counting processes

Nt =n1

1[Tn,)(t), t R+,

in which k = Tk+1 Tk, k N, is a sequence of independent identicallydistributed random variables. In particular, Poisson processes are renewalprocesses.

14.2 Compound Poisson Processes

The Poisson process itself appears to be too limited to develop realistic pricemodels as its jumps are of constant size. Therefore there is some interest inconsidering jump processes that can have random jump sizes.

Let (Zk)k1 denote an i.i.d. sequence of square-integrable random vari-ables with probability distribution (dy) on R, independent of the Poissonprocess (Nt)tR+ . We have

P(Zk [a, b]) = ([a, b]) =w b

a(dy), < a b

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is called a compound Poisson process.

The next figure represents a sample path of a compound Poisson process,with here Z1 = 0.9, Z2 = 0.7, Z3 = 1.4, Z4 = 0.6, Z5 = 2.5, Z6 = 1.5,Z7 = 1.2.

-0.5

0

0.5

1

1.5

2

2.5

0 2 4 6 8 10

Y t

t

Fig. 14.2: Sample path of a compound Poisson process (Yt)tR+ .

Given that {NT = n}, the n jump sizes of (Yt)tR+ on [0, T ] are independentrandom variables which are distributed on R according to (dx). Based onthis fact, the next proposition allows us to compute the characteristic functionof the increment YT Yt.

Proposition 14.3. For any t [0, T ] we have

IE [exp (i(YT Yt))] = exp((T t)

w

(eiy 1)(dy)

),

R.

Proof. Since Nt has a Poisson distribution with parameter t > 0 and isindependent of (Zk)k1, for all R we have by conditioning:

IE [exp (i(YT Yt))] = IE

[exp

(i

NTk=Nt+1

Zk

)]

= IE

[exp

(i

NTNtk=1

Zk

)]

=n=0

IE

[exp

(i

nk=1

Zk

)]P(NT Nt = n)

= e(Tt)n=0

n

n!(T t)n IE

[exp

(i

nk=1

Zk

)]

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= e(Tt)n=0

n

n!(T t)n (IE [exp (iZ1)])n

= exp ((T t) IE [exp (iZ1)])

= exp((T t)

w

(eiy 1)(dy)

),

since (dy) is the probability distribution of Z1 andr (dy) = 1.

From the characteristic function we can compute the expectation and vari-ance of Yt for fixed t, as

IE[Yt] = t IE[Z1] and Var [Yt] = t IE[|Z1|2].

For the expectation we have

IE[Yt] = id

dIE[eiYt ]|=0 = t

w

y(dy) = t IE[Z1].

This relation can also be directly recovered as

IE[Yt] = IE

[IE

[Ntk=1

Zk

Nt]]

= etn=0

ntn

n!IE

[nk=1

Zk

Nt = n]

= etn=0

ntn

n!IE

[nk=1

Zk

]

= tet IE[Z1]

n=1

(t)n1

(n 1)!

= t IE[Z1].

More generally one can show that for all 0 t0 t1 tn and1, . . . , n R we have

IE

[nk=1

eik(YtkYtk1 )]

= exp

(

nk=1

(tk tk1)w

(eiky 1)(dy)

)

=nk=1

exp((tk tk1)

w

(eiky 1)(dy)

)=

nk=1

IE[ei(YtkYtk1 )

].

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This shows in particular that the compound Poisson process (Yt)tR+ hasindependent increments, as the standard Poisson process (Nt)tR+ .

Since the compensated Poisson process also has centered increments by(14.5), we have the following proposition.

Proposition 14.4. The compensated compound Poisson process

Mt := Yt t IE[Z1], t R+,

is a martingale.

By construction, compound Poisson processes only have a finite numberof jumps on any interval. They belong to the family of Levy processes whichmay have an infinite number of jumps on any finite time interval, cf. [13].

14.3 Stochastic Integrals with Jumps

Given (t)tR+ a stochastic process we let the stochastic integral of (t)tR+with respect to (Yt)tR+ be defined by

w T

0tdYt :=

NTk=1

TkZk.

Note that this expressionw T

0tdYt has a natural financial interpretation as

the value at time T of a portfolio containing a (possibly fractional) quantityt of a risky asset at time t, whose price evolves according to random returnsZk at random times Tk.

In particular the compound Poisson process (Yt)tR+ in (14.1) admits thestochastic integral representation

Yt = Y0 +w t

0ZNsdNs.

Next, given (Wt)tR+ a standard Brownian motion independent of (Yt)tR+and (Xt)tR+ a jump-diffusion process of the form

Xt =w t

0usdWs +

w t

0vsds+ Yt, t R+,

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where (t)tR+ is a process which is adapted to the filtration (Ft)tR+ gen-erated by (Wt)tR+ and (Yt)tR+ , and such that

IE[w

02s|us|2ds

]

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Xt = X0 +w t

0usdWs +

w t

0vsds+

w t

0sdYs, t R+,

the stochastic integral of (t)tR+ with respect to (Xt)tR+ satisfies

w T

0sdXs :=

w T

0susdWs +

w T

0svsds+

w T

0ssdYs

=w T

0susdWs +

w T

0svsds+

NTk=1

TkTkZk, T > 0.

14.4 Ito Formula with Jumps

Let us first consider the case of a standard Poisson process (Nt)tR+ withintensity . We have the telescoping sum

f(Nt) = f(0) +

Ntk=1

(f(k) f(k 1))

= f(0) +w t

0(f(1 +Ns) f(Ns))dNs

= f(0) +w t

0(f(Ns) f(Ns))dNs.

Here, Ns denotes the left limit of the Poisson process at time s, i.e.

Ns = limh0

Nsh.

In particular we have

k = NTk = 1 +NTk, k 1.

By the same argument we find, in the case of the compound Poisson process(Yt)tR+ ,

f(Yt) = f(0) +

Ntk=1

(f(YTk+ Zk) f(YTk ))

= f(0) +w t

0(f(ZNs + Ys) f(Ys))dNs

= f(0) +w t

0(f(Ys) f(Ys))dNs,

which can be decomposed using a compensated Poisson stochastic integralas

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f(Yt) = f(0) +w t

0(f(Ys) f(Ys))(dNs ds) +

w t

0(f(Ys) f(Ys))ds.

More generally, for a process of the form

Xt = X0 +w t

0usdWs +

w t

0vsds+

w t

0sdYs, t R+,

we find, by combining the Ito formula for Brownian motion with the aboveargument we get

f(Xt) = f(X0) +w t

0usf

(Xs)dWs +1

2

w t

0f (Xs)|us|2ds

+w t

0vsf(Xs)ds+

NTk=1

(f(XTk+ TkZk) f(XTk ))

= f(X0) +w t

0usf

(Xs)dWs +1

2

w t

0f (Xs)|us|2ds+

w t

0vsf(Xs)ds

+w t

0(f(Xs + sZNs) f(Xs))dNs t R+.

i.e.

f(Xt) = f(X0) +w t

0usf

(Xs)dWs +1

2

w t

0f (Xs)|us|2ds+

w t

0vsf(Xs)ds

+w t

0(f(Xs) f(Xs))dNs, t R+. (14.9)

For example, in case

Xt =w t

0usdWs +

w t

0vsds+

w t

0sdNs, t R+,

we get

f(Xt) = f(0) +w t

0usf

(Xs)dWs +1

2

w t

0|us|2f (Xs)dWs

+w t

0vsf(Xs)ds+

w t

0(f(Xs + s) f(Xs))dNs

= f(0) +w t

0usf

(Xs)dWs +1

2

w t

0|us|2f (Xs)dWs (14.10)

+w t

0vsf(Xs)ds+

w t

0(f(Xs) f(Xs))dNs.

Given two processes (Xt)tR+ and (Yt)tR+ written as

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Xt =w t

0usdWs +

w t

0vsds+

w t

0sdNs, t R+,

and

Yt =w t

0asdWs +

w t

0bsds+

w t

0csdNs, t R+,

the Ito formula for jump processes also shows that

d(XtYt) = XtdYt + YtdXt + dXt dYt

where the product dXt dYt is computed according to the extension

dt dBt dNtdt 0 0 0

dBt 0 dt 0dNt 0 0 dNt

of the Ito multiplication table (4.19), i.e. we have

dXt dYt = (vtdt+ utdBt + tdNt)(btdt+ atdBt + ctdNt)= btvt(dt)

2 + btutdt dBt + bttdt dNt+atvtdtdBt + atut(dBt)

2 + attdBt dNt+ctvtdNt dBt + ctut(dBt)2 + cttdNt dNt

= atutdt+ cttdNt,

and in particular

(dXt)2 = (vtdt+ utdBt + tdNt)

2 = u2tdt+ 2t dNt.

For a process of the form

Xt = X0 +w t

0usdWs +

w t

0sdYt, t R+,

the Ito formula with jumps (14.10) can be rewritten as

f(Xt) = f(X0) +w t

0vsf(Xs)ds+

w t

0usf

(Xs)dWs

+1

2

w t

0f (Xs)|us|2ds+

w t

0sf(Xs)dYs

+w t

0(f(Xs) f(Xs)Xsf (Xs)) d(Ns s)

+w t

0(f(Xs) f(Xs)Xsf (Xs)) ds, t R+,

where we used the relation dYs = Xsf(Xs)dNs, which implies

w t

0sf(Xs)dYs =

w t

0Xsf

(Xs)dNs, t 0.

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This above formulation is at the basis of the extension of Itos formula toLevy processes with an infinite number of jumps on any interval, using thebound

|f(x+ y) f(x) yf (x)| Cy2,

for f a C2b (R) function. Such processes, also called infinite activity Levyprocesses [13] are also useful in financial modeling and include the gammaprocess, stable processes, variance gamma processes, inverse Gaussian pro-cesses, etc, as in the following illustrations.

1. Gamma process, d = 1.

0

t

Fig. 14.3: Sample trajectories of a gamma process.

2. Stable process, d = 1.

0

t

Fig. 14.4: Sample trajectories of a stable process.

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3. Variance Gamma process, d = 1.

0

t

Fig. 14.5: Sample trajectories of a variance gamma process.

4. Inverse Gaussian process, d = 1.

0

t

Fig. 14.6: Sample trajectories of an inverse Gaussian process.

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5. Negative Inverse Gaussian process, d = 1.

0

t

Fig. 14.7: Sample trajectories of a negative inverse Gaussian process.

14.5 Stochastic Differential Equations with Jumps

Let us start with the simplest example

dSt = StdNt, (14.11)

of a stochastic differential equation with respect to the standard Poisson pro-cess, with constant coefficient R.

WhenNt = Nt Nt = 1,

i.e. when the Poisson process has a jump at time t, the equation (14.11) reads

dSt = St St = St , t > 0.

which can be solved to yield

St = (1 + )St , t > 0.

By induction, applying this procedure for each jump time gives us the solution

St = S0(1 + )Nt , t R+.

Next, consider the case where is time-dependent, i.e.

dSt = tStdNt. (14.12)

At each jump time Tk, Relation (14.12) reads

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dSTk = STk STk = TkSTk ,

i.e.STk = (1 + Tk)STk

,

and repeating this argument for all k = 1, . . . , Nt yields the product solution

St = S0

Ntk=1

(1 + Tk) = S0

Ns=10st

(1 + s), t R+.

The equationdSt = tStdt+ tSt(dNt dt), (14.13)

is then solved as

St = S0 exp

(w t0sds

w t

0sds

) Ntk=1

(1 + Tk), t R+.

A random simulation of the numerical solution of the above equation (14.13)is given in Figure 14.8.

Fig. 14.8: Geometric Poisson process.

The above simulation can be compared to the real sales ranking data ofFigure 14.9.

The animation works in Acrobat reader on the entire file.

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Fig. 14.9: Ranking data.

A random simulation of the geometric compound Poisson process

St = S0 exp

(w t0sds IE[Z1]

w t

0sds

) Ntk=1

(1 + TkZk) t R+,

solution ofdSt = tStdt+ tSt(dYt IE[Z1]dt),

is given in Figure 14.10.

Fig. 14.10: Geometric compound Poisson process.

In the case of a jump-diffusion stochastic differential equation of the form

dSt = tStdt+ tSt(dYt IE[Z1]dt) + tStdWt,


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we get

St = S0 exp

(w t0sds IE[Z1]

w t

0sds+

w t

0sdWs

1

2

w t

0|s|2ds

)

Ntk=1

(1 + TkZk),

t R+. A random simulation of the geometric Brownian motion with com-pound Poisson jumps is given in Figure 14.11.

Fig. 14.11: Geometric Brownian motion with compound Poisson jumps.

By rewriting St as

St = S0 exp

(w t0sds+

w t

0s(dYs IE[Z1]ds) +

w t

0sdWs

1

2

w t

0|s|2ds

)

Ntk=1

(eTk (1 + TkZk)),

t R+, one can extend this jump model to processes with an infinite numberof jumps on any finite time interval, cf. [13]. The next Figure 14.12 showsa number of downward and upward jumps occuring in the historical priceof the SMRT stock, with a typical geometric Brownian behavior in betweenjumps.


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Fig. 14.12: SMRT Stock price.

14.6 Girsanov Theorem for Jump Processes

Recall that in its simplest form, the Girsanov theorem for Brownian motionfollows from the calculation

IE[f(WT T )] =12T

w

f(x T )ex

2/(2T )dx

=12T

w

f(x)e(x+T )

2/(2T )dx

=12T

w

f(x)ex

2T/2ex2/(2T )dx

= IE[f(WT )eWT2T/2]

= IE[f(WT )], (14.14)

for any bounded measurable function f on R, which shows that WT is aGaussian random variable with mean T under the probability measure Pdefined by

dP = eWT2T/2dP,

cf. Section 6.2. Equivalently we have

IE[f(WT )] = IE[f(WT + T )], (14.15)

hence

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under the probability measure

dP = eWT2T/2dP,

the random variable WT +T has a centered Gaussian distribution.

More generally, the Girsanov theorem states that (Wt + t)t[0,T ] is a stan-

dard Brownian motion under P.

When Brownian motion is replaced with a standard Poisson process(Nt)tR+ , the above space shift

Wt 7Wt + t

may not be used because Nt + t cannot be a Poisson process, whatever thechange of probability applied, since by construction, the paths of the stan-dard Poisson process has jumps of unit size and remain constant betweenjump times.

The correct way to proceed in order to extend (14.15) to the Poisson caseis to replace the space shift with a time contraction (or dilation) by a certainfactor 1 + c with c > 1, i.e.

Nt 7 Nt/(1+c).

By analogy with (14.14) we write

IE[f(NT (1+c))] =

k=0

f(k)P(NT (1+c) = k) (14.16)

= e(1+c)Tk=0

f(k)(T (1 + c))k

k!

= eT ecTk=0

f(k)(1 + c)k(T )k

k!

= ecTk=0

f(k)(1 + c)kP(NT = k)

= ecT IE[f(NT )(1 + c)NT ]

= ecTw

(1 + c)NT f(NT )dP

=w

f(NT )dP

= IE[f(NT )],

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for f any bounded function on N, where the probability measure P is definedby

dP = ecT (1 + c)NT dP.

Consequently,


dP = ecT (1 + c)NT dP,

the law of the random variable NT is that of NT (1+c) under P, i.e. it is aPoisson random variable with intensity (1 + c)T .

Equivalently we have

IE[f(NT )] = IE[f(NT/(1+c))],

i.e. under P the law of NT/(1+c) is that of a standard Poisson random variablewith parameter T .

In addition we have

Nt/(1+c) =n1

1[Tn,)(t/(1 + c))

=n1

1[(1+c)Tn,)(t), t R+,

which shows that under P, the jump times of (Nt/(1+c))t[0,T ] are given by

((1 + c)Tn)n1,

and we know that they are distributed as the jump times of a Poisson processwith intensity .

Next taking > 0 and

c := 1 + ,

we can rewrite the above by saying that

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dP = ecT (1 + c)NT dP = e()T

(

)NTdP,

the law of NT is that of a Poisson random variable with intensity

T = (1 + c)T.

Consequently, since both (Ntt)tR+ and (Nt(1+c)t)tR+ are processeswith independent increments, the compensated Poisson process

Nt (1 + c)t = Nt t

is a martingale under P by (6.2), although when c 6= 0 it is not a martingaleunder P.

In the case of compound Poisson processes the Girsanov theorem can beextended to variations in jump sizes in addition to time variations, and wehave the following more general result.

Theorem 14.2. Let (Yt)t0 be a compound Poisson process with inten-sity > 0 and jump distribution (dx). Consider another jump distribution(dx), and let

(x) =

d

d(x) 1, x R.

Then,


dP, := e()T

NTk=1

(1 + (Zk))dP, ,

the process

Yt =

Ntk=1

Zk, t R+,

is a compound Poisson process with

- modified intensity > 0, and

- modified jump distribution (dx).

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Proof. For any bounded measurable function f on R, we extend (14.16) tothe following change of variable

IE, [f(YT )] = e()T IE,

[f(YT )

NTi=1

(1 + (Zi))

]

= e()Tk=0

IE,

[f

(ki=1

Zi

)ki=1

(1 + (Zi))NT = k]P(NT = k)

= eTk=0

(T )k

k!IE,

[f

(ki=1

Zi

)ki=1

(1 + (Zi))

]

= eTk=0

(T )k

k!

w

w

f(z1 + + zk)

ki=1

(1 + (zi))(dz1) (dzk)

= eTk=0

(T )k

k!

w

w

f(z1 + + zk)

(ki=1

d

d(zi)

)(dz1) (dzk)

= eTk=0

(T )k

k!

w

w

f(z1 + + zk)(dz1) (dzk).

This shows that under P, , YT has the distribution of a compound Poissonprocess with intensity and jump distribution . We refer to Proposition 9.6of [13] for the independence of increments of (Yt)tR+ under P, .

Note that the compound Poisson process with intensity > 0 and jumpdistribution can be built as

Xt :=

Nt/k=1

h(Zk),

provided is the image measure of by the function h : R R, i.e.

P(h(Zk) A) = P(Zk h1(A)) = (h1(A)) = (A),

for all measurable subset A of R.

Compensated Compound Poisson Martingale

As a consequence of Theorem 14.2, the compensated process

Yt t IE [Z1]

becomes a martingale under the probability measure P, defined by

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N. Privault

dP, = e()T

NTk=1

(1 + (Zk))dP, .

Finally, the Girsanov theorem can be extended to the linear combinationof a standard Brownian motion (Wt)tR+ and an independent compoundPoisson process (Yt)tR+ , as in the following result which is a particular caseof Theorem 33.2 of [75].

Theorem 14.3. Let (Yt)t0 be a compound Poisson process with intensity > 0 and jump distribution (dx). Consider another jump distribution (dx)and intensity parameter > 0, and let

(x) =

d

d(x) 1, x R,

and let (ut)tR+ be a bounded adapted process. Then the process(Wt +

w t

0usds+ Yt IE [Z1]t

)tR+

is a martingale under the probability measure

dPu,, = exp(( )T

w T

0usdWs

1

2

w T

0|us|2ds

) NTk=1

(1+(Zk))dP, .

(14.17)

As a consequence of Theorem 14.3, if

Wt +w t

0vsds+ Yt

is not a martingale under P, , it will become a martingale under Pu,,provided u, and are chosen in such a way that

vs = us IE [Z1], s R, (14.18)

in which case we will have the martingale decomposition

dWt + utdt+ dYt IE [Z1]dt,

in which both

(Wt +

w t

0usds

)tR+

and(Yt t IE [Z1]

)tR+

are both mar-

tingales under Pu,,

When = = 0, Theorem 14.3 coincides with the usual Girsanov theoremfor Brownian motion, in which case (14.18) admits only one solution givenby u = v and there is uniqueness of Pu,0,0. Note that uniqueness occurs also

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when u = 0 in the absence of Brownian motion with Poisson jumps of fixedsize a (i.e. (dx) = (dx) = a(dx)) since in this case (14.18) also admitsonly one solution = v and there is uniqueness of P0,,a . These remarks willbe of importance for arbitrage pricing in jump models in Chapter 15.

Exercises

Exercise 14.1 Let (Nt)tR+ be a standard Poisson process with intensity > 0, started at N0 = 0.

1. Solve the stochastic differential equation

dSt = StdNt Stdt = St(dNt dt).

2. Using the first Poisson jump time T1, solve the stochastic differentialequation

dSt = Stdt+ dNtfor t (0, T2).

Exercise 14.2 Consider the compound Poisson process Yt :=

Ntk=1

Zk, where

(Nt)tR+ is a standard Poisson process with intensity > 0, (Zk)k1 is ani.i.d. sequence of N (0, 1) Gaussian random variables. Solve the stochasticdifferential equation

dSt = rStdt+ StdYt,

where , r R.

Exercise 14.3 Show, by direct computation or using the characteristic func-tion, that the variance of the compound Poisson process Yt with intensity > 0 satisfies

Var [Yt] = t IE[|Z1|2] = tw

x2(dx).

Exercise 14.4 Consider an exponential compound Poisson process of the form

St = S0et+Wt+Yt , t R+,

where (Yt)tR+ is a compound Poisson process of the form (14.6).

1. Derive the stochastic differential equation with jumps satisfied by (St)tR+ .

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N. Privault

2. Let r > 0. Find a family (Pu,,) of probability measures under whichthe discounted asset price ertSt is a martingale.

Exercise 14.5 Consider (Nt)tR+ a standard Poisson process with intensity > 0, independent of (Wt)tR+ , under a probability measure P. Let (St)tR+be defined by the stochastic differential equation

dSt = Stdt+ YNtStdNt, (14.19)

where (Yk)k1 is an i.i.d. sequence of random variables of the form Yk =eXk 1 where Xk ' N (0, 2), k 1.

1. Solve the equation (14.19).2. We assume that and the risk-free rate r > 0 are chosen such that the

discounted process (ertSt)tR+ is a martingale under P. What relationdoes this impose on and r ?

3. Under the relation of Question (2), compute the price at time t of aEuropean call option on ST with strike and maturity T , using a seriesexpansion of Black-Scholes functions.

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http://www.ntu.edu.sg/home/nprivault/indext.htmlStochastic Calculus for Jump ProcessesThe Poisson ProcessCompound Poisson ProcessesStochastic Integrals with JumpsIt Formula with JumpsStochastic Differential Equations with JumpsGirsanov Theorem for Jump ProcessesExercises

jump

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