jump
Content
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 continuity assumption does not seem reasonable in view of the possibility of
sudden price variations (jumps) resulting of market crashes. Secondly, the
modeling of risky asset prices by Brownian motion relies on the use of the
Gaussian distribution which tends to underestimate the probabilities of extreme events.
A solution is to use stochastic processes with jumps, that will account for
sudden variations of the asset prices. On the other hand, such jump models
are generally based on the Poisson distribution which has a slower tail decay
than the Gaussian distribution. This allows one to assign higher probabilities
to 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 process
which is a stochastic process (Nt )t∈R+ with jumps of size +1 only, and whose
paths are constant in between two jumps, i.e. at time t, the value Nt of the
process is given by∗
Nt =
∞
X
1[Tk ,∞) (t),
t ∈ R+ ,
(14.1)
k=1
where
∗
The notation Nt is not to be confused with the same notation used for num´
eraire
processes in Chapter 10.
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1[Tk ,∞) (t) =
1 if t ≥ Tk ,
0 if 0 ≤ t < Tk ,
k ≥ 1, and (Tk )k≥1 is the increasing family of jump times of (Nt )t∈R+ such
that
lim Tk = +∞.
k→∞
In addition, (Nt )t∈R+ satisfies the following conditions:
1. Independence of increments: for all 0 ≤ t0 < t1 < · · · < tn and n ≥ 1 the
random variables
Nt1 − Nt0 , . . . , Ntn − Ntn−1 ,
are independent.
2. Stationarity of increments: Nt+h − Ns+h has the same distribution as
Nt − Ns for all h > 0 and 0 ≤ s ≤ t.
The meaning of the above stationarity condition is that for all fixed k ∈ N
we 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.
7
6
5
Nt
4
3
2
1
0
0
2
4
6
8
10
t
Fig. 14.1: Sample path of a Poisson process (Nt )t∈R+ .
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Notes on Stochastic Finance
Based on the above assumption, given a time value T > 0 a natural question
arises:
what is the probability distribution of the random variable NT ?
We already know that Nt takes values in N and therefore it has a discrete
distribution for all t ∈ R+ .
It is a remarkable fact that the distribution of the increments of (Nt )t∈R+ ,
can be completely determined from the above conditions, as shown in the
following theorem.
As seen in the next result, cf. [6], Nt − Ns has the Poisson distribution
with parameter λ(t − s).
Theorem 14.1. Assume that the counting process (Nt )t∈R+ satisfies the
above Conditions 1 and 2. Then for all fixed 0 ≤ s ≤ t we have
P (Nt − Ns = k) = e−λ(t−s)
(λ(t − s))k
,
k!
k ∈ N,
(14.2)
for some constant λ > 0.
The parameter λ > 0 is called the intensity of the Poisson process (Nt )t∈R+
and it is given by
1
λ := lim P(Nh = 1).
(14.3)
h→0 h
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 )t∈R+ with intensity λ > 0 as being a process defined by (14.1), which is
assumed to have independent increments distributed according to the Poisson
distribution, in the sense that for all 0 ≤ t0 ≤ t1 < · · · < tn ,
(Nt1 − Nt0 , . . . , Ntn − Ntn−1 )
is a vector of independent Poisson random variables with respective parameters
(λ(t1 − t0 ), . . . , λ(tn − tn−1 )).
In particular, Nt has the Poisson distribution with parameter λt, i.e.
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P(Nt = k) =
(λt)k −λt
e ,
k!
t > 0.
The expected value E[Nt ] of Nt can be computed as
IE[Nt ] = λt,
(14.4)
cf. Exercise 16.1.
Short Time Behaviour
∗
From (14.3) above we deduce the short time asymptotics
P(Nh = 1) = hλe−hλ ' hλ,
h → 0,
P(Nh = 0) = e−hλ ' 1 − hλ,
h → 0.
and
By stationarity of the Poisson process we find more generally that
P(Nt+h − Nt = 1) = hλe−hλ ' hλ,
h → 0,
P(Nt+h − Nt = 0) = e−hλ ' 1 − hλ,
h → 0,
and
for all t > 0.
This means that within a “short” interval [t, t + h] of length h, the increment Nt+h − Nt behaves like a Bernoulli random variable with parameter
λh. This fact can be used for the random simulation of Poisson process paths.
We also find that
P(Nt+h − Nt = 2) ' h2
λ2
,
2
h → 0,
t > 0,
λk
,
k!
h → 0,
t > 0.
and more generally
P(Nt+h − Nt = k) ' hk
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 limh→0 f (h)/hk = 1.
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k
w
r t λ(u)du
t
s
P(Nt − Ns = k) = exp − λ(u)du
,
s
k!
k ≥ 0.
In particular,
−λ(t)dt
' 1 − λ(t)dt,
e
P(Nt+dt − Nt = k) = λ(t)e−λ(t)dt dt ' λ(t)dt,
o(dt),
k = 0,
k = 1,
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 (λ(t))t∈R+ can also be made random
in 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, also
known as the Erlang distribution in queueing theory.
Proposition 14.1. For all n ≥ 1 the probability distribution of Tn has the
density function
tn−1
t 7−→ λn e−λt
(n − 1)!
on R+ , i.e. for all t > 0 the probability P(Tn ≥ t) is given by
P(Tn ≥ t) = λn
w∞
t
e−λs
sn−1
ds.
(n − 1)!
Proof. We have
P(T1 > t) = P(Nt = 0) = e−λt ,
t ∈ R+ ,
and by induction, assuming that
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P(Tn−1 > t) = λ
w∞
t
e−λs
(λs)n−2
ds,
(n − 2)!
n ≥ 2,
we obtain
P(Tn > t) = P(Tn > t ≥ Tn−1 ) + P(Tn−1 > t)
= P(Nt = n − 1) + P(Tn−1 > t)
w∞
(λs)n−2
(λt)n−1
e−λs
+λ
ds
= e−λt
t
(n − 1)!
(n − 2)!
w∞
(λs)n−1
ds,
t ∈ R+ ,
=λ
e−λs
t
(n − 1)!
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) = e−λt
(λt)n
,
n!
i.e. pn−1 : R+ → R+ , n ≥ 1, is the density function of Tn .
Similarly we could show that the time
τk := Tk+1 − Tk
spent in state k ∈ N, with T0 = 0, forms a sequence of independent identically distributed random variables having the exponential distribution with
parameter λ > 0, i.e.
P(τ0 > t0 , . . . , τn > tn ) = e−λ(t0 +t1 +···+tn ) ,
t0 , . . . , tn ∈ R+ .
Since the expectation of the exponentially distributed random variable τk
with parameter λ > 0 is given by
IE[τk ] =
1
,
λ
we can check that the higher the intensity λ (i.e. the higher the probability
of having a jump within a small interval), the smaller is the time spent in
each state k ∈ N on average.
In addition, given that {NT = n}, the n jump times on [0, T ] of the Poisson
process (Nt )t∈R+ 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 − λt] = 0,
(14.5)
i.e. the compensated Poisson process (Nt − λt)t∈R+ has centered increments.
Since in addition (Nt − λt)t∈R+ also has independent increments we get
the following proposition.
Proposition 14.2. The compensated Poisson process
(Nt − λt)t∈R+
is a martingale with respect to its own filtration (Ft )t∈R+ .
Extensions of the Poisson process include Poisson processes with timedependent intensity, and with random time-dependent intensity (Cox processes). Renewal processes are counting processes
X
Nt =
1[Tn ,∞) (t),
t ∈ R+ ,
n≥1
in which τk = Tk+1 − Tk , k ∈ N, is a sequence of independent identically
distributed random variables. In particular, Poisson processes are renewal
processes.
14.2 Compound Poisson Processes
The Poisson process itself appears to be too limited to develop realistic price
models as its jumps are of constant size. Therefore there is some interest in
considering jump processes that can have random jump sizes.
Let (Zk )k≥1 denote an i.i.d. sequence of square-integrable random variables with probability distribution ν(dy) on R, independent of the Poisson
process (Nt )t∈R+ . We have
P(Zk ∈ [a, b]) = ν([a, b]) =
wb
a
ν(dy),
−∞ < a ≤ b < ∞.
Definition 14.1. The process
Yt =
Nt
X
Zk ,
t ∈ R+ ,
(14.6)
k=1
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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.
2.5
2
Yt
1.5
1
0.5
0
-0.5
0
2
4
6
8
10
t
Fig. 14.2: Sample path of a compound Poisson process (Yt )t∈R+ .
Given that {NT = n}, the n jump sizes of (Yt )t∈R+ on [0, T ] are independent
random variables which are distributed on R according to ν(dx). Based on
this fact, the next proposition allows us to compute the characteristic function
of the increment YT − Yt .
Proposition 14.3. For any t ∈ [0, T ] we have
w∞
IE [exp (iα(YT − Yt ))] = exp λ(T − t)
(eiyα − 1)ν(dy) ,
−∞
α ∈ R.
Proof. Since Nt has a Poisson distribution with parameter t > 0 and is
independent of (Zk )k≥1 , for all α ∈ R we have by conditioning:
"
IE [exp (iα(YT − Yt ))] = IE exp iα
!#
NT
X
Zk
k=Nt +1
"
= IE exp iα
NT
−Nt
X
!#
Zk
k=1
=
∞
X
"
IE exp iα
n=0
= e−λ(T −t)
n
X
!#
Zk
P(NT − Nt = n)
k=1
"
!#
∞
n
X
X
λn
(T − t)n IE exp iα
Zk
n!
n=0
k=1
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= e−λ(T −t)
∞
X
λn
n
(T − t)n (IE [exp (iαZ1 )])
n!
n=0
= exp (λ(T − t) IE [exp (iαZ1 )])
w∞
= exp λ(T − t)
(eiαy − 1)ν(dy) ,
−∞
since ν(dy) is the probability distribution of Z1 and
r∞
−∞
ν(dy) = 1.
From the characteristic function we can compute the expectation and variance of Yt for fixed t, as
IE[Yt ] = λt IE[Z1 ]
and
Var [Yt ] = λt IE[|Z1 |2 ].
For the expectation we have
IE[Yt ] = −i
w∞
d
IE[eiαYt ]|α=0 = λt
yν(dy) = λt IE[Z1 ].
−∞
dα
This relation can also be directly recovered as
##
" "N
t
X
Zk Nt
IE[Yt ] = IE IE
k=1
" n
#
∞
X
X
λn tn
=e
IE
Zk Nt = n
n!
n=0
k=1
" n
#
∞
X
X
λn tn
−λt
=e
IE
Zk
n!
n=0
−λt
k=1
∞
X
(λt)n−1
= λte−λt IE[Z1 ]
(n − 1)!
n=1
= λt IE[Z1 ].
More generally one can show that for all 0 ≤ t0 ≤ t1 · · · ≤ tn and
α1 , . . . , αn ∈ R we have
" n
#
!
n
w∞
Y iα (Y −Y
X
iαk y
tk
tk−1 )
k
(e
− 1)ν(dy)
IE
e
= exp λ
(tk − tk−1 )
k=1
−∞
k=1
=
=
n
Y
k=1
n
Y
exp λ(tk − tk−1 )
w∞
−∞
(eiαk y − 1)ν(dy)
h
i
IE eiα(Ytk −Ytk−1 ) .
k=1
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This shows in particular that the compound Poisson process (Yt )t∈R+ has
independent increments, as the standard Poisson process (Nt )t∈R+ .
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 − λt IE[Z1 ],
t ∈ R+ ,
is a martingale.
By construction, compound Poisson processes only have a finite number
of jumps on any interval. They belong to the family of L´evy processes which
may have an infinite number of jumps on any finite time interval, cf. [13].
14.3 Stochastic Integrals with Jumps
Given (φt )t∈R+ a stochastic process we let the stochastic integral of (φt )t∈R+
with respect to (Yt )t∈R+ be defined by
wT
0
φt dYt :=
NT
X
φTk Zk .
k=1
wT
Note that this expression
φt dYt has a natural financial interpretation as
0
the value at time T of a portfolio containing a (possibly fractional) quantity
φt of a risky asset at time t, whose price evolves according to random returns
Zk at random times Tk .
In particular the compound Poisson process (Yt )t∈R+ in (14.1) admits the
stochastic integral representation
wt
ZNs dNs .
Yt = Y0 +
0
Next, given (Wt )t∈R+ a standard Brownian motion independent of (Yt )t∈R+
and (Xt )t∈R+ a jump-diffusion process of the form
Xt =
wt
0
us dWs +
wt
0
vs ds + Yt ,
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t ∈ R+ ,
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where (φt )t∈R+ is a process which is adapted to the filtration (Ft )t∈R+ generated by (Wt )t∈R+ and (Yt )t∈R+ , and such that
hw ∞
i
hw ∞
i
IE
φ2s |us |2 ds < ∞ and IE
|φs vs |ds < ∞,
0
0
we let the stochastic integral of (φs )s∈R+ with respect to (Xs )s∈R+ be defined
by
wT
0
φs dXs :=
wT
0
φs us dWs +
wT
0
φs vs ds +
NT
X
φ Tk Z k ,
T > 0.
k=1
The coumpound Poisson compensated stochastic integral can be shown to
satisfy the Itˆo isometry
IE
"
w
T
0
2 #
w
T
φs− (dYs − λ IE[Z1 ]dt)
= λ IE[|Z1 |2 ] IE
|φ|2s ds ,
0
(14.7)
provided the process (φt )t∈R+ is adapted to the filtration generated by
(Yt )t∈R+ , which makes the left limit process (φs− )s∈R+ predictable. In the
case of simple predictable processes, the proof of (14.7) is similar to that of
Proposition 4.2.
For the mixed continuous-jump martingale
wt
Xt =
us dWs + Yt − λt IE[Z1 ],
0
t ∈ R+ ,
we have the isometry
IE
"
wT
0
2 #
φs− dXs
= IE
w
T
0
w
T
|φs |2 ds .
|φs |2 |us |2 ds + λ IE[|Z1 |2 ] IE
0
(14.8)
provided (φs )s∈R+ is adapted to the filtration (Ft )t∈R+ generated by (Wt )t∈R+
and (Yt )t∈R+ .
This isometry formula will be used in Section 15.5 for the computation of
hedging strategies in jump models.
When (Xt )t∈R+ takes the form
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Xt = X0 +
wt
0
us dWs +
wt
0
vs ds +
wt
0
t ∈ R+ ,
ηs dYs ,
the stochastic integral of (φt )t∈R+ with respect to (Xt )t∈R+ satisfies
wT
0
φs dXs :=
wT
φs us dWs +
wT
φs vs ds +
=
wT
φs us dWs +
wT
φs vs ds +
0
0
0
0
wT
0
NT
X
ηs φs dYs
φTk ηTk Zk ,
T > 0.
k=1
14.4 Itˆ
o Formula with Jumps
Let us first consider the case of a standard Poisson process (Nt )t∈R+ with
intensity λ. We have the telescoping sum
f (Nt ) = f (0) +
= f (0) +
= f (0) +
Nt
X
(f (k) − f (k − 1))
k=1
wt
0
wt
0
(f (1 + Ns− ) − f (Ns− ))dNs
(f (Ns ) − f (Ns− ))dNs .
Here, Ns− denotes the left limit of the Poisson process at time s, i.e.
Ns− = lim Ns−h .
h&0
In particular we have
k = NTk = 1 + NT − ,
k ≥ 1.
k
By the same argument we find, in the case of the compound Poisson process
(Yt )t∈R+ ,
f (Yt ) = f (0) +
= f (0) +
= f (0) +
Nt
X
(f (YT − + Zk ) − f (YT − ))
k
k
k=1
wt
0
wt
0
(f (ZNs + Ys− ) − f (Ys− ))dNs
(f (Ys ) − f (Ys− ))dNs ,
which can be decomposed using a compensated Poisson stochastic integral
as
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f (Yt ) = f (0) +
wt
0
(f (Ys ) − f (Ys− ))(dNs − λds) + λ
wt
0
(f (Ys ) − f (Ys− ))ds.
More generally, for a process of the form
wt
wt
wt
Xt = X0 +
us dWs +
vs ds +
ηs dYs ,
0
0
t ∈ R+ ,
0
we find, by combining the Itˆo formula for Brownian motion with the above
argument we get
f (Xt ) = f (X0 ) +
+
wt
0
wt
0
us f 0 (Xs )dWs +
vs f 0 (Xs )ds +
wt
NT
X
1 w t 00
f (Xs )|us |2 ds
2 0
(f (XT − + ηTk Zk ) − f (XT − ))
k
k
k=1
wt
1 w t 00
us f 0 (Xs )dWs +
f (Xs )|us |2 ds +
vs f 0 (Xs )ds
0
2 0
= f (X0 ) +
0
wt
+ (f (Xs− + ηs ZNs ) − f (Xs− ))dNs
0
t ∈ R+ .
i.e.
wt
wt
1 w t 00
f (Xt ) = f (X0 ) +
us f 0 (Xs )dWs +
f (Xs )|us |2 ds +
vs f 0 (Xs )ds
0
0
2 0
wt
+ (f (Xs ) − f (Xs− ))dNs ,
t ∈ R+ .
(14.9)
0
For example, in case
wt
wt
wt
vs ds +
ηs dNs ,
Xt =
us dWs +
0
0
0
t ∈ R+ ,
we get
wt
1wt
|us |2 f 00 (Xs )dWs
f (Xt ) = f (0) +
us f 0 (Xs )dWs +
0
2 0
wt
wt
+ vs f 0 (Xs )ds + (f (Xs− + ηs ) − f (Xs− ))dNs
0
0
wt
1wt
= f (0) +
us f 0 (Xs )dWs +
|us |2 f 00 (Xs )dWs
(14.10)
0
2 0
wt
wt
+ vs f 0 (Xs )ds + (f (Xs ) − f (Xs− ))dNs .
0
0
Given two processes (Xt )t∈R+ and (Yt )t∈R+ written as
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Xt =
wt
us dWs +
wt
vs ds +
wt
ηs dNs ,
t ∈ R+ ,
Yt =
wt
as dWs +
wt
bs ds +
wt
cs dNs ,
t ∈ R+ ,
and
0
0
0
0
0
0
the Itˆo formula for jump processes also shows that
d(Xt Yt ) = Xt dYt + Yt dXt + dXt · dYt
where the product dXt · dYt is computed according to the extension
·
dt
dBt
dNt
dt
0
0
0
dBt
0
dt
0
dNt
0
0
dNt
of the Itˆo multiplication table (4.19), i.e. we have
dXt · dYt = (vt dt + ut dBt + ηt dNt )(bt dt + at dBt + ct dNt )
= bt vt (dt)2 + bt ut dt · dBt + bt ηt dt · dNt
+at vt dtdBt + at ut (dBt )2 + at ηt dBt · dNt
+ct vt dNt · dBt + ct ut (dBt )2 + ct ηt dNt · dNt
= at ut dt + ct ηt dNt ,
and in particular
(dXt )2 = (vt dt + ut dBt + ηt dNt )2 = u2t dt + ηt2 dNt .
For a process of the form
Xt = X0 +
wt
0
us dWs +
wt
0
ηs dYt ,
t ∈ R+ ,
the Itˆo formula with jumps (14.10) can be rewritten as
wt
wt
f (Xt ) = f (X0 ) +
vs f 0 (Xs− )ds +
us f 0 (Xs− )dWs
0
0
w
w
t
1 t 00
+
f (Xs )|us |2 ds +
ηs f 0 (Xs− )dYs
0
2 0
wt
+ (f (Xs ) − f (Xs− ) − ∆Xs f 0 (Xs− )) d(Ns − s)
0
wt
+ (f (Xs ) − f (Xs− ) − ∆Xs f 0 (Xs− )) ds,
t ∈ R+ ,
0
where we used the relation dYs = ∆Xs f 0 (Xs− )dNs , which implies
wt
0
ηs f 0 (Xs− )dYs =
wt
0
∆Xs f 0 (Xs− )dNs ,
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This above formulation is at the basis of the extension of Itˆ
o’s formula to
L´evy processes with an infinite number of jumps on any interval, using the
bound
|f (x + y) − f (x) − yf 0 (x)| ≤ Cy 2 ,
for f a Cb2 (R) function. Such processes, also called “infinite activity L´evy
processes” [13] are also useful in financial modeling and include the gamma
process, stable processes, variance gamma processes, inverse Gaussian processes, 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 = ηSt− dNt ,
(14.11)
of a stochastic differential equation with respect to the standard Poisson process, with constant coefficient η ∈ R.
When
∆Nt = Nt − Nt− = 1,
i.e. when the Poisson process has a jump at time t, the equation (14.11) reads
dSt = St − St− = ηSt− ,
t > 0.
which can be solved to yield
St = (1 + η)St− ,
t > 0.
By induction, applying this procedure for each jump time gives us the solution
St = S0 (1 + η)Nt ,
t ∈ R+ .
Next, consider the case where η is time-dependent, i.e.
dSt = ηt St− dNt .
(14.12)
At each jump time Tk , Relation (14.12) reads
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dSTk = STk − ST − = ηTk ST − ,
k
k
i.e.
STk = (1 + ηTk )ST − ,
k
and repeating this argument for all k = 1, . . . , Nt yields the product solution
St = S0
Nt
Y
(1 + ηTk ) = S0
Y
t ∈ R+ .
(1 + ηs ),
∆Ns =1
0≤s≤t
k=1
The equation
dSt = µt St dt + ηt St− (dNt − λdt),
(14.13)
is then solved as
St = S0 exp
w
t
0
µs ds − λ
wt
0
Y
Nt
(1 + ηTk ),
ηs ds
t ∈ R+ .
k=1
A random simulation of the numerical solution of the above equation (14.13)
is given in Figure 14.8.
2
St
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.8: Geometric Poisson process.∗
The above simulation can be compared to the real sales ranking data of
Figure 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
t
0
µs ds − λ IE[Z1 ]
wt
0
Y
Nt
(1 + ηTk Zk )
ηs ds
t ∈ R+ ,
k=1
solution of
dSt = µt St dt + ηt St− (dYt − λ IE[Z1 ]dt),
is given in Figure 14.10.
2
St
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.10: Geometric compound Poisson process.∗
In the case of a jump-diffusion stochastic differential equation of the form
dSt = µt St dt + ηt St− (dYt − λ IE[Z1 ]dt) + σt St dWt ,
∗
The animation works in Acrobat reader on the entire file.
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we get
St = S0 exp
×
Nt
Y
w
t
0
µs ds − λ IE[Z1 ]
wt
0
ηs ds +
wt
0
σs dWs −
1wt
|σs |2 ds
2 0
(1 + ηTk Zk ),
k=1
t ∈ R+ . A random simulation of the geometric Brownian motion with compound Poisson jumps is given in Figure 14.11.
3
2.5
St
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.11: Geometric Brownian motion with compound Poisson jumps.∗
By rewriting St as
w
wt
wt
t
1wt
|σs |2 ds
µs ds +
ηs (dYs − λ IE[Z1 ]ds) +
σs dWs −
St = S0 exp
0
0
0
2 0
×
Nt
Y
(e−ηTk (1 + ηTk Zk )),
k=1
t ∈ R+ , one can extend this jump model to processes with an infinite number
of jumps on any finite time interval, cf. [13]. The next Figure 14.12 shows
a number of downward and upward jumps occuring in the historical price
of the SMRT stock, with a typical geometric Brownian behavior in between
jumps.
∗
The animation works in Acrobat reader on the entire file.
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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 motion
follows from the calculation
2
1 w∞
f (x − µT )e−x /(2T ) dx
IE[f (WT − µT )] = √
2πT −∞
2
1 w∞
= √
f (x)e−(x+µT ) /(2T ) dx
2πT −∞
2
2
1 w∞
= √
f (x)e−µx−µ T /2 e−x /(2T ) dx
2πT −∞
2
= IE[f (WT )e−µWT −µ
˜ (WT )],
= IE[f
T /2
]
(14.14)
for any bounded measurable function f on R, which shows that WT is a
˜
Gaussian random variable with mean −µT under the probability measure P
defined by
˜ = e−µWT −µ2 T /2 dP,
dP
cf. Section 6.2. Equivalently we have
˜ (WT + µT )],
IE[f (WT )] = IE[f
(14.15)
hence
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under the probability measure
2
˜ = e−µWT −µ
dP
T /2
dP,
the random variable WT + µT has a centered Gaussian distribution.
More generally, the Girsanov theorem states that (Wt + µt)t∈[0,T ] is a stan˜
dard Brownian motion under P.
When Brownian motion is replaced with a standard Poisson process
(Nt )t∈R+ , the above space shift
Wt 7−→ Wt + µt
may not be used because Nt + µt cannot be a Poisson process, whatever the
change of probability applied, since by construction, the paths of the standard Poisson process has jumps of unit size and remain constant between
jump times.
The correct way to proceed in order to extend (14.15) to the Poisson case
is to replace the space shift with a time contraction (or dilation) by a certain
factor 1 + c with c > −1, i.e.
Nt 7−→ Nt/(1+c) .
By analogy with (14.14) we write
IE[f (NT (1+c) )] =
∞
X
f (k)P(NT (1+c) = k)
(14.16)
k=0
= e−λ(1+c)T
=e
∞
X
f (k)
k=0
∞
X
−λT −λcT
e
(λT (1 + c))k
k!
f (k)(1 + c)k
k=0
=e
−λcT
∞
X
(λT )k
k!
f (k)(1 + c)k P(NT = k)
k=0
= e−λcT IE[f (NT )(1 + c)NT ]
w
= e−λcT
(1 + c)NT f (NT )dP
Ω
w
˜
=
f (NT )dP
Ω
˜ (NT )],
= IE[f
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˜ is defined
for f any bounded function on N, where the probability measure P
by
˜ ˜ = e−λcT (1 + c)NT dP.
dP
λ
Consequently,
under the probability measure
˜ = e−λcT (1 + c)NT dP,
dP
the law of the random variable NT is that of NT (1+c) under P, i.e. it is a
Poisson random variable with intensity λ(1 + c)T .
Equivalently we have
˜ (NT /(1+c) )],
IE[f (NT )] = IE[f
˜ the law of NT /(1+c) is that of a standard Poisson random variable
i.e. under P
with parameter λT .
In addition we have
Nt/(1+c) =
X
1[Tn ,∞) (t/(1 + c))
n≥1
=
X
1[(1+c)Tn ,∞) (t),
t ∈ R+ ,
n≥1
˜ the jump times of (Nt/(1+c) )t∈[0,T ] are given by
which shows that under P,
((1 + c)Tn )n≥1 ,
and we know that they are distributed as the jump times of a Poisson process
with intensity λ.
˜ > 0 and
Next taking λ
c := −1 +
˜
λ
,
λ
we can rewrite the above by saying that
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under the probability measure
˜
˜ ˜ = e−λcT (1 + c)NT dP = e−(λ−λ)T
dP
λ
˜
λ
λ
!NT
dP,
the law of NT is that of a Poisson random variable with intensity
˜ = λ(1 + c)T.
λT
Consequently, since both (Nt −λt)t∈R+ and (Nt −(1+c)λt)t∈R+ are processes
with independent increments, the compensated Poisson process
˜
Nt − (1 + c)λt = Nt − λt
˜ ˜ by (6.2), although when c 6= 0 it is not a martingale
is a martingale under P
λ
under P.
In the case of compound Poisson processes the Girsanov theorem can be
extended to variations in jump sizes in addition to time variations, and we
have the following more general result.
Theorem 14.2. Let (Yt )t≥0 be a compound Poisson process with intensity λ > 0 and jump distribution ν(dx). Consider another jump distribution
ν˜(dx), and let
˜ d˜
ν
λ
(x) − 1,
x ∈ R.
φ(x) =
λ dν
Then,
under the probability measure
˜
˜ ˜ := e−(λ−λ)T
dP
λ,˜
ν
NT
Y
˜ λ,ν ,
(1 + φ(Zk ))dP
k=1
the process
Yt =
Nt
X
Zk ,
t ∈ R+ ,
k=1
is a compound Poisson process with
˜ > 0, and
- modified intensity λ
- modified jump distribution ν˜(dx).
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Proof. For any bounded measurable function f on R, we extend (14.16) to
the following change of variable
#
"
NT
Y
˜
−(λ−λ)T
IEλ,˜
IEλ,ν f (YT ) (1 + φ(Zi ))
˜ ν [f (YT )] = e
i=1
˜
= e−(λ−λ)T
∞
X
"
k
X
IEλ,ν f
=e
˜
−λT
k=0
˜
= e−λT
˜
˜
= e−λT
"
IEλ,ν f
k!
−∞
∞ ˜
X
(λT )k w ∞
k=0
∞
X
k=0
k!
k
Y
−∞
···
···
w∞
−∞
w∞
#
(1 + φ(Zi )) NT = k P(NT = k)
i=1
k
X
!
Zi
i=1
∞
X
(λT )k w ∞
k=0
= e−λT
k!
Zi
i=1
k=0
∞
X
(λT )k
!
−∞
k
Y
#
(1 + φ(Zi ))
i=1
f (z1 + · · · + zk )
k
Y
(1 + φ(zi ))ν(dz1 ) · · · ν(dzk )
i=1
f (z1 + · · · + zk )
k
Y
d˜
ν
i=1
dν
!
(zi ) ν(dz1 ) · · · ν(dzk )
w∞
˜ )k w ∞
(λT
···
f (z1 + · · · + zk )˜
ν (dz1 ) · · · ν˜(dzk ).
−∞
−∞
k!
This shows that under Pλ,˜
˜ ν , YT has the distribution of a compound Poisson
˜ and jump distribution ν˜. We refer to Proposition 9.6
process with intensity λ
˜˜ .
of [13] for the independence of increments of (Yt )t∈R+ under P
λ,˜
ν
˜ > 0 and jump
Note that the compound Poisson process with intensity λ
distribution ν˜ can be built as
Nλt/λ
˜
Xt :=
X
h(Zk ),
k=1
provided ν˜ is the image measure of ν by the function h : R → R, i.e.
P(h(Zk ) ∈ A) = P(Zk ∈ h−1 (A)) = ν(h−1 (A)) = ν˜(A),
for all measurable subset A of R.
Compensated Compound Poisson Martingale
As a consequence of Theorem 14.2, the compensated process
˜ IEν˜ [Z1 ]
Yt − λt
˜ ˜ defined by
becomes a martingale under the probability measure P
λ,˜
ν
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˜
˜ ˜ = e−(λ−λ)T
dP
λ,˜
ν
NT
Y
˜ λ,ν .
(1 + φ(Zk ))dP
k=1
Finally, the Girsanov theorem can be extended to the linear combination
of a standard Brownian motion (Wt )t∈R+ and an independent compound
Poisson process (Yt )t∈R+ , as in the following result which is a particular case
of Theorem 33.2 of [75].
Theorem 14.3. Let (Yt )t≥0 be a compound Poisson process with intensity
λ > 0 and jump distribution ν(dx). Consider another jump distribution ν˜(dx)
˜ > 0, and let
and intensity parameter λ
φ(x) =
˜ d˜
λ
ν
(x) − 1,
λ dν
x ∈ R,
and let (ut )t∈R+ be a bounded adapted process. Then the process
wt
˜ IEν˜ [Z1 ]t
Wt +
us ds + Yt − λ
0
t∈R+
is a martingale under the probability measure
Y
NT
wT
1wT
˜ − λ)T −
˜ ˜ = exp −(λ
˜ λ,ν .
dP
us dWs −
(1+φ(Zk ))dP
|us |2 ds
u,λ,˜
ν
0
2 0
k=1
(14.17)
As a consequence of Theorem 14.3, if
wt
Wt +
vs ds + Yt
0
˜ λ,ν , it will become a martingale under P
˜ ˜
is not a martingale under P
u,λ,˜
ν
˜
provided u, λ and ν˜ are chosen in such a way that
˜ IEν˜ [Z1 ],
vs = us − λ
s ∈ R,
(14.18)
in which case we will have the martingale decomposition
in which both
˜ IEν˜ [Z1 ]dt,
dWt + ut dt + dYt − λ
wt
˜ IEν˜ [Z1 ]
Wt +
us ds
and Yt − λt
0
t∈R+
are both mart∈R+
˜ ˜
tingales under P
u,λ,˜
ν
˜ = λ = 0, Theorem 14.3 coincides with the usual Girsanov theorem
When λ
for Brownian motion, in which case (14.18) admits only one solution given
˜ u,0,0 . Note that uniqueness occurs also
by u = v and there is uniqueness of P
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when u = 0 in the absence of Brownian motion with Poisson jumps of fixed
size a (i.e. ν˜(dx) = ν(dx) = δa (dx)) since in this case (14.18) also admits
˜ = v and there is uniqueness of P
˜ ˜ . These remarks will
only one solution λ
0,λ,δa
be of importance for arbitrage pricing in jump models in Chapter 15.
Exercises
Exercise 14.1 Let (Nt )t∈R+ be a standard Poisson process with intensity
λ > 0, started at N0 = 0.
1. Solve the stochastic differential equation
dSt = ηSt− dNt − ηλSt dt = ηSt− (dNt − λdt).
2. Using the first Poisson jump time T1 , solve the stochastic differential
equation
dSt = −ηSt dt + dNt
for t ∈ (0, T2 ).
Exercise 14.2 Consider the compound Poisson process Yt :=
Nt
X
Zk , where
k=1
(Nt )t∈R+ is a standard Poisson process with intensity λ > 0, (Zk )k≥1 is an
i.i.d. sequence of N (0, 1) Gaussian random variables. Solve the stochastic
differential equation
dSt = rSt dt + ηSt− dYt ,
where η, r ∈ R.
Exercise 14.3 Show, by direct computation or using the characteristic function, that the variance of the compound Poisson process Yt with intensity
λ > 0 satisfies
w∞
Var [Yt ] = λt IE[|Z1 |2 ] = λt
x2 ν(dx).
−∞
Exercise 14.4 Consider an exponential compound Poisson process of the form
St = S0 eµt+σWt +Yt ,
t ∈ R+ ,
where (Yt )t∈R+ is a compound Poisson process of the form (14.6).
1. Derive the stochastic differential equation with jumps satisfied by (St )t∈R+ .
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˜ ˜ ) of probability measures under which
2. Let r > 0. Find a family (P
u,λ,˜
ν
the discounted asset price e−rt St is a martingale.
Exercise 14.5 Consider (Nt )t∈R+ a standard Poisson process with intensity
λ > 0, independent of (Wt )t∈R+ , under a probability measure P. Let (St )t∈R+
be defined by the stochastic differential equation
dSt = µSt dt + YNt St− dNt ,
(14.19)
where (Yk )k≥1 is an i.i.d. sequence of random variables of the form Yk =
eXk − 1 where Xk ' N (0, σ 2 ), k ≥ 1.
1. Solve the equation (14.19).
2. We assume that µ and the risk-free rate r > 0 are chosen such that the
discounted process (e−rt St )t∈R+ is a martingale under P. What relation
does this impose on µ and r ?
3. Under the relation of Question (2), compute the price at time t of a
European call option on ST with strike κ and maturity T , using a series
expansion of Black-Scholes functions.
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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 continuity assumption does not seem reasonable in view of the possibility of
sudden price variations (jumps) resulting of market crashes. Secondly, the
modeling of risky asset prices by Brownian motion relies on the use of the
Gaussian distribution which tends to underestimate the probabilities of extreme events.
A solution is to use stochastic processes with jumps, that will account for
sudden variations of the asset prices. On the other hand, such jump models
are generally based on the Poisson distribution which has a slower tail decay
than the Gaussian distribution. This allows one to assign higher probabilities
to 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 process
which is a stochastic process (Nt )t∈R+ with jumps of size +1 only, and whose
paths are constant in between two jumps, i.e. at time t, the value Nt of the
process is given by∗
Nt =
∞
X
1[Tk ,∞) (t),
t ∈ R+ ,
(14.1)
k=1
where
∗
The notation Nt is not to be confused with the same notation used for num´
eraire
processes in Chapter 10.
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1[Tk ,∞) (t) =
1 if t ≥ Tk ,
0 if 0 ≤ t < Tk ,
k ≥ 1, and (Tk )k≥1 is the increasing family of jump times of (Nt )t∈R+ such
that
lim Tk = +∞.
k→∞
In addition, (Nt )t∈R+ satisfies the following conditions:
1. Independence of increments: for all 0 ≤ t0 < t1 < · · · < tn and n ≥ 1 the
random variables
Nt1 − Nt0 , . . . , Ntn − Ntn−1 ,
are independent.
2. Stationarity of increments: Nt+h − Ns+h has the same distribution as
Nt − Ns for all h > 0 and 0 ≤ s ≤ t.
The meaning of the above stationarity condition is that for all fixed k ∈ N
we 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.
7
6
5
Nt
4
3
2
1
0
0
2
4
6
8
10
t
Fig. 14.1: Sample path of a Poisson process (Nt )t∈R+ .
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Based on the above assumption, given a time value T > 0 a natural question
arises:
what is the probability distribution of the random variable NT ?
We already know that Nt takes values in N and therefore it has a discrete
distribution for all t ∈ R+ .
It is a remarkable fact that the distribution of the increments of (Nt )t∈R+ ,
can be completely determined from the above conditions, as shown in the
following theorem.
As seen in the next result, cf. [6], Nt − Ns has the Poisson distribution
with parameter λ(t − s).
Theorem 14.1. Assume that the counting process (Nt )t∈R+ satisfies the
above Conditions 1 and 2. Then for all fixed 0 ≤ s ≤ t we have
P (Nt − Ns = k) = e−λ(t−s)
(λ(t − s))k
,
k!
k ∈ N,
(14.2)
for some constant λ > 0.
The parameter λ > 0 is called the intensity of the Poisson process (Nt )t∈R+
and it is given by
1
λ := lim P(Nh = 1).
(14.3)
h→0 h
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 )t∈R+ with intensity λ > 0 as being a process defined by (14.1), which is
assumed to have independent increments distributed according to the Poisson
distribution, in the sense that for all 0 ≤ t0 ≤ t1 < · · · < tn ,
(Nt1 − Nt0 , . . . , Ntn − Ntn−1 )
is a vector of independent Poisson random variables with respective parameters
(λ(t1 − t0 ), . . . , λ(tn − tn−1 )).
In particular, Nt has the Poisson distribution with parameter λt, i.e.
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P(Nt = k) =
(λt)k −λt
e ,
k!
t > 0.
The expected value E[Nt ] of Nt can be computed as
IE[Nt ] = λt,
(14.4)
cf. Exercise 16.1.
Short Time Behaviour
∗
From (14.3) above we deduce the short time asymptotics
P(Nh = 1) = hλe−hλ ' hλ,
h → 0,
P(Nh = 0) = e−hλ ' 1 − hλ,
h → 0.
and
By stationarity of the Poisson process we find more generally that
P(Nt+h − Nt = 1) = hλe−hλ ' hλ,
h → 0,
P(Nt+h − Nt = 0) = e−hλ ' 1 − hλ,
h → 0,
and
for all t > 0.
This means that within a “short” interval [t, t + h] of length h, the increment Nt+h − Nt behaves like a Bernoulli random variable with parameter
λh. This fact can be used for the random simulation of Poisson process paths.
We also find that
P(Nt+h − Nt = 2) ' h2
λ2
,
2
h → 0,
t > 0,
λk
,
k!
h → 0,
t > 0.
and more generally
P(Nt+h − Nt = k) ' hk
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 limh→0 f (h)/hk = 1.
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k
w
r t λ(u)du
t
s
P(Nt − Ns = k) = exp − λ(u)du
,
s
k!
k ≥ 0.
In particular,
−λ(t)dt
' 1 − λ(t)dt,
e
P(Nt+dt − Nt = k) = λ(t)e−λ(t)dt dt ' λ(t)dt,
o(dt),
k = 0,
k = 1,
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 (λ(t))t∈R+ can also be made random
in 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, also
known as the Erlang distribution in queueing theory.
Proposition 14.1. For all n ≥ 1 the probability distribution of Tn has the
density function
tn−1
t 7−→ λn e−λt
(n − 1)!
on R+ , i.e. for all t > 0 the probability P(Tn ≥ t) is given by
P(Tn ≥ t) = λn
w∞
t
e−λs
sn−1
ds.
(n − 1)!
Proof. We have
P(T1 > t) = P(Nt = 0) = e−λt ,
t ∈ R+ ,
and by induction, assuming that
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P(Tn−1 > t) = λ
w∞
t
e−λs
(λs)n−2
ds,
(n − 2)!
n ≥ 2,
we obtain
P(Tn > t) = P(Tn > t ≥ Tn−1 ) + P(Tn−1 > t)
= P(Nt = n − 1) + P(Tn−1 > t)
w∞
(λs)n−2
(λt)n−1
e−λs
+λ
ds
= e−λt
t
(n − 1)!
(n − 2)!
w∞
(λs)n−1
ds,
t ∈ R+ ,
=λ
e−λs
t
(n − 1)!
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) = e−λt
(λt)n
,
n!
i.e. pn−1 : R+ → R+ , n ≥ 1, is the density function of Tn .
Similarly we could show that the time
τk := Tk+1 − Tk
spent in state k ∈ N, with T0 = 0, forms a sequence of independent identically distributed random variables having the exponential distribution with
parameter λ > 0, i.e.
P(τ0 > t0 , . . . , τn > tn ) = e−λ(t0 +t1 +···+tn ) ,
t0 , . . . , tn ∈ R+ .
Since the expectation of the exponentially distributed random variable τk
with parameter λ > 0 is given by
IE[τk ] =
1
,
λ
we can check that the higher the intensity λ (i.e. the higher the probability
of having a jump within a small interval), the smaller is the time spent in
each state k ∈ N on average.
In addition, given that {NT = n}, the n jump times on [0, T ] of the Poisson
process (Nt )t∈R+ 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 − λt] = 0,
(14.5)
i.e. the compensated Poisson process (Nt − λt)t∈R+ has centered increments.
Since in addition (Nt − λt)t∈R+ also has independent increments we get
the following proposition.
Proposition 14.2. The compensated Poisson process
(Nt − λt)t∈R+
is a martingale with respect to its own filtration (Ft )t∈R+ .
Extensions of the Poisson process include Poisson processes with timedependent intensity, and with random time-dependent intensity (Cox processes). Renewal processes are counting processes
X
Nt =
1[Tn ,∞) (t),
t ∈ R+ ,
n≥1
in which τk = Tk+1 − Tk , k ∈ N, is a sequence of independent identically
distributed random variables. In particular, Poisson processes are renewal
processes.
14.2 Compound Poisson Processes
The Poisson process itself appears to be too limited to develop realistic price
models as its jumps are of constant size. Therefore there is some interest in
considering jump processes that can have random jump sizes.
Let (Zk )k≥1 denote an i.i.d. sequence of square-integrable random variables with probability distribution ν(dy) on R, independent of the Poisson
process (Nt )t∈R+ . We have
P(Zk ∈ [a, b]) = ν([a, b]) =
wb
a
ν(dy),
−∞ < a ≤ b < ∞.
Definition 14.1. The process
Yt =
Nt
X
Zk ,
t ∈ R+ ,
(14.6)
k=1
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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.
2.5
2
Yt
1.5
1
0.5
0
-0.5
0
2
4
6
8
10
t
Fig. 14.2: Sample path of a compound Poisson process (Yt )t∈R+ .
Given that {NT = n}, the n jump sizes of (Yt )t∈R+ on [0, T ] are independent
random variables which are distributed on R according to ν(dx). Based on
this fact, the next proposition allows us to compute the characteristic function
of the increment YT − Yt .
Proposition 14.3. For any t ∈ [0, T ] we have
w∞
IE [exp (iα(YT − Yt ))] = exp λ(T − t)
(eiyα − 1)ν(dy) ,
−∞
α ∈ R.
Proof. Since Nt has a Poisson distribution with parameter t > 0 and is
independent of (Zk )k≥1 , for all α ∈ R we have by conditioning:
"
IE [exp (iα(YT − Yt ))] = IE exp iα
!#
NT
X
Zk
k=Nt +1
"
= IE exp iα
NT
−Nt
X
!#
Zk
k=1
=
∞
X
"
IE exp iα
n=0
= e−λ(T −t)
n
X
!#
Zk
P(NT − Nt = n)
k=1
"
!#
∞
n
X
X
λn
(T − t)n IE exp iα
Zk
n!
n=0
k=1
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= e−λ(T −t)
∞
X
λn
n
(T − t)n (IE [exp (iαZ1 )])
n!
n=0
= exp (λ(T − t) IE [exp (iαZ1 )])
w∞
= exp λ(T − t)
(eiαy − 1)ν(dy) ,
−∞
since ν(dy) is the probability distribution of Z1 and
r∞
−∞
ν(dy) = 1.
From the characteristic function we can compute the expectation and variance of Yt for fixed t, as
IE[Yt ] = λt IE[Z1 ]
and
Var [Yt ] = λt IE[|Z1 |2 ].
For the expectation we have
IE[Yt ] = −i
w∞
d
IE[eiαYt ]|α=0 = λt
yν(dy) = λt IE[Z1 ].
−∞
dα
This relation can also be directly recovered as
##
" "N
t
X
Zk Nt
IE[Yt ] = IE IE
k=1
" n
#
∞
X
X
λn tn
=e
IE
Zk Nt = n
n!
n=0
k=1
" n
#
∞
X
X
λn tn
−λt
=e
IE
Zk
n!
n=0
−λt
k=1
∞
X
(λt)n−1
= λte−λt IE[Z1 ]
(n − 1)!
n=1
= λt IE[Z1 ].
More generally one can show that for all 0 ≤ t0 ≤ t1 · · · ≤ tn and
α1 , . . . , αn ∈ R we have
" n
#
!
n
w∞
Y iα (Y −Y
X
iαk y
tk
tk−1 )
k
(e
− 1)ν(dy)
IE
e
= exp λ
(tk − tk−1 )
k=1
−∞
k=1
=
=
n
Y
k=1
n
Y
exp λ(tk − tk−1 )
w∞
−∞
(eiαk y − 1)ν(dy)
h
i
IE eiα(Ytk −Ytk−1 ) .
k=1
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This shows in particular that the compound Poisson process (Yt )t∈R+ has
independent increments, as the standard Poisson process (Nt )t∈R+ .
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 − λt IE[Z1 ],
t ∈ R+ ,
is a martingale.
By construction, compound Poisson processes only have a finite number
of jumps on any interval. They belong to the family of L´evy processes which
may have an infinite number of jumps on any finite time interval, cf. [13].
14.3 Stochastic Integrals with Jumps
Given (φt )t∈R+ a stochastic process we let the stochastic integral of (φt )t∈R+
with respect to (Yt )t∈R+ be defined by
wT
0
φt dYt :=
NT
X
φTk Zk .
k=1
wT
Note that this expression
φt dYt has a natural financial interpretation as
0
the value at time T of a portfolio containing a (possibly fractional) quantity
φt of a risky asset at time t, whose price evolves according to random returns
Zk at random times Tk .
In particular the compound Poisson process (Yt )t∈R+ in (14.1) admits the
stochastic integral representation
wt
ZNs dNs .
Yt = Y0 +
0
Next, given (Wt )t∈R+ a standard Brownian motion independent of (Yt )t∈R+
and (Xt )t∈R+ a jump-diffusion process of the form
Xt =
wt
0
us dWs +
wt
0
vs ds + Yt ,
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t ∈ R+ ,
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Notes on Stochastic Finance
where (φt )t∈R+ is a process which is adapted to the filtration (Ft )t∈R+ generated by (Wt )t∈R+ and (Yt )t∈R+ , and such that
hw ∞
i
hw ∞
i
IE
φ2s |us |2 ds < ∞ and IE
|φs vs |ds < ∞,
0
0
we let the stochastic integral of (φs )s∈R+ with respect to (Xs )s∈R+ be defined
by
wT
0
φs dXs :=
wT
0
φs us dWs +
wT
0
φs vs ds +
NT
X
φ Tk Z k ,
T > 0.
k=1
The coumpound Poisson compensated stochastic integral can be shown to
satisfy the Itˆo isometry
IE
"
w
T
0
2 #
w
T
φs− (dYs − λ IE[Z1 ]dt)
= λ IE[|Z1 |2 ] IE
|φ|2s ds ,
0
(14.7)
provided the process (φt )t∈R+ is adapted to the filtration generated by
(Yt )t∈R+ , which makes the left limit process (φs− )s∈R+ predictable. In the
case of simple predictable processes, the proof of (14.7) is similar to that of
Proposition 4.2.
For the mixed continuous-jump martingale
wt
Xt =
us dWs + Yt − λt IE[Z1 ],
0
t ∈ R+ ,
we have the isometry
IE
"
wT
0
2 #
φs− dXs
= IE
w
T
0
w
T
|φs |2 ds .
|φs |2 |us |2 ds + λ IE[|Z1 |2 ] IE
0
(14.8)
provided (φs )s∈R+ is adapted to the filtration (Ft )t∈R+ generated by (Wt )t∈R+
and (Yt )t∈R+ .
This isometry formula will be used in Section 15.5 for the computation of
hedging strategies in jump models.
When (Xt )t∈R+ takes the form
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Xt = X0 +
wt
0
us dWs +
wt
0
vs ds +
wt
0
t ∈ R+ ,
ηs dYs ,
the stochastic integral of (φt )t∈R+ with respect to (Xt )t∈R+ satisfies
wT
0
φs dXs :=
wT
φs us dWs +
wT
φs vs ds +
=
wT
φs us dWs +
wT
φs vs ds +
0
0
0
0
wT
0
NT
X
ηs φs dYs
φTk ηTk Zk ,
T > 0.
k=1
14.4 Itˆ
o Formula with Jumps
Let us first consider the case of a standard Poisson process (Nt )t∈R+ with
intensity λ. We have the telescoping sum
f (Nt ) = f (0) +
= f (0) +
= f (0) +
Nt
X
(f (k) − f (k − 1))
k=1
wt
0
wt
0
(f (1 + Ns− ) − f (Ns− ))dNs
(f (Ns ) − f (Ns− ))dNs .
Here, Ns− denotes the left limit of the Poisson process at time s, i.e.
Ns− = lim Ns−h .
h&0
In particular we have
k = NTk = 1 + NT − ,
k ≥ 1.
k
By the same argument we find, in the case of the compound Poisson process
(Yt )t∈R+ ,
f (Yt ) = f (0) +
= f (0) +
= f (0) +
Nt
X
(f (YT − + Zk ) − f (YT − ))
k
k
k=1
wt
0
wt
0
(f (ZNs + Ys− ) − f (Ys− ))dNs
(f (Ys ) − f (Ys− ))dNs ,
which can be decomposed using a compensated Poisson stochastic integral
as
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f (Yt ) = f (0) +
wt
0
(f (Ys ) − f (Ys− ))(dNs − λds) + λ
wt
0
(f (Ys ) − f (Ys− ))ds.
More generally, for a process of the form
wt
wt
wt
Xt = X0 +
us dWs +
vs ds +
ηs dYs ,
0
0
t ∈ R+ ,
0
we find, by combining the Itˆo formula for Brownian motion with the above
argument we get
f (Xt ) = f (X0 ) +
+
wt
0
wt
0
us f 0 (Xs )dWs +
vs f 0 (Xs )ds +
wt
NT
X
1 w t 00
f (Xs )|us |2 ds
2 0
(f (XT − + ηTk Zk ) − f (XT − ))
k
k
k=1
wt
1 w t 00
us f 0 (Xs )dWs +
f (Xs )|us |2 ds +
vs f 0 (Xs )ds
0
2 0
= f (X0 ) +
0
wt
+ (f (Xs− + ηs ZNs ) − f (Xs− ))dNs
0
t ∈ R+ .
i.e.
wt
wt
1 w t 00
f (Xt ) = f (X0 ) +
us f 0 (Xs )dWs +
f (Xs )|us |2 ds +
vs f 0 (Xs )ds
0
0
2 0
wt
+ (f (Xs ) − f (Xs− ))dNs ,
t ∈ R+ .
(14.9)
0
For example, in case
wt
wt
wt
vs ds +
ηs dNs ,
Xt =
us dWs +
0
0
0
t ∈ R+ ,
we get
wt
1wt
|us |2 f 00 (Xs )dWs
f (Xt ) = f (0) +
us f 0 (Xs )dWs +
0
2 0
wt
wt
+ vs f 0 (Xs )ds + (f (Xs− + ηs ) − f (Xs− ))dNs
0
0
wt
1wt
= f (0) +
us f 0 (Xs )dWs +
|us |2 f 00 (Xs )dWs
(14.10)
0
2 0
wt
wt
+ vs f 0 (Xs )ds + (f (Xs ) − f (Xs− ))dNs .
0
0
Given two processes (Xt )t∈R+ and (Yt )t∈R+ written as
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Xt =
wt
us dWs +
wt
vs ds +
wt
ηs dNs ,
t ∈ R+ ,
Yt =
wt
as dWs +
wt
bs ds +
wt
cs dNs ,
t ∈ R+ ,
and
0
0
0
0
0
0
the Itˆo formula for jump processes also shows that
d(Xt Yt ) = Xt dYt + Yt dXt + dXt · dYt
where the product dXt · dYt is computed according to the extension
·
dt
dBt
dNt
dt
0
0
0
dBt
0
dt
0
dNt
0
0
dNt
of the Itˆo multiplication table (4.19), i.e. we have
dXt · dYt = (vt dt + ut dBt + ηt dNt )(bt dt + at dBt + ct dNt )
= bt vt (dt)2 + bt ut dt · dBt + bt ηt dt · dNt
+at vt dtdBt + at ut (dBt )2 + at ηt dBt · dNt
+ct vt dNt · dBt + ct ut (dBt )2 + ct ηt dNt · dNt
= at ut dt + ct ηt dNt ,
and in particular
(dXt )2 = (vt dt + ut dBt + ηt dNt )2 = u2t dt + ηt2 dNt .
For a process of the form
Xt = X0 +
wt
0
us dWs +
wt
0
ηs dYt ,
t ∈ R+ ,
the Itˆo formula with jumps (14.10) can be rewritten as
wt
wt
f (Xt ) = f (X0 ) +
vs f 0 (Xs− )ds +
us f 0 (Xs− )dWs
0
0
w
w
t
1 t 00
+
f (Xs )|us |2 ds +
ηs f 0 (Xs− )dYs
0
2 0
wt
+ (f (Xs ) − f (Xs− ) − ∆Xs f 0 (Xs− )) d(Ns − s)
0
wt
+ (f (Xs ) − f (Xs− ) − ∆Xs f 0 (Xs− )) ds,
t ∈ R+ ,
0
where we used the relation dYs = ∆Xs f 0 (Xs− )dNs , which implies
wt
0
ηs f 0 (Xs− )dYs =
wt
0
∆Xs f 0 (Xs− )dNs ,
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t ≥ 0.
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Notes on Stochastic Finance
This above formulation is at the basis of the extension of Itˆ
o’s formula to
L´evy processes with an infinite number of jumps on any interval, using the
bound
|f (x + y) − f (x) − yf 0 (x)| ≤ Cy 2 ,
for f a Cb2 (R) function. Such processes, also called “infinite activity L´evy
processes” [13] are also useful in financial modeling and include the gamma
process, stable processes, variance gamma processes, inverse Gaussian processes, 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 = ηSt− dNt ,
(14.11)
of a stochastic differential equation with respect to the standard Poisson process, with constant coefficient η ∈ R.
When
∆Nt = Nt − Nt− = 1,
i.e. when the Poisson process has a jump at time t, the equation (14.11) reads
dSt = St − St− = ηSt− ,
t > 0.
which can be solved to yield
St = (1 + η)St− ,
t > 0.
By induction, applying this procedure for each jump time gives us the solution
St = S0 (1 + η)Nt ,
t ∈ R+ .
Next, consider the case where η is time-dependent, i.e.
dSt = ηt St− dNt .
(14.12)
At each jump time Tk , Relation (14.12) reads
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dSTk = STk − ST − = ηTk ST − ,
k
k
i.e.
STk = (1 + ηTk )ST − ,
k
and repeating this argument for all k = 1, . . . , Nt yields the product solution
St = S0
Nt
Y
(1 + ηTk ) = S0
Y
t ∈ R+ .
(1 + ηs ),
∆Ns =1
0≤s≤t
k=1
The equation
dSt = µt St dt + ηt St− (dNt − λdt),
(14.13)
is then solved as
St = S0 exp
w
t
0
µs ds − λ
wt
0
Y
Nt
(1 + ηTk ),
ηs ds
t ∈ R+ .
k=1
A random simulation of the numerical solution of the above equation (14.13)
is given in Figure 14.8.
2
St
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.8: Geometric Poisson process.∗
The above simulation can be compared to the real sales ranking data of
Figure 14.9.
∗
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Fig. 14.9: Ranking data.
A random simulation of the geometric compound Poisson process
St = S0 exp
w
t
0
µs ds − λ IE[Z1 ]
wt
0
Y
Nt
(1 + ηTk Zk )
ηs ds
t ∈ R+ ,
k=1
solution of
dSt = µt St dt + ηt St− (dYt − λ IE[Z1 ]dt),
is given in Figure 14.10.
2
St
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.10: Geometric compound Poisson process.∗
In the case of a jump-diffusion stochastic differential equation of the form
dSt = µt St dt + ηt St− (dYt − λ IE[Z1 ]dt) + σt St dWt ,
∗
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we get
St = S0 exp
×
Nt
Y
w
t
0
µs ds − λ IE[Z1 ]
wt
0
ηs ds +
wt
0
σs dWs −
1wt
|σs |2 ds
2 0
(1 + ηTk Zk ),
k=1
t ∈ R+ . A random simulation of the geometric Brownian motion with compound Poisson jumps is given in Figure 14.11.
3
2.5
St
2
1.5
1
0.5
0
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
2.0
t
Fig. 14.11: Geometric Brownian motion with compound Poisson jumps.∗
By rewriting St as
w
wt
wt
t
1wt
|σs |2 ds
µs ds +
ηs (dYs − λ IE[Z1 ]ds) +
σs dWs −
St = S0 exp
0
0
0
2 0
×
Nt
Y
(e−ηTk (1 + ηTk Zk )),
k=1
t ∈ R+ , one can extend this jump model to processes with an infinite number
of jumps on any finite time interval, cf. [13]. The next Figure 14.12 shows
a number of downward and upward jumps occuring in the historical price
of the SMRT stock, with a typical geometric Brownian behavior in between
jumps.
∗
The animation works in Acrobat reader on the entire file.
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Notes on Stochastic Finance
Fig. 14.12: SMRT Stock price.
14.6 Girsanov Theorem for Jump Processes
Recall that in its simplest form, the Girsanov theorem for Brownian motion
follows from the calculation
2
1 w∞
f (x − µT )e−x /(2T ) dx
IE[f (WT − µT )] = √
2πT −∞
2
1 w∞
= √
f (x)e−(x+µT ) /(2T ) dx
2πT −∞
2
2
1 w∞
= √
f (x)e−µx−µ T /2 e−x /(2T ) dx
2πT −∞
2
= IE[f (WT )e−µWT −µ
˜ (WT )],
= IE[f
T /2
]
(14.14)
for any bounded measurable function f on R, which shows that WT is a
˜
Gaussian random variable with mean −µT under the probability measure P
defined by
˜ = e−µWT −µ2 T /2 dP,
dP
cf. Section 6.2. Equivalently we have
˜ (WT + µT )],
IE[f (WT )] = IE[f
(14.15)
hence
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under the probability measure
2
˜ = e−µWT −µ
dP
T /2
dP,
the random variable WT + µT has a centered Gaussian distribution.
More generally, the Girsanov theorem states that (Wt + µt)t∈[0,T ] is a stan˜
dard Brownian motion under P.
When Brownian motion is replaced with a standard Poisson process
(Nt )t∈R+ , the above space shift
Wt 7−→ Wt + µt
may not be used because Nt + µt cannot be a Poisson process, whatever the
change of probability applied, since by construction, the paths of the standard Poisson process has jumps of unit size and remain constant between
jump times.
The correct way to proceed in order to extend (14.15) to the Poisson case
is to replace the space shift with a time contraction (or dilation) by a certain
factor 1 + c with c > −1, i.e.
Nt 7−→ Nt/(1+c) .
By analogy with (14.14) we write
IE[f (NT (1+c) )] =
∞
X
f (k)P(NT (1+c) = k)
(14.16)
k=0
= e−λ(1+c)T
=e
∞
X
f (k)
k=0
∞
X
−λT −λcT
e
(λT (1 + c))k
k!
f (k)(1 + c)k
k=0
=e
−λcT
∞
X
(λT )k
k!
f (k)(1 + c)k P(NT = k)
k=0
= e−λcT IE[f (NT )(1 + c)NT ]
w
= e−λcT
(1 + c)NT f (NT )dP
Ω
w
˜
=
f (NT )dP
Ω
˜ (NT )],
= IE[f
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˜ is defined
for f any bounded function on N, where the probability measure P
by
˜ ˜ = e−λcT (1 + c)NT dP.
dP
λ
Consequently,
under the probability measure
˜ = e−λcT (1 + c)NT dP,
dP
the law of the random variable NT is that of NT (1+c) under P, i.e. it is a
Poisson random variable with intensity λ(1 + c)T .
Equivalently we have
˜ (NT /(1+c) )],
IE[f (NT )] = IE[f
˜ the law of NT /(1+c) is that of a standard Poisson random variable
i.e. under P
with parameter λT .
In addition we have
Nt/(1+c) =
X
1[Tn ,∞) (t/(1 + c))
n≥1
=
X
1[(1+c)Tn ,∞) (t),
t ∈ R+ ,
n≥1
˜ the jump times of (Nt/(1+c) )t∈[0,T ] are given by
which shows that under P,
((1 + c)Tn )n≥1 ,
and we know that they are distributed as the jump times of a Poisson process
with intensity λ.
˜ > 0 and
Next taking λ
c := −1 +
˜
λ
,
λ
we can rewrite the above by saying that
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N. Privault
under the probability measure
˜
˜ ˜ = e−λcT (1 + c)NT dP = e−(λ−λ)T
dP
λ
˜
λ
λ
!NT
dP,
the law of NT is that of a Poisson random variable with intensity
˜ = λ(1 + c)T.
λT
Consequently, since both (Nt −λt)t∈R+ and (Nt −(1+c)λt)t∈R+ are processes
with independent increments, the compensated Poisson process
˜
Nt − (1 + c)λt = Nt − λt
˜ ˜ by (6.2), although when c 6= 0 it is not a martingale
is a martingale under P
λ
under P.
In the case of compound Poisson processes the Girsanov theorem can be
extended to variations in jump sizes in addition to time variations, and we
have the following more general result.
Theorem 14.2. Let (Yt )t≥0 be a compound Poisson process with intensity λ > 0 and jump distribution ν(dx). Consider another jump distribution
ν˜(dx), and let
˜ d˜
ν
λ
(x) − 1,
x ∈ R.
φ(x) =
λ dν
Then,
under the probability measure
˜
˜ ˜ := e−(λ−λ)T
dP
λ,˜
ν
NT
Y
˜ λ,ν ,
(1 + φ(Zk ))dP
k=1
the process
Yt =
Nt
X
Zk ,
t ∈ R+ ,
k=1
is a compound Poisson process with
˜ > 0, and
- modified intensity λ
- modified jump distribution ν˜(dx).
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Proof. For any bounded measurable function f on R, we extend (14.16) to
the following change of variable
#
"
NT
Y
˜
−(λ−λ)T
IEλ,˜
IEλ,ν f (YT ) (1 + φ(Zi ))
˜ ν [f (YT )] = e
i=1
˜
= e−(λ−λ)T
∞
X
"
k
X
IEλ,ν f
=e
˜
−λT
k=0
˜
= e−λT
˜
˜
= e−λT
"
IEλ,ν f
k!
−∞
∞ ˜
X
(λT )k w ∞
k=0
∞
X
k=0
k!
k
Y
−∞
···
···
w∞
−∞
w∞
#
(1 + φ(Zi )) NT = k P(NT = k)
i=1
k
X
!
Zi
i=1
∞
X
(λT )k w ∞
k=0
= e−λT
k!
Zi
i=1
k=0
∞
X
(λT )k
!
−∞
k
Y
#
(1 + φ(Zi ))
i=1
f (z1 + · · · + zk )
k
Y
(1 + φ(zi ))ν(dz1 ) · · · ν(dzk )
i=1
f (z1 + · · · + zk )
k
Y
d˜
ν
i=1
dν
!
(zi ) ν(dz1 ) · · · ν(dzk )
w∞
˜ )k w ∞
(λT
···
f (z1 + · · · + zk )˜
ν (dz1 ) · · · ν˜(dzk ).
−∞
−∞
k!
This shows that under Pλ,˜
˜ ν , YT has the distribution of a compound Poisson
˜ and jump distribution ν˜. We refer to Proposition 9.6
process with intensity λ
˜˜ .
of [13] for the independence of increments of (Yt )t∈R+ under P
λ,˜
ν
˜ > 0 and jump
Note that the compound Poisson process with intensity λ
distribution ν˜ can be built as
Nλt/λ
˜
Xt :=
X
h(Zk ),
k=1
provided ν˜ is the image measure of ν by the function h : R → R, i.e.
P(h(Zk ) ∈ A) = P(Zk ∈ h−1 (A)) = ν(h−1 (A)) = ν˜(A),
for all measurable subset A of R.
Compensated Compound Poisson Martingale
As a consequence of Theorem 14.2, the compensated process
˜ IEν˜ [Z1 ]
Yt − λt
˜ ˜ defined by
becomes a martingale under the probability measure P
λ,˜
ν
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˜
˜ ˜ = e−(λ−λ)T
dP
λ,˜
ν
NT
Y
˜ λ,ν .
(1 + φ(Zk ))dP
k=1
Finally, the Girsanov theorem can be extended to the linear combination
of a standard Brownian motion (Wt )t∈R+ and an independent compound
Poisson process (Yt )t∈R+ , as in the following result which is a particular case
of Theorem 33.2 of [75].
Theorem 14.3. Let (Yt )t≥0 be a compound Poisson process with intensity
λ > 0 and jump distribution ν(dx). Consider another jump distribution ν˜(dx)
˜ > 0, and let
and intensity parameter λ
φ(x) =
˜ d˜
λ
ν
(x) − 1,
λ dν
x ∈ R,
and let (ut )t∈R+ be a bounded adapted process. Then the process
wt
˜ IEν˜ [Z1 ]t
Wt +
us ds + Yt − λ
0
t∈R+
is a martingale under the probability measure
Y
NT
wT
1wT
˜ − λ)T −
˜ ˜ = exp −(λ
˜ λ,ν .
dP
us dWs −
(1+φ(Zk ))dP
|us |2 ds
u,λ,˜
ν
0
2 0
k=1
(14.17)
As a consequence of Theorem 14.3, if
wt
Wt +
vs ds + Yt
0
˜ λ,ν , it will become a martingale under P
˜ ˜
is not a martingale under P
u,λ,˜
ν
˜
provided u, λ and ν˜ are chosen in such a way that
˜ IEν˜ [Z1 ],
vs = us − λ
s ∈ R,
(14.18)
in which case we will have the martingale decomposition
in which both
˜ IEν˜ [Z1 ]dt,
dWt + ut dt + dYt − λ
wt
˜ IEν˜ [Z1 ]
Wt +
us ds
and Yt − λt
0
t∈R+
are both mart∈R+
˜ ˜
tingales under P
u,λ,˜
ν
˜ = λ = 0, Theorem 14.3 coincides with the usual Girsanov theorem
When λ
for Brownian motion, in which case (14.18) admits only one solution given
˜ u,0,0 . Note that uniqueness occurs also
by u = v and there is uniqueness of P
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when u = 0 in the absence of Brownian motion with Poisson jumps of fixed
size a (i.e. ν˜(dx) = ν(dx) = δa (dx)) since in this case (14.18) also admits
˜ = v and there is uniqueness of P
˜ ˜ . These remarks will
only one solution λ
0,λ,δa
be of importance for arbitrage pricing in jump models in Chapter 15.
Exercises
Exercise 14.1 Let (Nt )t∈R+ be a standard Poisson process with intensity
λ > 0, started at N0 = 0.
1. Solve the stochastic differential equation
dSt = ηSt− dNt − ηλSt dt = ηSt− (dNt − λdt).
2. Using the first Poisson jump time T1 , solve the stochastic differential
equation
dSt = −ηSt dt + dNt
for t ∈ (0, T2 ).
Exercise 14.2 Consider the compound Poisson process Yt :=
Nt
X
Zk , where
k=1
(Nt )t∈R+ is a standard Poisson process with intensity λ > 0, (Zk )k≥1 is an
i.i.d. sequence of N (0, 1) Gaussian random variables. Solve the stochastic
differential equation
dSt = rSt dt + ηSt− dYt ,
where η, r ∈ R.
Exercise 14.3 Show, by direct computation or using the characteristic function, that the variance of the compound Poisson process Yt with intensity
λ > 0 satisfies
w∞
Var [Yt ] = λt IE[|Z1 |2 ] = λt
x2 ν(dx).
−∞
Exercise 14.4 Consider an exponential compound Poisson process of the form
St = S0 eµt+σWt +Yt ,
t ∈ R+ ,
where (Yt )t∈R+ is a compound Poisson process of the form (14.6).
1. Derive the stochastic differential equation with jumps satisfied by (St )t∈R+ .
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˜ ˜ ) of probability measures under which
2. Let r > 0. Find a family (P
u,λ,˜
ν
the discounted asset price e−rt St is a martingale.
Exercise 14.5 Consider (Nt )t∈R+ a standard Poisson process with intensity
λ > 0, independent of (Wt )t∈R+ , under a probability measure P. Let (St )t∈R+
be defined by the stochastic differential equation
dSt = µSt dt + YNt St− dNt ,
(14.19)
where (Yk )k≥1 is an i.i.d. sequence of random variables of the form Yk =
eXk − 1 where Xk ' N (0, σ 2 ), k ≥ 1.
1. Solve the equation (14.19).
2. We assume that µ and the risk-free rate r > 0 are chosen such that the
discounted process (e−rt St )t∈R+ is a martingale under P. What relation
does this impose on µ and r ?
3. Under the relation of Question (2), compute the price at time t of a
European call option on ST with strike κ and maturity T , using a series
expansion of Black-Scholes functions.
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