I have a stochastic process $N(t)$ which is equal to $n$ with probability

$P\{N(t) = n\}=\frac{\left(\lambda t \right)^{n}}{n!}e^{-\lambda t }$

where $t$ represents the time period. In other words, the corresponding process for a fixed $t$, is a random variable $N(t) \equiv N$ which is a (homogeneous) Poisson (point) process with the following Poisson distribution:

$P\{N = n\}=\frac{\lambda^{n}}{n!}e^{-\lambda}$

$P$ is the probability measure defined with respect to the sample space $\Omega$. Together with the sigma-algebra $A$, these three elements define my probability space.

Now I introduce the equivalent martingale measure (EMM) $Q$, which is equivalent to $P$ and has the property that under $Q$ each process becomes a martingale. To be more clear, this is the typical setup for the Black-Scholes type of pricing. For example, a stock which has a process under $P$ defined by

$dS_t=S_t\mu+S_t\sigma dW_t$

where $W_t$ is a Wiener process, has a drift given by the so called "risk-free rate" $r_f$ under $Q$ after also changing to the corresponding Wiener process under $Q$, i.e. $W^{Q}_t$

$dS_t=S_t r_f+S_t\sigma dW^{Q}_t$

The change of measure requires $W^{Q}_t=W_t+((\mu-r_f) / \sigma)t$

My question: is a Poisson process like the one presented above affected at all by the change of measure? My guess is that since it depends only on a certain parameter $\lambda$ and it is a counting process this is not the case, but I would like to hear further opinions. Any suggestion is well accepted and well taken

  • $\begingroup$ Are you 1. defining a Poisson process, and then changing to some equivalent measure, or 2. defining a Poisson and a stock price process, and changing to a measure which makes the stock price a martingale (the one you explicitly write out)? $\endgroup$
    – Arshdeep
    Jun 4, 2021 at 13:37
  • $\begingroup$ Number 1, i.e. I define a Poisson Process and then I change it to an equivalent measure. If there is a difference between the two options you pointed out, I would like to hear what both options involve $\endgroup$ Jun 4, 2021 at 13:45
  • 1
    $\begingroup$ Nah I'm just trying to understand the problem before I take a shot at it. BTW, you might want to look into basic option pricing theory with jumps, you'll probably find something useful/ related there. I haven't read much of it but I can roughly recall that the Poisson parameter changes under change of measure, don't quote me though. $\endgroup$
    – Arshdeep
    Jun 4, 2021 at 13:49
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    $\begingroup$ If you can find a measure such that $N(T)+k*T$ is a Poisson process with parameter k in this new measure - this would be your EMM. Whether one exists or not would probably be answered by some analogy to Girsanov's theorem in this case. $\endgroup$
    – Arshdeep
    Jun 4, 2021 at 14:17
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    $\begingroup$ For the inner workings, Privault’s lecture notes have a lot of details on Girsanov for jumps: personal.ntu.edu.sg/nprivault/MA5182/… $\endgroup$ Jun 5, 2021 at 3:57

2 Answers 2


Consider a radon nikodym derivative, the Random variable: $Z(w)= 1_{N(T,w)=1}+1-Pr(N(T)=1)$. It is admissible since it is always positive and has an expectation 1. This will lead us to the formation of an equivalent measure, which I will denote by $'$.

We start with the easy theorem that $E'(X)=E(XZ)$ for any random variable X, and RND $Z$



$Pr'(N(T)=1)=2Pr(N(T)=1)-[Pr(N(T)=1)]^2 > Pr(N(T)=1)$

Note also

$Pr'(N(T)=2)=Pr(N(T)=2)[1-Pr(N(T)=1)] < Pr[N(T)=2]$

So the new measure is providing more probability mass at $1$, while taking it away from all the other points (like $2$, as shown above). This is no longer a poisson process, which you can verify with a bit of algebra.

Comment: I mistook your question as to finding a EMM for the poisson process, which was admittedly hard. But establishing that it is sensitive to a change of equivalent measure is not hard. Almost all non degenerate process are sensitive to a change of measure.

  • $\begingroup$ Nice! 2 things though: how do you get the two lines after "thus" in your answer? Also, do you have an idea of how to at least attacke the problem of finding the EMM for a poisson process? I know this is another thing, but it would still be very interesting $\endgroup$ Jun 5, 2021 at 7:55
  • $\begingroup$ Expectation of the indicator variable is just the probability of the event. Expectation of $1_A*1_B*$ is 0 if $A$ and $B$ are mutually exclusive is 0. $\endgroup$
    – Arshdeep
    Jun 5, 2021 at 11:53
  • $\begingroup$ For EMM, you usually work back a variable $Y=:f(N)$, which becomes a Poisson process under a certain measure, the proof of which usually reduces to $N$ becoming a martingale. Like I found $Y(t)=:N(t)+kt$, and thought of a measure under which this becomes Poisson. It follows that $N$ must be a martingale under that measure. Whether it is 'equivalent' follows iff the density of $N$ is non 0 only in the set {0,1,2...} in this new measure, which is not the case for the above measure. $\endgroup$
    – Arshdeep
    Jun 5, 2021 at 12:05
  • 1
    $\begingroup$ So I find it hard to think about an explicit representation $Y=f(N)$ which would satistfy 'equivalence' (recall that if $Y=:N+kt$ is always an integer (as it's Poisson), $N$ is taking non integer values now with positive chance), so I couldn't proceed further. $\endgroup$
    – Arshdeep
    Jun 5, 2021 at 12:09

$N_t$ process comes with its own Poisson law (probability measure) $P$ defined via intensity $\lambda$. Under it, $N_t-\lambda t$ is a martingale wrt ${\cal F}_t =\sigma(N_u | u\in [0,t])$ (as $E^P[N_t]=\lambda t$ and $N_t-\lambda t$ has independent increments).

Any other equivalent Poisson law, $Q$, defined via a given intensity $\gamma$, can be built using Radon-Nikodym density

$$ \frac{dQ}{dP}{\bigg|}_{{\cal F}_t} = \exp\left( \ln(\gamma/\lambda)N_t-(\gamma -\lambda)t \right), $$ by noticing that

$$ Q(N_t=n) = \exp\left( \ln(\gamma/\lambda)n-(\gamma -\lambda)t \right) P(N_t=n)$$ $$= \exp\left( \ln(\gamma/\lambda)n-(\gamma -\lambda)t \right) (n!)^{-1}(\lambda t)^n\exp(-\lambda t) = (n!)^{-1}(\gamma t)^n\exp(-\gamma t).$$

Under $Q$, $N_t-\gamma t$ is a ${\cal F}_t$-martingale.

In the finance context, if we only have a simple asset modeled by process (an independent Brownian motion can be added - see Merton's model, for example, but we will keep it about Poisson processes here):

$$ dS_t= S_{t^-} (\mu dt + \zeta d(N_t-\lambda t)),$$

under $P$, $\zeta>0$, then (with constant interest rate $r$)

$$ d(e^{-rt}S_t)/(e^{-rt}S_{t^-}) = (\mu -r) dt + \zeta d(N_t -\lambda t) $$

which shows that $e^{-rt}S_t$ is not a martingale when $\mu \not= r$.

The discounted asset is a martingale under Poisson $Q$ built as above for intensity:

$$\gamma = \lambda - \frac{\mu -r}{\zeta}, $$

making $Q$ an EMM.

(See notes suggested in the comments and also these ones.)


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