Changing Poisson Intensities Under an Equivalent Martingale Measure

# Poisson process under equivalent martingale measure


# Poisson process under equivalent martingale measure












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

## Answer by Arshdeep (score 1)

https://quant.stackexchange.com/a/65407

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 $Z$

$E'(1_{N(T)=1})=E(1_{N(T)=1}*(1_{N(T)=1}+1-Pr(N(T)=1)))$

Thus,

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

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.

## Answer by ir7 (score 1)

https://quant.stackexchange.com/a/65418

$N_t$ 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.)