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