When is compound Poisson process a martingale?

53 Views Asked by At

here's my proof of a claim that the Compound Poisson Process (CPP for short) is a martingale $\iff$ the expected value of the iid random variables we are summing is $0$.

A stochastic process $X_t$ adapted to the filtration $(\mathcal{F}_t)_{t\geq0}$ is a martingale if the following condition holds: $$ E\left[X_t\mid\mathcal{F}_s\right] = X_s $$ for every $0\leq s < t$.

We want to show that, in general, a Compound Poisson Process (CPP) is not a martingale.

Claim:

Let $(S_t)_{t\geq0}$ be a CPP with intensity $\lambda>0$, and let $X_i$ be independent and identically distributed random variables with $E\left[X_i\right] = \mu$ for each $i$. Then, $S_t$ is a martingale if and only if $\mu = 0$.

Proof:

Consider $0\leq s\leq t$. We have \begin{align*} E\left[S_t\mid\mathcal{F}_s\right] &= E\left[S_t - S_s + S_s\mid \mathcal{F}_s\right] \\ &= E\left[S_t - S_s\right] + E\left[S_s\mid \mathcal{F}_s\right] \\ & \text{recall that} \ E\left[S_t\right] = \mu\lambda t \\ &= \mu\lambda(t-s) + S_s \end{align*} The equality $\mu\lambda(t-s) + S_s = S_s$ holds $\iff$ $\mu\lambda(t-s) = 0 \iff \mu = 0$.

I'm wondering if there is any fault in my proof, or is it fine?