How exactly can this martingale be bounded?

Question

We define $X_t:=e^{(\lambda-\kappa)t}(\frac{\kappa}{\lambda})^{N_t}$, where $N_t$ is a Poisson process with parameter $\lambda$ and both parameters $\lambda,\kappa>0$. I want to find an upper bound of

$$E[\sup_{0\le t\le T}X_t]$$

I thought the following:

• if $\lambda\ge \kappa$ then we have $E[\sup_{0\le t\le T}X_t]\le e^{(\lambda-\kappa)T}E[\sup_{0\le t\le T}\big(\frac{\kappa}{\lambda}\big)^{N_t}]\le e^{(\lambda-\kappa)T}$
• if $\lambda < \kappa$ then we have $E[\sup_{0\le t\le T}X_t]\le E[\sup_{0\le t\le T}\big(\frac{\kappa}{\lambda}\big)^{N_t}]\le E[\big(\frac{\kappa}{\lambda}\big)^{N_T}]$

I know that $N_T$ is Poisson distributed with parameter $\lambda T$. But how can I get a bound? Thanks for the help

hulik

Answer 1

Just to summarize what have been given in comments.

1. You can compute the expectation explicitly since this just involves some exponential series that are easy to deal with.

2. Note that for $x>0$ it holds that $\mathsf E[x^{N_T}] = \mathsf E[\exp(\xi\cdot\ln x)] = \mathrm {MGF}_\xi(\ln x)$ where $\xi\sim \mathrm{Poi}(\lambda T)$ and its MGF is hence finite.