# How exactly can this martingale be bounded?

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

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Can't you just calculate the expectation $E\left[\left(\frac{\kappa}{\lambda}\right)^{N_T}\right]$? – Stefan Hansen Jan 18 '13 at 8:40
Can't you just use Doob's inequality for martingales? – Ilya Jan 18 '13 at 8:52
@Ilya This is from an exercise sheet and a priori you should not know that $S_t$ is a martingale. You just know that it is a local martingale. Showing that the $\sup$ is integrable you can conclude that it is a true martingale. And this is the aim of this exercise. I just calculated "for fun" $E[S_t|\mathcal{F_s}]$ and recognized already that $S$ has to be a martingale. However I want to do the exercise, as they suggest, therefore I can not use Doob. I edited my question. – user20869 Jan 18 '13 at 8:58
Well, can't you then just use the fact that the latter expectation $E(\kappa/\lambda)^{N_T}$ is bounded for any $T$? – Ilya Jan 18 '13 at 9:01
Well, this is just a moment generating function of $\mathrm{Poi}(\cdot)$. Perhaps, its finetness is not 100% obvious, but as Stefan mentioned it is almost obvious. Nevertheless, by no mean it is said to discourage you - if you have any doubts there are no words too careful. – Ilya Jan 18 '13 at 9:10

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.