# 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

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.

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Thanks for your answer, especially for the second point –  user20869 Jan 19 '13 at 12:41
@hulik: welcome! –  Ilya Jan 19 '13 at 12:44