# Expected value of an exponential martingale

Consider the symmetric random walk $$(S_n)_{n \ge 0}$$ on $$\mathbb{Z}$$. For real $$u \ne 0$$ consider the exponential martingale $$Z_n = e^{uS_n}cosh(u)^{-n}$$ of the symmetric random walk $$(S_n)_{n \ge 0}$$ on $$Z$$.

Show that the martingale $$Z$$ is not $$L^p$$-bounded for any $$p > 1$$ by computing $$\mathbb{E}[Z_n^p]$$.

I do not see how to start with this. I guess that we should use the optional stopping theorem, which yields

$$\mathbb{E}[Z_n^p] = \mathbb{E}[Z_0^p] = \int_{- \infty}^{\infty} u \cdot e^{uS_0p}cosh(u)^{-0 \cdot p} du.$$

However, I do not see how to go on from here.

The step $$\mathbb{E}[Z_n^p] = \mathbb{E}[Z_0^p]$$ is not valid: it is true that $$(Z_n)$$ is a martingale, but $$(Z_n^p)$$ is not a martingale.
It is possible to compute $$\mathbb E\left[Z_n^p\right]$$ explicitely by using independence of the increment of the random walk, denoted $$X_j$$, namely, $$\mathbb E\left[e^{upS_n}\right]=\prod_{j=1}^n\mathbb E\left[e^{upX_j}\right]=\left(\cosh(pu)\right)^n$$, where the last equality follows from the fact that $$\mathbb P(X_i=1)=\mathbb P(X_i=-1)=1/2$$. You will get that $$\mathbb E\left[Z_n^p\right]=a^n$$ for some $$a$$ that you will have to show that it is greater than one, by strict convexity of $$t\mapsto t^p$$.
• Thanks for your answer, but I do not get why $\prod_{j=1}^n\mathbb E\left[e^{upX_i}\right]=\left(\cosh(pu)\right)^n$. Could you please explain this? Nov 10, 2022 at 9:54
• $X_i$ is $+1$ with probability one half and $-1$ with probability one half. So the expectation $E[e^{upX_i}]$ is... Nov 10, 2022 at 10:07
• Sorry, but I still do not see it. I get that $\mathbb{E}[X_i] = 0$, but why should this imply that $\mathbb{E}[e^{upX_i}] = \cosh(pu)$, shouldn't it be $\mathbb{E}[e^{upX_i}] = 1$? Nov 10, 2022 at 10:48
• The random variable $e^{upX_i}$ takes the value $e^{up}$ with probability $1/2$ and $e^{-up}$ with probability $1/2$. Nov 12, 2022 at 9:30