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.


1 Answer 1


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 $t\mapsto t^p$.

  • $\begingroup$ 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? $\endgroup$
    – 3nondatur
    Nov 10, 2022 at 9:54
  • 2
    $\begingroup$ $X_i$ is $+1$ with probability one half and $-1$ with probability one half. So the expectation $E[e^{upX_i}]$ is... $\endgroup$
    – justt
    Nov 10, 2022 at 10:07
  • $\begingroup$ 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 $\mathbb{E}[e^{upX_i}] = 1$? $\endgroup$
    – 3nondatur
    Nov 10, 2022 at 10:48
  • $\begingroup$ The random variable $e^{upX_i}$ takes the value $e^{up}$ with probability $1/2$ and $e^{-up}$ with probability $1/2$. $\endgroup$ Nov 12, 2022 at 9:30

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