I'm stuck in the following problem! Would be great if you can help.

Suppose you have $(Y_n)_{n\in \mathbb N}$, a sequence of independent and identically distributed random variables, $Y_1 \sim \mathcal N (0,\sigma^2)$. Define $(\mathcal F_n)_{n\in \mathbb N}$ as the natural filtration, $\mathcal F_n = \sigma(Y_1,...,Y_n)$.

We define now two processes: $X_n = Y_1 + ... + Y_n$ and $Z_n^u = \exp\left(\displaystyle uX_n - \frac{nu^2\sigma^2}2 \right)$.

Since $(Z_n^u)$ is a martingale bounded in $\mathcal L^1$ (i.e. $\sup_n E(|Z_n^u|) < \infty$), converges almost surely to a random variable $Z_\infty^u \in \mathcal L^1$ $-$for every $u\in \mathbb R$$-$ in virtue of the Doob's supermartingale convergence theorem.

My main question is: how one can find the limit? For which values of $u$ is true that $Z_n^u = \mathbb E (Z_\infty^u | \mathcal F_n)$?

A known theorem says that $Z_n^u = \mathbb E (Z_\infty^u | \mathcal F_n)$ is true when $Z_n^u \to Z_\infty^u$ in $L^1$, or the sequence $(Z_n)$ is uniformly integrable, but I don't know how to show neither of these conditions.


2 Answers 2


Note that $T_n=uX_n - \frac12nu^2\sigma^2$ defines a random walk $(T_n)$ whose steps have mean $E(uY_1- \frac12u^2\sigma^2)=- \frac12u^2\sigma^2\lt0$ if $u\ne0$ hence, for every $u\ne0$, $T_n\to-\infty$ almost surely, $Z^u_\infty=0$ almost surely and $Z^u_n\ne E(Z^u_\infty\mid\mathcal F_n)$. The case $u=0$ is direct.


You should say the hipothesis about the sequence $(Y_n)$. If $Y ~ ber(1/2)$ the result is false.

  • $\begingroup$ They are normal random variables. Thank you for the observation! $\endgroup$
    – jpvigneaux
    Mar 1, 2014 at 4:49
  • $\begingroup$ Please clarify your statement regarding "$Y ber(1/2)$" as this is not meaningful. $\endgroup$
    – MPW
    Mar 1, 2014 at 5:14
  • $\begingroup$ I forgot the symbol $ \sim $... But I had thought it could be clear in the context... And sorry, I'm writing from my phone and that shouldn't be an answer but a comment $\endgroup$ Mar 1, 2014 at 7:09

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