Good day,

I have the problem of an old exam without solutions. It is the following:

Let $X_1,X_2,...$ be i.i.d. (independent and identically distributed) random variables on the same probability space such stat $$P\left(X_1=-\frac{1}{4}\right)=P\left(X_1=\frac{1}{4}\right)=\frac{1}{2}$$ For all $n\in \mathbb{N}$ define $M_n:=\prod_{k=1}^n (1+X_k)$.

  • Show that $(M_n)_{n\in\mathbb{N}} $ is a martingale w.r.t. a suitably chosen filtation (This was no problem for me)
  • Use the law of large numbers to show that $(M_n)_{n\in\mathbb{N}}$ converges almost surely and determine the limit. (Here is the problem for me.)

This is the strong law of large numbers (SLLN) we had in the lecture:

Let $X_i$ be uncorrelated random variables, $E(X_i)=\mu, E(X_i^2)<\infty$, $\exists C$ such that $\text{Var}(X_i)\leq C$ for all $i$ and define $S_n=\sum_{i=1}^n X_i$. Then $$\frac{S_n}{n} \to \mu \text{ almost surely.}$$

Therefore I have to get this $M_n$ into the form of $\frac{S_n}{n}$. I had a problems in the class where I had $$ A_n:=\left( \prod_{i=1}^n X_i \right)^{1/n}.$$ This was pretty elegant since we get the wished form via $$\log A_n = \frac{1}{n} \sum_{i=1}^n \log(X_i)=:\frac{1}{n} \sum_{i=1}^n Y_i.$$

I wanted to use this here:

$$\log M_n = \sum_{k=1}^n \log(1+X_k) = \frac{1}{n} \sum_{k} \log(1+X_k)^n $$

But now this "new" random variable $Y_{n,k}:=\log(1+X_k)^n$ is dependent of $n$. I don't know if this is a problem (now I think this is a huge problem, see below.) Let's try and use the SLLN for this: We would get $\log M_n \to E(Y_{n,k})$ if the assumptions are fulfilled but the limit should not be dependent on $n$. $$E(Y_{n,k})=E(\log(1+X_k)^n)=\frac{1}{2} \left( \log(5/4)^n +\log(3/4)^n \right) = \frac{1}{2} \log (15/16)^n \to -\infty$$

I don't know what to do. I think I could use the Martingale Convergence Theorem to get that $M_n$ converges a.s. but via this method I don't get the limit. Further the exam states that I have to use the SLLN explicitly.

I am thankful for every help/hint.


1 Answer 1


Consider $\log M_n = \sum_{i=1}^n \log(1+X_i)$. Let $\mathbb E[\log(1+X_i)] = \mu$ and notice that $\mu < 0$. As $n \to \infty$, $\dfrac{1}{n} \log M_n$ converges almost surely to $\mu$, which implies $\log M_n$ converges almost surely to $-\infty$, and thus $M_n$ converges almost surely to $0$.

  • $\begingroup$ Thanks a lot for your answer. I should have stopped at the intermediate step and not forced this $\frac{1}{n}$ into it. One Question: How do you get this step? $$\frac{1}{n} \log M_n \to \mu<0 \Rightarrow \log M_n \to -\infty$$ $\endgroup$
    – Cahn
    Commented Sep 23, 2016 at 17:19
  • 1
    $\begingroup$ If $\dfrac{1}{n} \log M_n \to \mu < 0$, for all sufficiently large $n$ we have $\dfrac{1}{n} \log M_n < \mu/2$, and then $\log M_n < n \mu/2$, but $n \mu/2 \to -\infty$. $\endgroup$ Commented Sep 23, 2016 at 18:06

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