Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

The question is:

Let $Y_1 , Y_2, \dots$ be nonnegative i.i.d. random variables with $\mathbb{E}Y_m = 1$ and $\mathbb{P} (Y_m = 1) < 1$. (i) Show that $X_n = \prod_{m \le n} Y_m$ defines a martingale. (ii) Use an argument by contradiction to show $X_n \to 0$ a.s.

(i) is easy to check. For (ii), by Martingale Convergence Theorem, we can show that $X_n$ converges to almost surely to some $X$ with $\mathbb{E}X \le \mathbb{E}X_0 = 1.$ ($X_0$ is not explicitly defined in the question, but to make $X_n$ a martingale, we need $X_0 = 1$.)

My guess is that $X = 0$ almost surely must comes from the fact $\mathbb{P} (Y_m = 1) < 1$. But I can't see how to continue from here.

share|cite|improve this question
Note that if $X_n$ converges to a positive value then $X_n/X_{n-1}=Y_n$ converges to $1$. – Colin McQuillan Feb 20 '13 at 17:42
@ColinMcQuillan You should post this as an answer. – Byron Schmuland Feb 20 '13 at 18:19
See also Theorem 13.2.3. here: – Byron Schmuland Feb 20 '13 at 18:41
@ColinMcQuillan But couldn't there exists a set $A = \{\omega:Y_n(\omega) \to 1\}$ such that $\mathbb P\{A\} > 0$? – ablmf Feb 20 '13 at 19:20
@ablmf: no: $\mathbb{P}[Y_n\to 1] = 0$. Try showing that there exists $\epsilon>0$ such that for all $N$ we have $\mathbb{P}[|Y_n-1| < \epsilon\text{ for all $n>N$}] = 0$. – Colin McQuillan Feb 20 '13 at 20:31
up vote 2 down vote accepted

Why an argument by contradiction? Note that $\log X_n$ is the sum of $n$ i.i.d. random variables with mean $m=\mathbb E(\log Y_1)$, hence, if $m\lt0$, by the strong law of large numbers, $\log X_n\to-\infty$.

But $m\leqslant\log\mathbb E(Y_1)=0$ by Jensen inequality and one knows that this convexity inequality is strict as soon as the random variable $Y_1$ is not almost surely constant. This is what the hypothesis $\mathbb P(Y_1=1)\lt1$ ensures. Hence, $m\lt0$, $\log X_n\to-\infty$ almost surely, and $X_n\to0$ almost surely.

share|cite|improve this answer
Do we need to worry about the case when $E(|\log(Y)|)=\infty$? – Byron Schmuland Feb 20 '13 at 18:23
@ByronSchmuland Good question. The answer is no because $\log^+Y_1\leqslant Y_1$ hence one would have $E(\log^+Y_1)$ finite and $E(\log^-Y_1)$ infinite, in which case the SLLN applies. – Did Feb 20 '13 at 18:26
Right you are!${ }$ – Byron Schmuland Feb 20 '13 at 18:29

The Hewitt-Savage zero one law says that $X$ is almost surely a constant. Also, $X=Y_1\cdot\prod_{i=2}^\infty Y_i$ has the same distribution as $Y_1\cdot X$. Since $Y_1$ is not constant almost surely, this forces $X=0$.

share|cite|improve this answer
Or $X=+\infty$? – Did Feb 20 '13 at 18:29
@Did Touche. But we already know that $\mathbb{E}(X)\leq 1$. – Byron Schmuland Feb 20 '13 at 18:32
Argh... of course (stoopid I am). – Did Feb 20 '13 at 18:42

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.