Probability convergence of a martingale defined as iid random variable product

Question

Let $(\beta_n)_{n \geq 1}$ be positive independent and identically distributed random variables with $\mathbb{E}[\beta_1] = 1$ and $\mathbb{P}[\beta_1 < 1 ] > 0$. Define the martingale $M_n = \beta_1 \cdot \beta_2 \cdots \beta_n$.

I have to prove that $M_n \to 0$ almost surely (a.s.).

Below what I've tried so far.

1. First attempt. Given that $(M_n)$ is bounded in $L^1$ the pointwise limit exist a.s.. In order to compute the limit let's study the convergence in probability: by the Paul Levy theorem it suffices to study the characteristic function pointwise convergence: $$ \varphi_{M_n}(t) = \mathbb{E}\left[ \exp(itM_n) \right] = \mathbb{E}\left[ \exp\left( it \prod_{i=1}^n \beta_i \right) \right] $$ but I don't know how to go ahead.

2. Second attempt. Let's try to use the Borel Cantelli lemma. It suffices to show that $$ \forall \epsilon > 0 \quad \mathbb{P}[M_n < \epsilon \text{ eventually}] = 1. $$ So $$ \mathbb{P}[M_n < \epsilon \text{ eventually}] = 1 - \mathbb{P}[M_n > \epsilon \text{ infinitely often}]. $$ Let's study $$ \sum_{n \geq 0} \mathbb{P}[M_n > \epsilon] $$ but I can't show that the serie converges.

Any hints would be appreciated. In particular I cannot figure out how to use the hypothesis $\mathbb{P}[\beta_1 < 1 ] > 0$.

Answer 1

Let $\mathcal{F}_n = \sigma (\beta_1,\dots, \beta_n)$. Then $M_n$ is a martingale with respect to $\mathcal{F}_n$ since $E[M_{n+1}|\mathcal{F}_n] = E[\beta_{n+1} M_n|\mathcal{F}_n] = M_nE [\beta_{n+1}|\mathcal{F}_n] = M_nE\beta_{n+1} = M_n$. The second equality is by $M_N\in \mathcal{F}_n$ and the third equality by independence of $\beta_{n+1}$ and $\mathcal{F}_n$.

Since $\beta_n$ are positive, $M_n$ is positive martingale. Therefore, the martingale convergence theorem implies $M_n$ converges to a limit $M$ almost surely with $EM < \infty$.

Since $M_n$ is a martingale, $E M_n = EM_1 = 1$. So, Fatou's lemma implies $EM \leq 1$.

Suppose $P(M>0)>0$. Let $\epsilon$ be any positive number. For each $\omega \in {M>0}$, there exists $N$ such that $m, n \geq N$ implies $|M_n - M_m|< \epsilon$ and $M_n \geq M/2$. Letting $m=n+1$, we have $|\beta_{n+1}-1|<\frac{\epsilon}{M_n}\leq \frac{2\epsilon}{M}$. This implies $\beta_n \to 1$ on $M>0$.

On the other hand, we have $P(\beta_n\neq 1)>0$. So there exists $\delta>0$ such that $P(|\beta_n-1|\geq \delta)>0$. Then $\sum_{n=1}^\infty P(|\beta_n-1|\geq \delta)=\infty$. Since $\beta_n$ are independent, the second Borel_Cantelli lemma implies $P(|\beta_n-1|\geq \delta \quad i.o.) = 1$.

However, this is a contradiction to $\beta_n \to 1$ on $M>0$.

Answer 2

Here's a reasonably straightforward approach without using too many theorems.

Note that $\log$ is strictly concave and $\beta_i$ has a non-degenerate distribution. Therefore, by Jensen's Theorem: $$\mathbb{E}[\log(\beta_i)]<\log(\mathbb{E}[\beta_i])=0$$

Let $\mu = \mathbb{E}[\log(\beta_i)]$, so the above states that $\mu<0$. Let $S_n=(\sum_{i=1}^n \log(\beta_i))/n$.

Then, the strong Law of Large numbers gives that $S_n \rightarrow \mu$ a.s.

Therefore, $n S_n \rightarrow -\infty$ a.s.

Therefore, $\prod_{i=1}^n \beta_i = e^{n S_n} \rightarrow 0$ a.s. as desired.