With $Y_n$ i.i.d, $EY_n = 1$, $X_n = \prod_{k=1}^n Y_k$, use Strong Law of Large Numbers to show $\frac{\log(X_n)}{n} \to c < 0$

Question

Let $Y_n$ be a sequence of non-negative i.i.d random variables with $EY_n = 1$ and $P(Y_n = 1) < 1$. Consider the martingale process formed by $X_n = \prod_{k=1}^n Y_k$. Use the strong law of large numbers to show that:

\begin{align*} \frac{\log(X_n)}{n} \to c < 0 \; \; \text{a.s.} \\ \end{align*}

The strong law of numbers says that:

\begin{align*} \frac{\log(X_n)}{n} &= \frac{1}{n} \sum_{k=1}^n \log Y_k \to E(\log Y_k) \\ \end{align*}

How do I show that $E(\log Y_k) = c < 0$?

Answer 1

You can use the equality condition for Jensens' inequality: https://en.wikipedia.org/wiki/Jensen's_inequality#Measure-theoretic_and_probabilistic_form, which tells us that the inequality $\varphi(\mathbb E[X])\leq \mathbb E[\varphi(X)]$ is an equality if and only if $\varphi$ is a linear function on some convex set $A$ s.t. $P(X\in A) = 1$. In our case, we use the concavity of log (so flip the inequality), and use that log is not linear on any open interval to get the strict equality (this is why it's important that the problem include $P(Y_n=1)<1$, since the only convex subsets of the real line are the intervals, and the only intervals not containing some open interval are the singleton points).

Answer 2

$\log x$ is a concave function. Jensen's inequality shows that $E \log Y_k \leq \log EY_k=0$.