I am currently working my way through Durett's Probability Theory, 4th edition and am a bit confused about what something means. It is problem 5.2.9.
Let $Y_{1},Y_{2},...$ be nonnegative i.i.d random variables with $\mathbb{E}Y_{m}=1$ and $\mathbb{P}\left(Y_{m}=1\right)<1$. Define $X_{n}=\prod_{k=1}^{n}Y_{n}$.
The first two parts of the problem I can do. First is to show that $X_{n}$ is a martingale. Second is to show that $X_{n}\to0$ a.s.
The part I am confused about is: Use the strong law of large numbers to conclude that $\frac{1}{n}\log X_{n}\to c<0$.
My solution uses Jensen's inequality with $\varphi(x)=\log(x)$ and then finds that $\mathbb{E}\log Y_{i}\in[-\infty,0]$. We then can use this in the strong law of large numbers since $\frac{1}{n}\log X_{n}=\frac{1}{n}\sum_{i=1}^{n}\log Y_{i}$ and this converges to $\mathbb{E}Y_{i}=\mathbb{E}Y_{1}$.
What I don't understand is what this means? What information about $X_{n}$ is this result giving us? I tried looking at this from the context of a non-negative distribution, such as considering $Y_{i}\sim Poisson(\lambda=1)$, but that didn't help. My other thought is that this is saying something about $\mathbb{E}\left[\sup_{n\in\mathbb{N}}X_{n}\right]$.
If someone could explain the meaning of this result, I would be very appreciative!
This says something about the rate at which $X_n \to 0$. Note that $\frac {\log X_n} n$ is an indeterminate form since $\log X_n \to -\infty$. If $\log x_n \to 0$ very fast then $\frac {\log x_n} n \to -\infty$ and if $\log x_n \to 0$ very slowly then $\frac {\log x_n} n \to 0$. The fact that the limit is $c<0$ says that $X_n$ behaves like $e^{nc}$ as $n \to \infty$.