Convergece of Uniform process

Question

Let $S_0=1$ and $S_n$ be uniformly distributed between $(0,S_{n-1})$. How do we show that $m^nS_n \rightarrow0$ almost surely if $0 < m < e$ and $m^nS_n \rightarrow \infty$ almost surely if $m > e $

This is a bit confusing because I know X_n goes to zero a.s. For it to go to infinity does it mean it's growing faster than the uniform process? What's the best way to approach the problem? I was trying Borel Cantelli and the definition of almost sure convergence but both yielded no results. Any hints or helpful resources will be appreciated.

Answer 1

Let $U_n=\frac{S_n}{S_{n-1}}$. This is uniformly distributed on $(0,1]$ and each of the $U_n$s is independent of the others.

Let $V_n = \log_e\left(U_n\right)$. Then both the CDF and density of $V_n$ is $e^t$ for $t \le 0$ and $\mathbb E[V_n]=-1$, and each of the $V_n$s is independent of the others.

Now let $W_n = \log_e(m)+V_n$ so $\mathbb E[W_n]=\log_e(m)-1$, and each of the $W_n$s is independent of the others. By the strong law of large numbers $\frac1n \sum\limits_{i=1}^n W_i \to \log_e(m)-1$ almost surely. Then (in effect multiplying both sides by $n$)

• if $\log_e(m) \lt 1$, i.e. $m \lt e$, then $\sum\limits_{i=1}^n W_i \to -\infty$ almost surely
• if $\log_e(m) \gt 1$, i.e. $m \gt e$, then $\sum\limits_{i=1}^n W_i \to +\infty$ almost surely

But $\sum\limits_{i=1}^n W_i = \log_e(m^n S_n)$ so (in effect taking anti-logarithms)

• if $m \lt e$, then $m^n S_n \to 0$ almost surely
• if $m \gt e$, then $m^n S_n \to +\infty$ almost surely