Computing a limit almost surely using the strong law of large numbers

Let $X_0=(1,0)$ and define $X_n \in \mathbb R^2$ recursively by declaring that $X_{n+1}$ is chosen at random uniform distribution from the ball $B(0,|X_n|)$ and $\frac{X_{n+1}}{|X_n|}$ is independent of $X_1,...,X_n$. Prove that $\frac{\log |X_n|}{n} \to c$ a.s. and compute $c$.

I have no idea what I can do in this exercise, and how I can use the strong law of large numbers here :/

EDITED: The book says $X_n$ but it's $|X_n|$. Thanks!

• $\log X_n$ is undefined since $X_n \in \mathbb{R}^2$. Did you mean $\log |X_n|$ instead? – gt6989b Oct 21 '13 at 18:11

By definition, $|X_0|=1$ and $|X_n|\ne 0$ a.s. for every $n\ge 1$, so $$Y_n:=\log \frac{|X_n|}{|X_{n-1}|}=\log|X_n|-\log|X_{n-1}|$$ is a.s. well defined for every $n\ge 1$ and $(Y_n)_{n\ge 1}$ are mutually independent. Moreover, $Y_n$'s distribution $F_n(x):=\Bbb P(Y_n\le x)$ satisfies that if $x\ge 0$, $F_n(x)=1$; if $x<0$,
$$F_n(x)=\Bbb P(Y_n\le x)=\Bbb P (|X_n|\le e^x|X_{n-1}|)=\frac{\pi\cdot(e^x|X_{n-1}|)^2}{\pi |X_{n-1}|^2}=e^{2x}.$$ Therefore, $(Y_n)_{n\ge 1}$ are i.i.d., and $\Bbb E|Y_1|=-\Bbb E Y_1=\frac{1}{2}<\infty$, so by Strong Law of Large Numbers, $$\frac{1}{n}\log |X_n|=\frac{1}{n}\sum_{k=1}^n Y_k\to \Bbb EY_1=-\frac{1}{2}\quad a.s.\,.$$