TITLE: The Star Trek Problem in Williams's Book
QUESTION [11 upvotes]: This problem is from the book Probability with martingales by Williams. It's numbered as Exercise 12.3 on page 236. It can be stated as follows:

The control system on the starship has gone wonky. All that one can do is to set a distance to be traveled. The spaceship will then move that distance in a randomly chosen direction and then stop. The object is to get into the Solar System, a ball of radius $r$. Initially, the starship is at a distance $R>r$ from the sun.
If the next hop-length is automatically set to be the current distance to the Sun.("next" and "current" being updated in the obvious way). Let $R_n$ be the distance from Sun to the starship after $n$ space-hops. Prove $$\sum_{n=1}^\infty \frac{1}{R^2_n}<\infty$$
holds almost everywhere.

It has puzzled me a long time. I tried to prove the series has a finite expectation, but in fact it's expectation is infinite. Does anyone has a solution?

REPLY [3 votes]: I think this way works basically, but there are some details to be completed.
Let $A_k$ be the event that the starship goes back to the Solar System just after the n th space jump. Then $\{A_k\}_{k=1}^\infty$ is a sequence of disjoint events. Obviously $A_k$ is adapt to $\mathcal{F}_n$, here $\mathcal{F}_n$ is the $\sigma$- field generated by the first  $n$ jumps.
Let $p_{i+1}=P(A_{i+1}|\mathcal{F}_{i})$, then$$p_{i+1}= \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$$ 
this follows from the area formula of a sphere cap.
(Here I am not sure because $ \frac{R_n-\sqrt{R^2_n-r^2}}{2R_n}$ is the conditional probability that given the ship is outside the solar system after n jumps, it will go back home after the next (n+1 th) jump, is it the same with $P(A_{i+1}|\mathcal{F}_{i})$?)
suppose it is, then we use the inequality
$$1-\sqrt{1-x}\geq \frac{x}{2}\quad (0<x<1)$$
to derive
$$p_{n+1}\geq \frac{r^2}{4R^2_n}$$
 so if $\sum_{n}p_n$ converges, so dose $\sum_{n}\frac{1}{R^2_n}$
Levy's extension of the Borel-Cantelli lemma says (see page 124 in Williams's book)
on the set $S=\{\omega:\sum_{n}p_n=\infty\}$, almost surely holds $$\frac{\sum_{k=1}^n \mathbf{1}_{A_k}}{\sum_{k=1}^n p_k}\rightarrow 1$$
but the numerator can only have values 0 and 1,(note the $A_k$s are disjoint) so the set $S$ must have measure 0.