The Star Trek Problem in Williams's Book

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?

Solution 1:

$R^2_{n+1}/R^2_n$ has the same distribution as $|u+V|^2$ where $u$ is a fixed unit vector in ${\mathbb R}^3$ and $V$ is a random unit vector (uniform on the unit sphere). I would guess (although I haven't computed it) that $E [\log (|u+V|^2)] > 0$, and the Central Limit Theorem will tell you that almost surely $R_n > c^n$ for some $c > 1$. (EDIT: for sufficiently large $n$; Strong Law of Large Numbers is actually more relevant).

EDIT: Yes, $E[\log(|u+V|^2)] = \int_0^{\pi} \sin(\theta) \log(2 - 2 \cos(\theta))\ d\theta/2 = 2 \log(2) - 1 > 0$.

In the two-dimensional version, however, $E[\log(|u+V|^2)] = \int_0^\pi \log(2 - 2 \cos(\theta))\ d\theta/\pi = 0$, so the Law of Large Numbers doesn't tell you whether $R_n$ will go to $0$ or $\infty$. However, the Central Limit Theorem will say $P(R_n < 1) \to 1/2$ as $n \to \infty$. Thus it's certainly not the case in the two-dimensional version that $R_n \to 0$ with probability $1$, nor does $R_n \to \infty$ with probability $1$.

Solution 2:

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.

Solution 3:

By a simple geometrical property : ${\mathbb P}(R_{n+1}^2>2R_n^2)=\frac{1}{2}$

In fact, we have ($0\le\theta\le\pi$): $${\mathbb P}(R_{n+1}<2\sin(\frac{\theta}{2})R_n)=\frac{1-\cos\theta}{2}$$

Let $L_n=\ln(R_n)$ and $I$ such that ${\mathbb P}(I<\ln(2\sin(\frac{\theta}{2})))=\frac{1-\cos\theta}{2}$

$E(I)>0$ and $V(I)$ is defined. As $L_n$ is the sum of $(n-1)$ independent $I$, you can use the central limit theorem, and $L_n$ will be approximated by a normal law with $E=n.E(I)$ and $V=n.V(I)$.

Hence ${\mathbb P}(L_n>2\ln(n)E(I))\rightarrow 1$, hence ${\mathbb P}(R_n>k.n^2)\rightarrow 1$, so the serie is convergent...

Solution 4:

in the book it seems to me that the sum terminates when they get home, so if they get home the sum is finite. williams also hints: it should be clear what thm to use here. i am sure he is referring to levy's extension of the borel-cantelli lemma, $\frac 1 {R_n^2}$ is the conditional probability of getting home at time $n+1$ given your position at time n is $R_n$, it is the proportion of the area of the sphere of radius etc.