Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 travelled. 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?

share|improve this question
1  
Would love to see why the problem is called as such. –  Sniper Clown Jul 26 '12 at 6:14
1  
Is it William T. Riker? –  draks ... Jul 26 '12 at 6:24
    
Is the fact that it's 3-dimensional relevant? Do you know anything about what happens in other dimensions? –  Harry Altman Jul 26 '12 at 6:25
1  
I can't think of any specific Star Trek episode that featured random jumps of this sort. It might be just as appropriate to call it the "Lost in Space" problem. –  Robert Israel Jul 26 '12 at 6:44
2  
Will they ever come home? @RobertIsrael Maybe it's referring to Star Trek: Voyager... –  draks ... Jul 26 '12 at 7:07
show 6 more comments

4 Answers

up vote 4 down vote accepted

$R^2_{n+1}/R^2_n$ 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 $c > 0$. (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 $1$.

share|improve this answer
    
how do you derive that $R^2_{n+1}/R^2_n$ has a distribution like $|u+V|^2$? –  zemora Jul 26 '12 at 6:52
1  
Let me suggest that the central limit theorem is not the relevant tool here, rather the law of large numbers. –  Did Jul 26 '12 at 7:52
1  
@zemora: $X_{n+1} = X_n + R_n V$ where $V$ is uniform on the sphere and independent of $X_n$, so $R_{n+1} = |X_n + R_n V| = R_n |U + V|$ where $U = X_n/R_n$. Now $U = M u$ for some orthogonal matrix $M$, so $|U+V| = |u + M^{-1} V|$. But $M^{-1} V$ has the same distribution as $V$, namely uniform on the unit sphere and independent of $X_n$. –  Robert Israel Jul 26 '12 at 16:21
    
@RobertIsrael: would you please check my answer? It's a different way posted by mike –  zemora Jul 28 '12 at 8:54
add comment

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.

share|improve this answer
    
that issue you are worried about isn't a problem if $A_n$ is the event of being inside, instead of the first time you are inside. Then you have $\sum \frac 1 {R_n} < \infty$ for 2 different reasons , 1) because they get home and the process stops and 2) because they don't get home in which case $R_n \rightarrow \infty $, as the same sort of borel cantelli argument shows . –  mike Jul 30 '12 at 15:17
    
How can we show that $R_n \to \infty$ a.s.? The problem states that the ship never stops, so we must prove that $R_n \to \infty$ otherwise this proof falls apart. –  nullUser Apr 13 '13 at 17:12
add comment

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...

share|improve this answer
1  
In fact $E(I)=\frac{2\ln(2)-1}{4}>0$ –  Xoff Jul 26 '12 at 7:13
    
Let me suggest that the central limit theorem is not the relevant tool here, rather the law of large numbers: (1.) $L_n/n\to E(I)$ almost surely hence for every $x\lt E(I)$, $P(L_n\gt nx)\to1$, (2.) since $E(I)\gt0$, this holds for some $x\gt0$. Done. –  Did Jul 26 '12 at 7:51
add comment

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.

share|improve this answer
    
How did you get the probability $1/R^2_n$? –  zemora Jul 26 '12 at 17:39
    
you have a sphere of radius $R_n$ touching zero, and you're going to move somewhere on its surface, and the probability that you go home is the prob that you move to the part that intersect the ball of radius r around the origin. You could probably compute that exactly, but is it clear the the since the surface area is propotional to $R_n^2$ the prob of a fixed patch on it is order of $\frac 1 {R_n^2}$ ? –  mike Jul 26 '12 at 21:00
    
The probability is $\frac{R-\sqrt{R^2-r^2}}{2R}$. not order of $1/R^2$. –  zemora Jul 27 '12 at 4:41
    
your expression is of order $\frac 1 {R^2}$ viz. $R - \sqrt{R^2 - r^2} = R - R \sqrt{1-\frac {r^2}{R^2}} \approx R ( 1-(1-\frac 12 (\frac rR)^2$ –  mike Jul 27 '12 at 11:46
    
would you please check my answer? –  zemora Jul 28 '12 at 8:51
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.