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I'm reading the course notes of a probability course about martingales currently and I'm trying to solve some of the exercises, however I'm very much stuck with the following exercise:

Let $\left\{ X_{n}\right\} _{n=1}^{\infty}$ be a simple random walk on $\mathbb{Z}$ started at $X_{0}=a$ . Given $b\in\mathbb{Z}$ denote by $T_{b}$ the first hitting time of $b$ , show that $\mathbb{E}_{a}\left[T_{b}|\, T_{b}<T_{0}\right]=\frac{b^{2}-a^{2}}{3}$ .

Hint: Show that $M_{n}:=X_{n}^{3}-3nX_{n}$ is a martingale and use the fact that $\left|M_{\min\left(T_{b},n\right)}\right|\leq b^{3}+3bT_{b}$

I've shown that indeed that is a martingale but I have no idea how to proceed, help would be appreciated.

Thanks!

LlamaMan
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  • Which uses have martingales, in general? – Did Jul 01 '14 at 19:12
  • The only intuition I got is that Martingales are supposed to represent a "fair game/gamble", beyond that I'm not very comfortable with the concept or its uses... It would be nice if people that down voted would at least explain the reasoning behind their down vote. Obviously I'm having difficulty or I wouldn't have asked the question. – LlamaMan Jul 01 '14 at 19:15
  • Honestly, what is a bit surprising is that you do not seem to be even aware of the mammoth result in this field, called the stopping time theorem. – Did Jul 01 '14 at 19:18
  • @Did I've looked up the theorem and I see it's indeed a very powerful result. But I'm not sure the condition here meets the criteria of the theorem, the third condition requires that $\left|M_{\min\left(T_{b},n\right)}\right|\leq c$ a.s for every $n$, but the bound I have depends on $T_{b}$.

    I apologize if the question seems banal to some but I've been glaring at this for some time now and I'm not making much progress myself, I would really appreciate it if you could help me.

     – LlamaMan Jul 01 '14 at 19:48
  • can you show that $T_b$ is integrable?, i.e. $\mathbb{E}aT_b<\infty$. If so, then option sampling theorem applies, because your martingale is bounded by a random variable which is integrable. $|M{\min(T_b,n)}|\leq c$ is sufficient but not necessarily (effectively, you need to apply the dominated convergent theorem, bounded by an $L^1$ R.V. is sufficient) – Lost1 Jul 01 '14 at 20:40
  • @lost1 I can show that $T_{b}$ is integrable but I'm not sure how to finish the proof. What I don't see is how to use the optional stopping theorem to reach the required result. – LlamaMan Jul 01 '14 at 20:57
  • okay, i will write something – Lost1 Jul 01 '14 at 20:58
  • Are you assuming that $a$ is between $0$ and $b$? If so, you should add that assumption to the statement of your problem. –  Jul 01 '14 at 21:33

1 Answers1

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Optional samping theorem holds for every bounded stopping time.

Consider the stopping time $\min(T_0,T_b,n)$ where $T_0$ is the first time the Markov Chain hits $0$, then

$|M_{\min(T_0,T_b,n)}|\leq b^3+3b\min(T_0,T_b)$, the right handside is integrable.

Optional Sampling theorem says

$\mathbb{E}_a M_{\min(T_0,T_b,n)} = \mathbb{E}_a M_0$

since the right hand side is bounded by an integrable random varaible, we take $n$ to infinity and get

$\mathbb{E}_a M_{\min(T_0,T_b) }=a^3$

so the left handside equals to

$\mathbb{P}_a(T_b<T_0)(b^3 - 3b \mathbb{E}_aT_b1_{\{T_b<T_0\}}) + \mathbb{P}_a(T_0<T_b)(0) = a^3$

Note that $X$ is also a martingale, so $\mathbb{P}_a(T_b<T_0)=a/b$ (apply OTS to $X$). This means

$$\frac{b^2-a^2}{3} = \mathbb{E}_aT_b1_{\{T_b<T_0\}}$$

Lost1
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  • this only gives $ET_b$, need a bit more than this. I am editting. – Lost1 Jul 01 '14 at 21:08
  • Doesn't $\min\left(T_{b},n\right)$ need to be bounded by a constant that doesn't depend on $n$ in order to use the OST in that fashion? – LlamaMan Jul 01 '14 at 21:09
  • @NewGuy $min(T_b,n)$ is bounded by $n$, for each $n$, then you take the limit because the RANDOM VARIABLES are bounded. Optional Sampling theorem only tells you $EX_T=EX_0$ when $T$ is a bounded stopping time for 1 bounded stopping time... very rarely are stopping times you are interested in are actually bounded... – Lost1 Jul 01 '14 at 21:10
  • @NewGuy something is wrong here, because $ET_b$ should be infinite for a random walk... also please check your question again and confirm that it is correct – Lost1 Jul 01 '14 at 21:13
  • @NewGuy so $T_b$ is not actually integrable... – Lost1 Jul 01 '14 at 21:15
  • Sorry about that, I seem to have gotten confused with another result about hitting times for a simple random walk.. I suppose you can just delete the answer now. If you have an alternative solution to the question it would be appreciated... – LlamaMan Jul 01 '14 at 21:20
  • @NewGuy see edit. – Lost1 Jul 01 '14 at 21:27
  • That is a nice trick with defining that stopping time, I've definitely learnt something. Thanks! – LlamaMan Jul 01 '14 at 21:42
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    @NewGuy btw, Could have use the optional stopping theorem version (2) on wiki http://en.wikipedia.org/wiki/Optional_stopping_theorem the increment is bounded by 1. Also in this question $T_b$ and $T_0$ both appear so it was the right to consider their min. Morever $E\min(T_0,T_b)$ is finite, whereas $ET_b$ is not. – Lost1 Jul 01 '14 at 21:45
  • Thank you very much Lost1 for the help and the patience. Again apologies for my silly mistake. – LlamaMan Jul 01 '14 at 21:49