# Thread: Martingale &amp;amp; Stopping time

1. ## Martingale &amp;amp; Stopping time

Hi !

Okay, this one has been bugging (and p***ing) me...

We have $\displaystyle (Y_i)_i$ iid rv's where $\displaystyle P(Y_1=1)=p$ and $\displaystyle P(Y_1=-1)=1-p$

Let $\displaystyle a\in\mathbb{N}^*$ and $\displaystyle S_n=a+\sum_{i=1}^n Y_i$

So for $\displaystyle p=1/2$, $\displaystyle S_n$ is a martingale (with the natural filtration)

Let $\displaystyle S_{n\wedge T}=S^T$ be the stopped martingale, where $\displaystyle T=\inf\{n>0 ~:~ S_n=0 \text{ or } S_n=a+b\}$ (b is another positive integer)

Then I have to prove, using the convergence of $\displaystyle S^T$, that $\displaystyle P(T<\infty)=1$.

And I don't know how to do it (I've been scrabbling on it, without finding a formal proof...).

Thanks for any hint...

2. I think you'll be very disappointed when you get it...

Hint: almost-surely, $\displaystyle S^T$ is an integer-valued sequence that has a finite limit.

3. I'm disappointed (of myself) because I still don't understand

1. How do we show that it converges ? The only theorem that we have is this one (I know there are others, though, but I learnt them in another course) :
Xn is a supermartingale such that sup E|Xn| is finite. Then there exists a rv X such that E|X| is finite and Xn converges to X a.s.
But here, the condition of $\displaystyle S^T$ to be bounded is unknown because $\displaystyle E|S^T|<T$ or that $\displaystyle E|S^T|<n$, ce qui ne fait pas avancer le schmilblick

2. Then I understand clearly why, if S^T converges, then it's constant from a certain rank. But the only way I can see to show this is : if it converges to $\displaystyle m\in(0,a+b)$, it's not possible because the following step is different from m, a.s.. So it must converge to 0 or a+b. Is that correct to write it this way ?

Thanks, as always

4. Originally Posted by Moo
1. How do we show that it converges ? The only theorem that we have is this one (I know there are others, though, but I learnt them in another course) :
Xn is a supermartingale such that sup E|Xn| is finite. Then there exists a rv X such that E|X| is finite and Xn converges to X a.s.
But here, the condition of $\displaystyle S^T$ to be bounded is unknown because $\displaystyle E|S^T|<T$ or that $\displaystyle E|S^T|<n$, ce qui ne fait pas avancer le schmilblick
Sorry I didn't understand your question also concerned this part. Notice that the martingale $\displaystyle S^T$ is, by construction, bounded: it is stopped when reaching either 0 or a+b, hence it can't leave [0,a+b]...

2. Then I understand clearly why, if S^T converges, then it's constant from a certain rank. But the only way I can see to show this is : if it converges to $\displaystyle m\in(0,a+b)$, it's not possible because the following step is different from m, a.s.. So it must converge to 0 or a+b. Is that correct to write it this way ?
This way, you don't really prove that the limit is an integer (I know, this is so obvious...). I would first recall the general result that a convergent integer-valued sequence is stationary (and give a short proof: the sequence is Cauchy, hence we have $\displaystyle |u_n-u_m|<1$ for $\displaystyle m,n\geq n_0$, hence $\displaystyle u_n=u_m$). And then conclude: almost-surely, $\displaystyle S^T$ is stationary; however, $\displaystyle S_{T\wedge n}=S_{T\wedge(n+1)}\pm 1$ if $\displaystyle n< T$; therefore almost-surely there exists $\displaystyle n\geq T$, which is to say that $\displaystyle T$ is finite.

5. Originally Posted by Laurent
Sorry I didn't understand your question also concerned this part. Notice that the martingale $\displaystyle S^T$ is, by construction, bounded: it is stopped when reaching either 0 or a+b, hence it can't leave [0,a+b]...
Nah, I'm sorry, I wasn't clear about that.
And yeah, I actually already thought of that, I just forgot to write it and hence forgot it ...

This way, you don't really prove that the limit is an integer (I know, this is so obvious...). I would first recall the general result that a convergent integer-valued sequence is stationary (and give a short proof: the sequence is Cauchy, hence we have $\displaystyle |u_n-u_m|<1$ for $\displaystyle m,n\geq n_0$, hence $\displaystyle u_n=u_m$). And then conclude: almost-surely, $\displaystyle S^T$ is stationary; however, $\displaystyle S_{T\wedge n}=S_{T\wedge(n+1)}\pm 1$ if $\displaystyle n< T$; therefore almost-surely there exists $\displaystyle n\geq T$, which is to say that $\displaystyle T$ is finite.
Nice one

Now it's all packed, solved, clear. Thanks !

Moral : don't try to do maths after a too short night

6. Oh, another question (just checking if it's correct) :

I've showed that $\displaystyle S_n'=(S_n-a)^2-n$ is a martingale.
And from this, I have to compute E[T].

Is it correct... ?

Since it's a martingale, $\displaystyle \left((S_{n\wedge T}-a)^2-n\wedge T\right)_n$ is too.
So in particular, $\displaystyle E[(S_{n\wedge T}-a)^2]-E[n\wedge T]=E[S_0-a]=0$

So $\displaystyle \lim_{n\to\infty}E[(S_{n\wedge T}-a)^2]=\lim_{n\to\infty} E[n\wedge T]$

By applying the dominated convergence theorem to the LHS and because T is finite a.s., we have $\displaystyle E[(S_T-a)^2]=\lim_{n\to\infty} E[n\wedge T]$

But for the RHS, I don't really know... We can't apply the dominated theorem because we only know that T is finite, not bounded ? I prefer asking, because I know I have many troubles with the bounded stuff...
But we can apply the monotone convergence theorem, can't we ?

That's just what I want to know

7. Originally Posted by Moo
We can't apply the dominated theorem because we only know that T is finite, not bounded ?
But we can apply the monotone convergence theorem, can't we ?
Twice correct!