I think you'll be very disappointed when you get it...
Hint: almost-surely, is an integer-valued sequence that has a finite limit.
Hi !
Okay, this one has been bugging (and p***ing) me...
We have iid rv's where and
Let and
So for , is a martingale (with the natural filtration)
Let be the stopped martingale, where (b is another positive integer)
Then I have to prove, using the convergence of , that .
And I don't know how to do it (I've been scrabbling on it, without finding a formal proof...).
Thanks for any hint...
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 to be bounded is unknown because or that , 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 , 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
Sorry I didn't understand your question also concerned this part. Notice that the martingale is, by construction, bounded: it is stopped when reaching either 0 or a+b, hence it can't leave [0,a+b]...
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 for , hence ). And then conclude: almost-surely, is stationary; however, if ; therefore almost-surely there exists , which is to say that is finite.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 , 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 ?
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 ...
Nice oneThis 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 for , hence ). And then conclude: almost-surely, is stationary; however, if ; therefore almost-surely there exists , which is to say that is finite.
Now it's all packed, solved, clear. Thanks !
Moral : don't try to do maths after a too short night
Oh, another question (just checking if it's correct) :
I've showed that is a martingale.
And from this, I have to compute E[T].
Is it correct... ?
Since it's a martingale, is too.
So in particular,
So
By applying the dominated convergence theorem to the LHS and because T is finite a.s., we have
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