# Thread: de Moivre's martingale

1. ## de Moivre's martingale

Hi !

Looks like it's the same as last year : I can't wait for the next class to solve a problem ^^'
We've studied martingales for only 3 hours already, so I'm still very new to it.

So we have a sequence $\displaystyle (Y_n)_{n\geq 1}$ of iid rv's such that $\displaystyle P(Y_n=-1)=q$ and $\displaystyle P(Y_n=1)=p$
where $\displaystyle p+q=1 \ , \ 0<p<q<1$
Let $\displaystyle X_0=0 \ , \ Z_0=1$... $\displaystyle X_n=Y_1+\dots+Y_n$ and $\displaystyle Z_n=\left(\tfrac qp\right)^{X_n}$

Let's consider the filtration $\displaystyle \mathcal{F}_n=\sigma(Z_0,\dots,Z_n)$

I have to show that $\displaystyle Z_n$ is a positive martingale.

3 points to show :
1/ $\displaystyle Z_n$ is $\displaystyle \mathcal{F}_n$-measurable
2/ $\displaystyle Z_n$ is integrable
3/ $\displaystyle E(Z_{n+1}\mid \mathcal{F}_n)=Z_n$

Point 1/ is okay.
Point 2/ is okay.
Point 3/ is totally unknown Which method to use ?

Thanks for any help/hints

2. Fine I'll do it. Conditioning on the Z's you can pull the first n Y's out of the expectation...

$\displaystyle \biggl({q\over p}\biggr)^{Y_1+\cdots +Y_n} E\biggl({q\over p}\biggr)^{Y_{n+1}}$

Now use your distribution of the Y's

$\displaystyle E\biggl({q\over p}\biggr)^{Y_{n+1}} =\biggl({q\over p}\biggr)^{-1}q +\biggl({q\over p}\biggr)^1p=p+q=1$

So your conditional expectation is

$\displaystyle \biggl({q\over p}\biggr)^{Y_1+\cdots +Y_n}=\biggl({q\over p}\biggr)^{X_n}=Z_n$

3. Darn ! That was really...

Thanks !

4. easy?
Everything is easy when you know what you're doing.
I used to say certain problems were easy (and also some were hard)
in class and that would annoy the students who didn't realize the difference.

But I don't know why they have the condition p<q here.
Maybe that's the next step.
That we expect more failures than successes.
This then goes to infinity.

and I saw the unedited post, lol.
I've found out that the cow sees everything.

I have exams to grade next week, if anyone is interested.

5. Originally Posted by matheagle
easy?
Everything is easy when you know what you're doing.
I used to say certain problems were easy (and also some were hard)
in class and that would annoy the students who didn't realize the difference.
Hmmm no, I would've said obvious or direct. Once you've done enough of conditional expectation of course.

But I don't know why they have the condition p<q here.
Maybe that's the next step.
Yep, we had to prove then that $\displaystyle X_n$ goes to -infinity. And if you're interested, the following questions are :

2. Let $\displaystyle T_k=\inf\{n\geq 0 : X_n\geq k\}$
Consider the martingale $\displaystyle Z_{T_k\wedge n}$ (it's true because Tk is a bounded stopping time and Zn is a martingale)
Then show that $\displaystyle P(T_k<\infty)=\left(\tfrac pq\right)^k$
(I think this is where the condition over p and q is the most important actually)

3. Finally show that $\displaystyle E(\sup_{n\geq 0} X_n)=\frac{p}{q-p}$

and I saw the unedited post, lol.
I've found out that the cow sees everything.
Huh ? which undedited post ?

6. finals are over!!!

### de moivre m

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