martingale proof

**willowtree** · May 5th 2010, 05:38 AM

I need to do the proof for the second condition but I'm not getting the right answer.

$\{S_n , n\geq 1\}$ } is a martingale if

1) E[| $S_n$ |]< $\infty \forall n$
2) E[ $S_{n+1}|S_1,...,S_n$ ] = $S_n \forall n$

$S_n$ = $X_1X_2...X_n$

Suppose E[| $X_n$ |] < $\infty \forall n$ and E[ $X_n$ ]=1

Prove that $\{S_n , n\geq 1\}$ } is a martingale

I can prove the first condition but for the second one I keep getting E[ $S_{n+1}|S_1,...,S_n$ ] = $S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

(ps sorry for any bad use of LaTex, this is the first time I have used it!)

**matheagle** · May 5th 2010, 07:38 AM

Originally Posted by willowtree

I need to do the proof for the second condition but I'm not getting the right answer.

$\{S_n , n\geq 1\}$ } is a martingale if

1) E[| $S_n$ |]< $\infty \forall n$
2) E[ $S_{n+1}|S_1,...,S_n$ ] = $S_n \forall n$

$S_n$ = $X_1X_2...X_n$

Suppose E[| $X_n$ |] < $\infty \forall n$ and E[ $X_n$ ]=1

Prove that $\{S_n , n\geq 1\}$ } is a martingale

I can prove the first condition but for the second one I keep getting E[ $S_{n+1}|S_1,...,S_n$ ] = $S_n+1$ which is wrong. Can anyone shine a light on this problem? Thanks

(ps sorry for any bad use of LaTex, this is the first time I have used it!)

If this is a product and not a sum, then

E[ $S_{n+1}|S_1,...,S_n] = E( X_1X_2\cdots X_{n+1}|S_1,...,S_n)=X_1X_2\cdots X_nE(X_{n+1}|S_n)$

and I'm guessing you need $E(X_1)=1$

YUP, I see it there.
I found your $E(X_1)=1$

$=X_1X_2\cdots X_nE(X_{n+1}) =S_n (1)=S_n$

**willowtree** · May 5th 2010, 11:40 AM

ahh i seee, thank you, i was doing something a bit backwards with the stuff inside the brackets!

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