# martingale proof

• May 5th 2010, 06:38 AM
willowtree
martingale proof
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!)
• May 5th 2010, 08:38 AM
matheagle
Quote:

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 $E(X_1)=1$

$=X_1X_2\cdots X_nE(X_{n+1}) =S_n (1)=S_n$
• May 5th 2010, 12:40 PM
willowtree
ahh i seee, thank you, i was doing something a bit backwards with the stuff inside the brackets!