# Martingales - Explaination.

• Apr 24th 2010, 12:40 AM
Redeemer_Pie
Martingales - Explaination.
Hi all!

I've got a problem with a question at the moment and i was wondering if you could clarify it? Thanks =)

Let $\displaystyle X_{n}$ = $\displaystyle X_{0}$ + $\displaystyle \sum^n_{i=1}$$\displaystyle Y_{i}$ be a random walk.

Suppose that $\displaystyle t^*$ > 0 such that $\displaystyle m(t^*) = 1$ where $\displaystyle m(t)=e^{tY_{i}}$ is the moment generating function of $\displaystyle Y_{i}$.

Show $\displaystyle e^{t^*X_{n}}$

The property of martingale:

$\displaystyle E|X_{n}| < \infty$

$\displaystyle E(X_{n+1}|X_{1},X_{1}...X_{n})$

so show that : $\displaystyle E(X_{n+1}|X_{1},X_{2}...X_{n})$ = $\displaystyle e^{t^*X_{n}}$

Now $\displaystyle E(e^{t^*(X_{0}+Y_{1}+...+Y_{n}} |X_{1}...X_{n})$

From that it becomes

$\displaystyle E(e^{t^*X_{0}+t^*Y_{1}...+t^*Y_{i}}|X_{1}...X_{n})$

Now the one thing i dont get is that the equation then becomes:

$\displaystyle E(e^{t^*X_{n}}e^{Y_{i}}|X_{1}...X_{n})$

=$\displaystyle e^{t^*X_{n}}E(e^{Y_{i}}|X_{1}...X_{n})$

why from suddenly $\displaystyle e^{t^*X_{0}}$ becomes $\displaystyle e^{t^*X_{n}}$?

Any explainations would be great. Thank you.

=)
• Apr 24th 2010, 01:25 AM
Moo
Hello,

Because $\displaystyle X_n=X_0+Y_1+\dots+Y_n$ ? :D

Quote:

Now $\displaystyle E(e^{t^*(X_{0}+Y_{1}+...+Y_{n{\color{red}+1}}} |X_{1}...X_{n})$

From that it becomes

$\displaystyle E(e^{t^*X_{0}+t^*Y_{1}...+t^*Y_{\color{red}n+1}}|X _{1}...X_{n})$

Now the one thing i dont get is that the equation then becomes:

$\displaystyle E(e^{t^*X_{n}}e^{{\color{red}t^*}Y_{\color{red}n+1 }}|X_{1}...X_{n})$

=$\displaystyle e^{t^*X_{n}}E(e^{{\color{red}t^*}Y_{\color{red}n+1 }}|X_{1}...X_{n})$
I think there are typos (in red)
• Apr 24th 2010, 01:34 AM
Redeemer_Pie
Oh k. Thanks. lol that was a bit embarassing but i forgot to add:

I thought via martingale property that it shouldnt we be solving for:

$\displaystyle E(e^{t^*X_{n+1}}|X_{1}...X_{n})$

rather than:

$\displaystyle E(e^{t^*X_{n}}|X_{1}...X_{n})$?
• Apr 24th 2010, 01:36 AM
Redeemer_Pie
Quote:

Originally Posted by Moo
Hello,

Because $\displaystyle X_n=X_0+Y_1+\dots+Y_n$ ? :D

I think there are typos (in red)

yep! sorry they meant to say $\displaystyle Y_{i}$

Sorry my first time using LaTeX
• Apr 24th 2010, 01:57 AM
Moo
I don't understand... Do you have any further question ? oO
• Apr 24th 2010, 06:58 PM
Redeemer_Pie
nope that'll be all thanks :)