Random Processes (Martingale)

**Redeemer_Pie** · April 25th 2010, 06:59 AM

Hello all, I have a bit of a problem at the moment solving this task. I done half of it already, just can solve it for the other half. Hope u can help! Thanks =)

Let $S_n = e^{\sum Y_i}$ where $Y_i, i = 1, 2,...,n$ all are independent with $N (\mu, \sigma)$ distribution, find a relation such that r can be expressed in terms of $\mu, \sigma^2$

Here's my working out so far:

Def of a martingale: $E(X_{n+1} | X_1,...,X_n) = X_n$

Hence: $E (S_{n+1} e^{-r(n+1)} | S_1,,,,,S_n) = S_ne^{-rn}$

Now because $S_n = e^{\sum Y_i}$ where $Y_i, i = 1, 2,...,n$

That becomes: $E(e^{Y_1 + ....+ Y_{n+1}} e^{-r(n+1)} |S_n)$

From that point im not too sure what to do. If anyone can help me or tell me what to do would be much appreciated. Thanks.

**Moo** · April 25th 2010, 08:11 AM

Hello,

What is r ?

**Redeemer_Pie** · April 25th 2010, 09:01 AM

"r" is just a constant variable.

**Moo** · April 25th 2010, 09:56 AM

Yes, but you're not clear at all !!!!!

I'll assume you want to find r such that $S_ne^{-rn}$ is a martingale. If that's not the case, then read again what you wrote...

$E(e^{Y_1 + ....+ Y_{n+1}} e^{-r(n+1)} |S_n)$
Okay with this.

Now this can also be written : $e^{-rn} e^{-r} \cdot E[e^{Y_1+\dots+Y_n}e^{Y_{n+1}}\mid S_n]$

Since the Yn are independent, we'll have a product of expectations :

$e^{-rn} \cdot E[e^{Y_1+\dots+Y_n}\mid S_n] \cdot e^{-r} \cdot E[e^{Y_{n+1}}\mid S_n]$

Since Y1,...,Yn are Sn-measurable, $E[e^{Y_1+\dots+Y_n}\mid S_n]=e^{Y_1+\dots+Y_n}$

Since $Y_{n+1}$ is independent with $S_n$ , $E[e^{Y_{n+1}}\mid S_n]=E[e^{Y_{n+1}}]$

So finally, you want r such that $e^{-rn} \cdot e^{Y_1+\dots+Y_n}\cdot e^{-r} \cdot E[e^{Y_{n+1}}]=S_n e^{-rn}$

which means to find r such that $e^{-r}\cdot E[e^{Y_{n+1}}]=1$

Have a look at the MGF of a normal distribution when t=1 and you're done...

(to simplify a bit all these things, you could've written $S_{n+1}$ in terms of $S_n$ , but it doesn't really matter, it's just more beautiful)

**Redeemer_Pie** · April 25th 2010, 09:47 PM

i think that should be it! Thanks alot!

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