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sub martingales and more

Can someone help me compute the expected value of $X_{n+1}$, that is : $E[X_{n+1}| X_0,\dots,X_n] $?

Given : $X_n = X_0 e^ {\mu S_n}$, $X_0 > 0$

where $S_n$ is a symmetric random walk and $\mu$ is greater than zero.

I am aware that the expected value of a given function is the mean. But i would like to know a method to compute the above. What is the right approach to get started on such problems on expected value computation.


I understand that $X_{n+1} = X_n \cdot e^{\mu (S_{n+1} - S_n)}$ ? How do I proceed with computing the expectation?

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marked as duplicate by Ilya, J. M., Asaf Karagila, Nate Eldredge, Zev Chonoles Feb 8 '12 at 14:11

This question was marked as an exact duplicate of an existing question.

It appeared that you have even elder and more general version of the question. I hope the sequence of duplicates stops here. Voting to close this version as well – Ilya Feb 7 '12 at 12:17

Hint: $X_n = e^{\mu Y_n} X_{n-1}$ where $Y_n = S_n - S_{n-1}$ is the $n$'th step of the random walk.

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so i understand that Xn+1 = Xn e^(mu * (Sn+1 - Sn) ? How do i proceed with computing the expectation? – Probabilityman Feb 7 '12 at 7:04
Also the next step would be Xn * E(e^ mu(Sn-1 - Sn)) – Probabilityman Feb 7 '12 at 7:06
Could someone guide me on how to proceed further? – Probabilityman Feb 7 '12 at 7:24
Since $Y_n$ and $X_{n-1}$ are independent, $E[X_n] = E[e^{\mu Y_n}] E[X_{n-1}]$. If this is a simple random walk ($Y_n = \pm 1$, each with probability $1/2$), $E[e^{\mu Y_n}] = (e^\mu + e^{-\mu})/2 = \cosh(\mu)$. – Robert Israel Feb 7 '12 at 16:59

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