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This is a problem on sub-martingales.

Given : $X_n = X_0 \mathrm{e}^{\mu S_n}$, $n= 1,2,3,\ldots$, where $X_0 > 0$ and where $S_n$ is a symmetric random walk and $\mu$ is greater than zero. We have to prove that $\{X_n\}$ is a sub martingale.

How do we go about this?

I understand that, if $Z_n$ is a stochastic process, with $\mathbb{E}(Z_n) < \infty$ , then it is a sub martingale if $\mathbb{E}(Z_{n+1} | Z_1,Z_2,Z_3,\ldots,Z_n) \geqslant Z_n$. So how do I proceed with the problem?

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First, let us denote $$ S_n = \sum\limits_{i=1}^n\xi_i $$ where $(\xi_i)_{i=1}^\infty$ is a sequence of iid, symmetrically disctributed random variables: $\mathsf E\xi_1 = 0$. Furthermore, $$ X_{n+1} = X_n\cdot \mathrm e^{\mu(S_{n+1} - S_n)} $$ where $S_{n+1}-S_n = \xi_{n+1}$, so $$ X_{n+1} = X_n\cdot \mathrm e^{\mu\xi_{n+1}}. $$ Note that $\xi_{n+1}$ is independent on $X_1,\dots,X_n$. You definition of the submartingale is correct and we are going to apply it.

We need to find $$ \mathsf E[X_{n+1}|X_n,\dots,X_1] = \mathsf E\left[X_n\cdot \mathrm e^{\mu\xi_{n+1}}|X_n,\dots,X_1\right]. $$

Now we apply the fact that $X_n$ is measurable w.r.t. to $\sigma$-algebra in conditional expectation and $\xi_{n+1}$ is independent of this $ \sigma$-algebra, so $$ \mathsf E[X_{n+1}|X_n,\dots,X_1] = X_n\mathsf E\mathrm e^{\mu \xi_{n+1}}\geq X_n $$ since $\mathrm e^x$ is a convex function and $\mathsf E\mu\xi_{n+1} = 0$, so $\mathsf E\mathrm e^{\mu \xi_{n+1}}\geq 1$ by Jensen's inequality. To complete the proof you only need to show that $\mathsf E|X_n|<\infty$.

By the way, where do we use the fact that $\mu>0$?

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It's not necessary because $\mu \xi \sim -\mu \xi$ due to simmetry. – Kolmo Feb 7 '12 at 23:19
@Kolmo: that's right, but the point for Porbabilityman to discover it :) – Ilya Feb 8 '12 at 8:43

Hint: $\mathbb{E}(X_{n+1} | X_n) = X_n \mathbb{E}\left( \mathrm{e}^{\mu \left( S_{n+1} - S_n \right)} \right)$. What is the distribution of $S_{n+1} - S_n$ ? Can you prove that the expectation is greater than 1, provided $\mu > 0$?

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Thanks for your response. I do know that the expectation of a random walk will always be zero. Will the distribution of Sn+1 - Sn be normal? I am new to the topic and I am making an effort to be on the same page as what you have asked me to think on – Probabilityman Feb 7 '12 at 6:52

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