sub martingales and more 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?
 A: 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 $\mu>0$?
A: 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 $\mu > 0$?
