An exercise I have to solve indicates the following: "Show that for any discrete random variable X: e^tE[X]<= E[e^tX] where t belongs to R and is fixed.
I think this is related with Jansen's inequality but I certainly don't know how to prove that. I would appreciate any help you can provide me.
Regards
On
The only difficulty here is that you have infinite convex combinations instead of finite ones. Let $g(x)=e^{tx}$. Then $g$ is a convex function. Fix a positive integer $N$ and consider $g(\sum_{k=1}^{N} \frac {p_k} {q_N} x_k)$ where $x_k$ are the values of the random variable $X$ and $p_k$'s are the corresponding probabilities,and $q_N =\sum_{k=1}^{N} p_k$. Convexity of $g$ gives $g(\sum_{k=1}^{N} \frac {p_k} {q_N} x_k) \leq \sum_{k=1}^{N} \frac {p_k} {q_N} g(x_k)$. Now let $N \to \infty$ and use the fact that $g$ is continuous. Since $q_N \to 1$ we get $e^{tEX} \leq Ee^{tX}$.
Well what do you think? (Are you sure that is the correct formulation of the task..?) The expectation is simply an integral, so just... use the definition of expectation, see that $$ \mathbb{E}[X]=\int x f(x)dx... $$