Expected value of $e^{tX}$ for $t > 0$

Question

I am trying to prove the following inequality: Given a random variable $X$ with $$ 0\leq X\leq 1 \\ E[e^{tX}] \leq (1 - E[X]) + E[X]e^t, t > 0$$

I have the following steps so far: Since $$X \in [0,1] \implies E[X] \in [0,1]$$ Define a convex function $$f(X) = e^{tX}$$ Then $$f(E[X](1) + (1 - E[X])0) \leq E[X] f(1) + (1-E[X])f(0)\\ \implies e^{tE[X]}\leq E[X]e^t + (1-E[X])$$ Now if I apply Jensen's inequality, I get $$f(E[X]) \leq E[f(X)] \\ \implies e^{tE[X]} \leq E[e^{tX}]$$

How to go from here?

Answer 1

$$e^{tx} \leq (1-x)+xe^{t}$$ for $0 \leq x \leq 1$ and $t \geq 0$.

To see this fix $x$ and note that equlaity holds when $t=0$; now check that the derivative of $$e^{tx} -(1-x)+xe^{t}$$ w.r.t. is non-positive for all $t \geq 0$. This gives above inequality.

Now we have $e^{tX} \leq (1-X)+Xe^{t}$ and taking expectation completes the proof.

Answer 2

Note that deterministically,

$$e^{tx}\leq (1-x)+xe^t\quad \forall x\in [0,1],t\in\mathbb{R} \quad (1).$$

To see this, consider a discrete random variable $Y$ having two-point support $\{0,t\}$ with respective probability masses $1-x,x.$ Since $z\rightarrow e^{z}$ is a convex function, Jensen's inequality tells us

$$e^{tx}=e^{E[Y]}\leq E[e^Y]=(1-x)+xe^t.$$

Applying $(1)$ to random variable $X$ in place of $x$ and taking the expectation of both sides gives the desired result.