Expectation inequality Let $X, Y$ be random variables with $0 \leq X \leq Y$ and $\mathbb E[Y]=1$. Let $t&gt;0$. Does the inequality

$$
\mathbb E[e^{tX}] \leq x\, \mathbb E[e^{tY}]+1-x
$$

where $x=\mathbb E[X]$ hold?
This inequality does not hold in general. For example, take $X$ and $Y$ such that:


with probability $\frac{2}{3}$, $X = 0$ and $Y = \dfrac{1}{2}$,
with probability $\frac{1}{3}$, $X = 2$ and $Y = 2$.


In this case, the inequality would write:
$$
\frac{2}{3}+\frac{e^{2t}}{3} \leq \frac{2}{3}\left(\frac{2e^{t/2}}{3} + \frac{e^{2t}}{3}\right) + 1 - \frac{2}{3}
$$
As $t$ tend to $+\infty$, this would lead to $1 \leq \frac{2}{3}$, wich is of course absurd.



However, it can be made true under certain hypotheses: namely, if we assume that $X = ZY$ (hence $0 \leq Z \leq 1$) where $Z$ is independent of $Y$.

The convexity of the function $\exp$ then gives:
$$
e^{tX} = e^{t(Z\times Y + (1-Z)\times 0)} \leq Z e^{tY} + (1-Z)e^0
$$
for every $t &gt; 0$. Taking the expectations, the independence gives
$$
E(e^{tX}) \leq x E(e^{tY}) + 1 - x
$$
since $x = E(YZ)=E(Z)\times 1$.
By series expansion $e^{tX} = 1 + tX + \frac{t^2X^2}{2!} + \frac{t^3X^3}{3!} + \cdots +\frac{t^nX^n}{n!} + \cdots.$

$M_X(t) = E(e^{tX}) = 1 + tm_1 + \frac{t^2m_2}{2!} + \frac{t^3m_3}{3!}+\cdots + \frac{t^nm_n}{n!}+\cdots$

Thus, $E(e^{tX}) &gt; 1+tE[X]$ and hence $E(e^{tX})&gt;1+tx$ since $E[X]=x$.

Similarly, $E(e^{tY})&gt;1+t$

Now,
\begin{eqnarray}
 xE(e^{tY})+1-x &amp;&gt;&amp; x(1+t)+1-x      \nonumber \\
   &amp;&gt;&amp; x+tx+1-x \nonumber \\
   &amp;&gt;&amp;1+tx \nonumber \\
   &amp;&gt;&amp; E(e^{tX})
\end{eqnarray} 

Since $X\leq Y$, thus, $E(e^{tX})\leq E(e^{tY})$ and hence the inequality follows.