Whether is there a bound of $\sigma^2$ such that $pe^t+qe^{-t}\leq e^{t^2\sigma^2/2}$? Given $0<p,q<1$, $p+q=1$ and $\forall t\in \mathbf{R} $, is there a lower bound of $\sigma^2=\sigma^2(p)$, such that
$$pe^t+qe^{-t}\leq \exp\left(\frac{\sigma^2}{2}t^2\right).$$
Thank you very much guys.
 A: If $p=q$, we have the inequality
$$
\frac12e^t+\frac12e^{-t}=\cosh t\le\exp(t^2/2)
$$
for all $t$.
If $p\neq q$, then no upper bound on $pe^t+qe^{-t}$ of the form $\exp(kt^2)$  can hold for all $t$. Intuitively, the reason is that $\exp(kt^2)$ has value $1$ and slope zero near $t=0$, while $pe^t+qe^{-t}$ has value $1$ and nonzero slope near $t=0$. So there will be a small region where $pe^t+qe^{-t}$ exceeds $\exp(kt^2)$.
More rigorously, suppose $p>q$, and let $k>0$. We have for all $t$
$$
pe^t + qe^{-t}\ge p(1+t) + q(1-t)=1 + (p-q)t,\tag1
$$
using the inequality $e^x\ge 1+x$. If $t$ is positive and sufficiently small, we have
$$
1+(p-q)t> 1+2kt^2\tag2
$$
and
$$
1+2kt^2>\exp(kt^2).\tag3
$$
The inequality in (3) follows from $e^x< 1+2x$ for all small positive $x$. The case $p<q$ is handled similarly.
A: $pe^t+qe^{-t}$$\leq$$e^{\frac{\sigma^2}{2}t^2}$
Apply the following unequality for the left side: 
if $a\gt0$ and $b\gt0$ real numbers then $\sqrt[]{ab}\le$${a+b}\over2$,
$a=pe^t$  and $b=qe^{-t}$, 
$2\sqrt[]{(pe^tqe^{-t}}$)$\leq $$e^{\frac{\sigma^2}{2}t^2}$
We have 
$2\sqrt[]{p(1-p)}$$\leq$$e^{\frac{\sigma^2}{2}t^2}$, 
that is valid for all $\sigma$ and $t$ $\in \mathbf{R}$.
