# 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.

• I'm assuming $t$ is not fixed? Feb 14 '17 at 23:31
• Yes, $t$ is arbitrary. Feb 14 '17 at 23:41

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.

$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 $t$ $\in \mathbf{R}$.