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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.

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  • $\begingroup$ I'm assuming $t$ is not fixed? $\endgroup$
    – Chee Han
    Feb 14 '17 at 23:31
  • $\begingroup$ Yes, $t$ is arbitrary. $\endgroup$
    – peter
    Feb 14 '17 at 23:41
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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.

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

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