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.

  • $\begingroup$ 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

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.



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.