Vtyjkyuk

While we can prove the following with AM-GM inequality, $$ fracp(1-p)e^t(1-p+p e^t)^2 le frac 1 4, t in mathbb R, p in [0,1]$$

I want to find a proof that involves probability or alternatively to give a probabilistic interpretation of the inequality. We are given that $t$ is a real number and $p,1-p in [0,1]$, so we have that $p(1-p) le frac 1 4$. Thus, if we can prove that

$$ frace^t(1-p+p e^t)^2 le 1, p in (0,1),$$

then we will have our result.

I was thinking to use Hoffeding's Lemma or Chernoff bound and to consider a binomially distributed random variable $X$ with parameters $n=2, p in (0,1)$. Then, we have

$$E[e^Xt] = (1-p+pe^t)^2$$

$$E[e^(X-2p)t] = e^-2tp(1-p+pe^t)^2$$

Hoffeding's Lemma gives

$$E[e^t(X-2p)] le e^fract^22$$

Chernoff bound gives for $a>0$ (the assumption in Markov inequality, of which the Chernoff bound is based)

$$e^taP(X-2p ge a) le E[e^t(X-2p)]$$

Hence,

$$e^taP(X-2p ge a) le e^-2tp(1-p+pe^t)^2 le e^fract^22$$

$$implies frac1e^taP(X-2p ge a) ge frace^2tp(1-p+pe^t)^2 ge e^frac-t^22$$

I guess Hoffeding's Lemma doesn't help (Btw, this problem is kind of a converse to Hoffeding's Lemma I guess), so let's use Chernoff bound to say

$$frace^2tp(1-p+pe^t)^2 le frac1e^taP(X-2p ge a)$$

$$implies frac1(1-p+pe^t)^2 le frac1e^2tpe^taP(X-2p ge a)$$

$$implies frac1(1-p+pe^t)^2 le frac1e^t(a-2p)P(X-2p ge a)$$

$$implies frace^t(1-p+pe^t)^2 le frace^te^t(a-2p)P(X-2p ge a) = frace^(t)(1-a+2p)P(X-2p ge a) = frace^(t)(1-a+2p)1-F_X(2p+a-1)$$

I guess either there's some special $a$ to pick or we take limits as $a to infty$ or $a to b^+$ or something. What $a$ or $b$ might I pick?

I also thought of something like for degenerate $Y=c=1$, we might show

$$ textmgf of degenerate = e^ct le (1-p+pe^t)^2 = textmgf of Bin(2,p) for c=1$$