Jtdylktuy

Problem. Let $X_1, . . . , X_n$ be independent random variables such that $mathbf{E}(X_k) = 0$ and
$|X_k| leq 1$ for $k = 1, . . . , n$. Let $X = X_1 + . . . + X_n$. Prove the following strengthening of
the Bernstein inequality:
$$P(X geq alpha n) leq left(frac{1}{(1 − alpha)^{1−alpha}cdot(1 + alpha)^{1+alpha}}right)^{n/2} quad text{ for all } 0 leq alpha < 1.$$

Hint: Follow the proof of the Bernstein inequality, only do not use that $frac{e^t + e^{−t}}{2} leq e^{frac{t^2}{2}}$

attempt: Following Probability and random processes by grimmett.

Say $mathbf{P}(X_k)=p$. For small positive values of $epsilon$,
$$mathbf{P}left(frac{1}{n}X_ngeq p +epsilonright)=Sigma_{k geq n(p+epsilon)} mathbf{P}(X_n=k) $$
hence
$$mathbf{P}(X_n=k)=Sigma_{k=m}^nbinom{n}{k}p^k(1-p)^{n-k}$$
where $m=[n(p+epsilon)]$ the least integer not less than $n(p+epsilon)$. Let $lambda>0$ and note that $exp(lambda k) geq exp(lambda n (p+epsilon))$ if $k geq m$. Allowing $q=1-p.$ We have:

$$begin{align} mathbf{P}left(frac{1}{n}X_n geq p+ epsilonright) leq & Sigma_{k=m}^n exp(lambda[k-n(p+epsilon)]) binom{n}{k} p^kq^{n-k} \ leq & exp{-lambda n epsilon}Sigma_{k=0}^n binom{n}{k} pexp{(lambda q)}^k(qexp{(-lambda p)})^{n-k} \ =& exp{(-lambda n epsilon)}(pexp{(lambda q)}+qexp{(-lambda p)})^nend{align}$$
By the binomial theorem. By the fact that $e^x leq x+e^{x^2}$ for $x in mathbb{R}$ we get:

$$begin{align} mathbf{P}left(frac{1}{n} X_n geq p +epsilonright) leq & exp{(-lambda n epsilon)}[pe^{lambda^2q^2}+qe^{lambda^2 p^2}]^n \ leq & e^{lambda^2 n-lambda n epsilon}.end{align}$$

We can pick $lambda$ to minimize the RHS, $lambda =frac{1}{2}epsilon$, to get:

$$mathbf{P}left(frac{1}{n}X_n geq p +epsilonright)leq e^{-frac{1}{4}nepsilon^2} quad text{ for } epsilon >0 $$

Not sure how to get the inequality asked for.