Let $X_{1}, \ldots, X_{n}$ be independent Poisson trials such that $\operatorname{Pr}\left(X_{i}=1\right)=$ $p$. Let $X=\sum_{i=1}^{n} X_{i}$, so that $\mathbf{E}[X]=p n$. Let
$$
F(x, p)=x \ln (x / p)+(1-x) \ln ((1-x) /(1-p)) .
$$
(a) Show that, for $1 \geq x>p$,
$$
\operatorname{Pr}(X \geq x n) \leq \mathrm{e}^{-n F(x, p)}
$$
(b) Show that, when $0<x, p<1$, we have $F(x, p)-2(x-p)^{2} \geq 0$. (Hint: Take the second derivative of $F(x, p)-2(x-p)^{2}$ with respect to $x$.)
(c) Using parts (a) and (b), argue that
$$
\operatorname{Pr}(X \geq(p+\varepsilon) n) \leq \mathrm{e}^{-2 n \varepsilon^{2}}
$$
(d) Use symmetry to argue that
$$
\operatorname{Pr}(X \leq(p-\varepsilon) n) \leq \mathrm{e}^{-2 n \varepsilon^{2}},
$$
and conclude that
$$
\operatorname{Pr}(|X-p n| \geq \varepsilon n) \leq 2 \mathrm{e}^{-2 n \varepsilon^{2}}
$$