Is the following tail bound on the Binomial distribution accurate?

Question

The Poisson distribution has the following tail-bound:

Given $X \leftarrow Po(\lambda)$, for any $t \geq 1$

$Pr(X \geq t) \leq (\frac{e \lambda}{t})^t$

(For $t > \lambda$, this follows directly from dropping the $e^{-\lambda}$ term from the tail bound presented here. For $t \leq \lambda$, $\frac{e \lambda}{t} \geq e$, so the bound is trivially true since $Pr(X \geq t) \leq 1 < e^t$.)

Does this bound apply to the Binomial distribution (setting $\lambda = np$)?

Given $X \leftarrow Bin(n, p)$, is it true that for all $t \geq 1$, $Pr(X \geq t) \leq (\frac{enp}{t})^t $? If so why?

Answer 1

This follows directly from Chernoff bounds.

$Pr(X \geq (1 + \delta) \mu) \leq \left( \frac{e^\delta}{(1+\delta)^{1 + \delta}}\right)^\mu = \left( \frac{e}{1+\delta} \right)^{(1 + \delta)\mu} e^{-\mu}$

So, setting $t = (1 + \delta) \mu$

$Pr(X \geq t) \leq \left( \frac{e \mu}{t} \right)^t e^{-\mu}$

Here $\mu = np$, so:

$$Pr(X \geq t) \leq \left( \frac{enp}{t} \right)^t e^{-np} \leq \left( \frac{enp}{t} \right)^t$$