Derivation of alternative form of Chernoff bound

I have encountered an alternative form for the Chernoff bound for the sum of $$n$$ coins which I have not been able to derive.
Specifically, let $$X_1,...,X_n$$ be independent Poisson trials, let $$X = \sum_{i=1}^{n} X_i$$ and let $$\mu = \mathbb{E}(X)$$. Then
$$\forall t > 0 . \mathbb{P}(X \geq \mu + t) \leq \exp\left(-2\frac{t^2}{n}\right)$$

I am familiar with how to arrive at the more common variant of this Chernoff bound, that is $$\forall \delta > 0. \mathbb{P}(X \geq (1+\delta)\mu) \leq \left(\frac{e^\delta}{(1+\delta)^{(1+\delta)}}\right)^\mu,$$ but I have not been able to derive the former from it. Any help would be greatly appreciated.

EDIT: parsiad's answer makes use of Hoeffding's inequality, which was introduced in the lecture notes I was reading well after this variant of the bound was presented. So if there is a way to derive this bound without using Hoeffding's inequality/lemma, I would be grateful to see it.

UPDATE: It is possible to show that the bound on the probability as given by the formula I was looking to derive is actually tighter than the other bound. Hence I don't think it is possible to derive it without Hoeffding's inequality.

For each $$n$$, let $$X_{n}$$ be a random variable bounded between $$a_{n}$$ and $$b_{n}$$. Let $$X\equiv X_{1}+\cdots+X_{n}$$ and $$\mu \equiv \mathbb{E}X$$. Hoeffding's inequality states that $$\mathbb{P}(X \geq \mu + t)\leq\exp\left(-\frac{2t^{2}}{\sum_{i=1}^{n}\left(b_{i}-a_{i}\right)^{2}}\right).$$ In your case, $$b_{i}=1$$ and $$a_{i}=0$$ and hence the right hand side above becomes $$\exp(-2t^{2}/n)$$, as desired.
• Perhaps I am not familiar with the definition of Poisson trials. I was under the impression that this referred to a sequence $(X_n)_n$ of independent Bernoulli random variables that do not necessarily have the same success probability. If this is indeed the case, then each one is bounded between $0$ and $1$. Aug 27, 2019 at 8:16