Question

It $X$is a Poisson random variable with a mean$\lambda$, showing that for$i<\lambda$,

$P\{X\le i\}\le \frac{e^{-\lambda }(e\lambda {)}^i}{i^i}$

Short answer

Therefore,

$\frac{d}{dt}\lambda \left(e^t-1\right)-it=0\Rightarrow e^t=\frac{i}{\lambda }$

Hence proved.

Step-by-step solution

Given Information.

$X$is a Poisson random variable with a mean$\lambda$,

Explanation.

We are going to be using Chernoff bounds. Specifically, it $X$is a random variable with$MGFM\left(t\right)$and$i>0,t<0$

$\Rightarrow P(X\le i)\le e^{-it}M\left(t\right)$

Explanation.

$X$$poiss$$\left(\lambda \right)$then the $MGF$is$M\left(t\right)=e^{\lambda \left(e^t-1\right)}$. Then using Chernoff bound:

$$\begin{gathered} i<\lambda ,P(X\le i)\le e^{-it}{e}^{\lambda \left(e^t-1\right)} \\ =e^{\lambda \left(e^t-1\right)-it} \end{gathered}$$

We want $t$to minimize this. Since $e^x$is a strictly increasing function, the minimizing argument is given by the first-order conditions of

$\lambda \left(e^t-1\right)-it$

So we derive:

$\frac{d}{dt}\lambda \left(e^t-1\right)-it=0\Rightarrow e^t=\frac{i}{\lambda }$

Since$i\le \lambda$this means that $t<0$and so the Chernoff bound is valid.