1. ## Markov inequality

If X is a Poisson random variable with mean $\displaystyle \lambda$ show that for $\displaystyle i < \lambda,$

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

I've tried all the inequalities, but I keep getting stuck. Any push in the right direction would be greatly appreciated.

2. Originally Posted by Anonymous1
If X is a Poisson random variable with mean $\displaystyle \lambda$ show that for $\displaystyle i < \lambda,$

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

I've tried all the inequalities, but I keep getting stuck. Any push in the right direction would be greatly appreciated.
Since $\displaystyle P(X \leq i) = \sum_{j=0}^i \frac{1}{j!} j^i e^{-\lambda}$,
the desired inequality is equivalent to
$\displaystyle \sum_{j=0}^i \frac{1}{j!} j^i \leq (\frac{e \lambda}{i})^i$.

To that end, note that
$\displaystyle (\frac{e \lambda}{i})^i = (\frac{\lambda}{i})^i e^i$
$\displaystyle =(\frac{\lambda}{i})^i \sum_{j=0}^{\infty}\frac{i^j}{j!}$
$\displaystyle > (\frac{\lambda}{i})^i \sum_{j=0}^i \frac{i^j}{j!}$ ...(truncating)
$\displaystyle =\sum_{j=0}^i \frac{\lambda^i}{j! i^{i-j}}$
$\displaystyle \geq \sum_{j=0}^i \frac{\lambda^i}{j! \lambda^{i-j}}$ ... since $\displaystyle i \leq \lambda$ and $\displaystyle i-j \geq 0$
$\displaystyle =\sum_{j=0}^i \frac{\lambda^j}{j!}$
and we are done.