Markov inequality

**Anonymous1** · February 9th 2010, 04:27 PM

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

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

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

**awkward** · February 12th 2010, 02:36 PM

Originally Posted by Anonymous1

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

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

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

Since $P(X \leq i) = \sum_{j=0}^i \frac{1}{j!} j^i e^{-\lambda}$ ,
the desired inequality is equivalent to
$\sum_{j=0}^i \frac{1}{j!} j^i \leq (\frac{e \lambda}{i})^i$ .

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

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