# Poisson Distribution

• Sep 14th 2008, 10:37 PM
lllll
Poisson Distribution
Let $\displaystyle X$ be a Poisson distribution. Show that $\displaystyle P\{X=k\}$ increases as it approaches to $\displaystyle \lambda$, reaching it's max when k is the largest integer less then $\displaystyle \lambda$ then decrease as $\displaystyle k > \lambda$

Hint: Consider $\displaystyle \frac{P\{X=k\}}{P\{X=k-1\}}$.

since the Poisson distribution is defined as:

$\displaystyle \frac{\lambda^k}{k!}e^{-k}$ I was thinking of using derivatives to max out the equation with respect to k, but realized that due to the factorial I can't since it's not continuous.

I'm not sure I'm doing this right, but if I apply the hint I would get:

$\displaystyle \left( \frac{\lambda^k}{k!}e^{-k} \right) \left( \frac{k!}{\lambda^k}e^{-k} \right) = \left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$

now I'm not sure what I have to do
• Sep 14th 2008, 11:17 PM
CaptainBlack
Quote:

Originally Posted by lllll
Let $\displaystyle X$ be a Poisson distribution. Show that $\displaystyle P\{X=k\}$ increases as it approaches to $\displaystyle \lambda$, reaching it's max when k is the largest integer less then $\displaystyle \lambda$ then decrease as $\displaystyle k > \lambda$

Hint: Consider $\displaystyle \frac{P\{X=k\}}{P\{X=k-1\}}$.

since the Poisson distribution is defined as:

$\displaystyle \frac{\lambda^k}{k!}e^{-k}$ I was thinking of using derivatives to max out the equation with respect to k, but realized that due to the factorial I can't since it's not continuous.

I'm not sure I'm doing this right, but if I apply the hint I would get:

$\displaystyle \left( \frac{\lambda^k}{k!}e^{-k} \right) \left( \frac{k!}{\lambda^k}e^{-k} \right) = \left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$

now I'm not sure what I have to do

$\displaystyle \frac{P\{X=k\}}{P\{X=k-1\}}=\left( \frac{(k-1)!}{k!} \right) \left( \frac{\lambda^k}{\lambda^{k-1}} \right) = \frac{\lambda}{k}$.

Therefore:

$\displaystyle \frac{P\{X=k\}}{P\{X=k-1\}}\ge 1$ when $\displaystyle k\le \lambda$

and:

$\displaystyle \frac{P\{X=k\}}{P\{X=k-1\}} < 1$ when $\displaystyle k > \lambda$

RonL