# Poisson distribution - help with question!

• Feb 5th 2011, 12:50 PM
AAZZ
Poisson distribution - help with question!
Hello,

I am having problems solving the following question:
Let the random variable X have a Poisson distribution with parameter [lambda]. Show that for every n>=1 one has E(X^n) = [lambda]E[(X +1)^(n-1)]: Note that, using this formula, we can nd E(X);E(X2), . . .recursively.

-Thanks
• Feb 5th 2011, 02:16 PM
theodds
Quote:

Originally Posted by AAZZ
Hello,

I am having problems solving the following question:
Let the random variable X have a Poisson distribution with parameter [lambda]. Show that for every n>=1 one has E(X^n) = [lambda]E[(X +1)^(n-1)]: Note that, using this formula, we can nd E(X);E(X2), . . .recursively.

-Thanks

$\displaystyle
\sum_{x = 0} ^ \infty x^n \frac{\lambda^x e^{-\lambda}}{x!} = \sum_{x = 0} ^ \infty (x + 1)^n \frac{\lambda^{x + 1} e^{-\lambda}}{(x + 1)!}.
$

Fill in the gaps and complete the problem.
• Feb 6th 2011, 02:31 PM
stats2010
What do we do with this formula??
• Feb 6th 2011, 07:31 PM
mr fantastic
Quote:

Originally Posted by stats2010
What do we do with this formula??

The expectation (ha ha, ... expectation) is that you realise that $\displaystyle \sum_{x = 0} ^ \infty (x + 1)^n \frac{\lambda^{x + 1} e^{-\lambda}}{(x + 1)!} = \lambda \sum_{x = 0} ^ \infty (x + 1)^{n-1} \frac{\lambda^{x} e^{-\lambda}}{x!}$ ....
• Feb 6th 2011, 07:45 PM
mr fantastic
Quote:

Originally Posted by AAZZ
Hello,

I am having problems solving the following question:
Let the random variable X have a Poisson distribution with parameter [lambda]. Show that for every n>=1 one has E(X^n) = [lambda]E[(X +1)^(n-1)]: Note that, using this formula, we can nd E(X);E(X2), . . .recursively.