# problem with calculating expected value

• Feb 16th 2009, 01:24 AM
problem with calculating expected value
Let X be Poisson with Parameter λ
Compute the mean of (1+X)^-1.

thanks
• Feb 16th 2009, 02:16 AM
mr fantastic
Quote:

Let X be Poisson with Parameter λ
Compute the mean of (1+X)^-1.

thanks

Start by noting that $E\left(\frac{1}{1+X} \right) = \sum_{x = 0}^{+\infty} \frac{1}{1 + x} \cdot \frac{\lambda^x e^{-\lambda}}{x!} = \sum_{x = 0}^{+\infty}\frac{\lambda^x e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$.

$\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$ will be useful in calculating the above sum ....
• Feb 16th 2009, 08:00 AM
thanks but...
first of all thanks for the answer but...
The answer (i have a only the final answer) is λ^-1(1- e^-λ) and

so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d

thanks again
• Feb 16th 2009, 08:21 AM
Moo
Quote:

first of all thanks for the answer but...
The answer (i have a only the final answer) is λ^-1(1- e^-λ) and

so isn't the sum above shoudnt be (1- e^-λ) instead of 1???d

thanks again

Try to look further than your nose ^^

$\frac{1}{\lambda} \sum_{x = 0}^{+\infty}\frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!}$
Make the change $k=x+1$

and $1=\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!}=e^{-\lambda}+ \sum_{k=1}^\infty \frac{\lambda^k e^{-\lambda}}{k!}$
• Feb 16th 2009, 10:52 AM
mr fantastic
Quote:

$\frac{1}{\lambda} \sum_{x = 0}^{+\infty} \frac{\lambda^{x+1} e^{-\lambda}}{(x+1)!} = \frac{1}{\lambda} \left( \frac{\lambda e^{-\lambda}}{1!} + \frac{\lambda^{2} e^{-\lambda}}{2!} + \, .... \right) = \frac{1}{\lambda} \left( 1 - e^{-\lambda}\right)$ using $\sum_{k = 0}^{+\infty}\frac{\lambda^{k} e^{-\lambda}}{k!} = 1$.