Find mu (mean); E [X (X - 1)] and sigma squared (variance) for each of the following

**Intsecxtanx** · September 27th 2009, 09:47 AM

Find mu (mean); E [X (X - 1)] and sigma squared (variance) for each of the following distributions:
1. f (x) = (1 / x! ) (1/2)^5 (5! / (5 - x)! ) ; for x = 0, 1, 2, 3, 4, 5
2. f (x) = (e^ -a) (a^x)/(x!) for x = 0, 1, 2,... (a > 0)

any help will be very much appreciated I'm so lost on this homework question.

**matheagle** · September 27th 2009, 07:23 PM

I'll do 2, since I mentioned that in class last week

Here you have a Poisson and your mean is a.

You calculate the mean via the Taylor Series for $e^a=\sum_{x=0}^{\infty} {a^x\over x!}$

so $\mu =\sum_{x=0}^{\infty} {xe^{-a}a^x\over x!} =ae^{-a}\sum_{x=1}^{\infty} {a^{x-1}\over (x-1)!}$

Let $y=x-1$ and you have

$=ae^{-a}\sum_{y=0}^{\infty} {a^y\over y!}= ae^{-a}e^a=a$

IN ORDER to get the variance NOTE that we need the second moment

BUT it's easier to get $E(X(X-1))=E(X^2)-E(X)=E(X^2)-a$ which is NOT the variance.

NOW do the same thing I did above with this and it's over...

$E(X(X-1))=\sum_{x=0}^{\infty} {x(x-1)e^{-a}a^x\over x!}=a^2e^{-a}\sum_{x=2}^{\infty} {a^{x-2}\over (x-2)!}$

NOW let $y=x-2$ and you have

$=a^2e^{-a}\sum_{y=0}^{\infty} {a^y\over y!}= a^2e^{-a}e^a=a^2$

Add this up and the variance is a.

**Intsecxtanx** · September 27th 2009, 07:26 PM

Thank you.

**matheagle** · September 27th 2009, 07:42 PM

Originally Posted by Intsecxtanx

Thank you for the advice but our professor hasn't discussed the poisson distribution and advised us not to go that route.

I have no idea what other route to take.

The idea is to solve for the second moment via $E(X(X-1))$

**Intsecxtanx** · September 27th 2009, 07:48 PM

I'm so sorry, I edited my last comment because I didn't realize how thorough your explanation was and how helpful it actually was. I really, really appreciate your help and will use this knowledge to complete my homework! Thank you

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