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

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

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.

2. 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.

3. Thank you.

4. 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 $E(X(X-1))$

5. 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

### find mu=e(x), e[x(x-1)], and e[x(x-1)] e(x)-mu^2

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