# Thread: Finding variance for a discrete random variable

1. ## Finding variance for a discrete random variable

Let W be a discrete rv with pmf f(w) = c|w-4| for w within {0,1,...,8}.
a) Find c.
b) Find E(W).
c) Find Var(W).
d) Find Var((W-4)^2).

I already found that c = 1/20, E(W) = 2.4, and Var(W) = 20.24; I'm just having problems with part (d). I split it up into:
Var((W-4)^2) = E((W-4)^4) - (E((W-4)^2))^2

I know that (W-4)^2 = W^2 - 8W + 16, but I can't really see how this would help. This there any way to go about this problem using the values that I have already found?

2. Originally Posted by uberbandgeek6
Let W be a discrete rv with pmf f(w) = c|w-4| for w within {0,1,...,8}.
a) Find c.
b) Find E(W).
c) Find Var(W).
d) Find Var((W-4)^2).

I already found that c = 1/20, E(W) = 2.4, and Var(W) = 20.24; I'm just having problems with part (d). I split it up into:
Var((W-4)^2) = E((W-4)^4) - (E((W-4)^2))^2

I know that (W-4)^2 = W^2 - 8W + 16, but I can't really see how this would help. This there any way to go about this problem using the values that I have already found?
Var(W^2 - 8W + 16) = Var(W^2) + 8 Var(W) and you should know how to calculate Var(W^2) from the given pmf.

3. Would't you have to square the -8 when you take it out though? Should it be Var(W^2) + 64*Var(W)?

4. Originally Posted by uberbandgeek6
Would't you have to square the -8 when you take it out though? Should it be Var(W^2) + 64*Var(W)?
Yes. My mistake.