# Thread: Possible to find p(X=x) given E(X) and E(X^2)?

1. ## Possible to find p(X=x) given E(X) and E(X^2)?

This practice problem gives us E(X) = 11 and E(X^2) = 121 and asks for us to find probability fn and cdf.

Initially I was thinking we could use some identities of the probability generating function like P"(1) = E(X^2) - E(X) and P'(1) = E(X), but I'm running into a mental wall where I'm positive that you can't work it out unless you know how X is distributed (as in is it a poisson, binomial, etc).

2. Originally Posted by drowned
This practice problem gives us E(X) = 11 and E(X^2) = 121 and asks for us to find probability fn and cdf.

Initially I was thinking we could use some identities of the probability generating function like P"(1) = E(X^2) - E(X) and P'(1) = E(X), but I'm running into a mental wall where I'm positive that you can't work it out unless you know how X is distributed (as in is it a poisson, binomial, etc).

If $\displaystyle \overline{x}=11$ and $\displaystyle \overline{x^2}=121$ what is the variance of $\displaystyle x$

CB

3. Originally Posted by CaptainBlack
If $\displaystyle \overline{x}=11$ and $\displaystyle \overline{x^2}=121$ what is the variance of $\displaystyle x$

CB
I see that as well where V(X) = E(X^2) - E(X)^2 = 121 - 121 = 0. Getting the vibe that a variance = 0 means something significant but I'm a bit lost still. Does 0 variance mean we're limited to one value of x being 100% probable and all other values or 0%?

4. Originally Posted by drowned
I see that as well where V(X) = E(X^2) - E(X)^2 = 121 - 121 = 0. Getting the vibe that a variance = 0 means something significant but I'm a bit lost still. Does 0 variance mean we're limited to one value of x being 100% probable and all other values or 0%?
Yes.

CB

5. since the mean is 11 and the second moment is 11 squared....

P(X=11)=1