W is the joint density of three i.i.d variates. What does this mean?
I had a similar problem and I was trying to solve this question based on reply (thank you so much!!) but im having a difficult time to do following question:
For Z distributed on R^3 with Z1, Z2, Z3 IID Z ~ exp(1)
If i have W = (Z1 ^2, Z2 ^2, Z1Z2) then how can I find expected value and variance?
Do I still use gamma distribution? I am so confused....
plz help me!!!
The variance/covariance matrix has in the position the covariance between Zi and Zj.
Note that the covariance between Z1 and Z1 is its variance.
It's a 3 by 3 symmetric matrix, where you have the 3 variances down the diagonal and the covariances in the other positions.
Thank you so much!!! so you mean that 3 by 3 symmetric matrix will look like
var (Z1^2) cov (Z1^2, Z2^2) cov (Z1^2, Z1Z2^2)
cov (Z2^2, Z1^2) var (Z2^2) cov (Z2^2, Z1Z2^2)
cov (Z1Z2^2, Z1^2) cov (Z1Z2^2, Z2^2) var (Z1Z2^2)
but aren't they all 1s for var(Z1^2) and var(Z2^2) and even var(Z1Z2^2)??? in this case??? im not sure....T-T
False
Var(Z^2)=E(Z^4)-[E(Z^2)]^2
And it's easy, the moments of an exp(1) are : E[X^k]=k!
So Var(Z^2)=4!-2^2=24-4=20
For the covariance, we have for example cov(Z2^2,Z1^2)=0, because Z1 and Z2 are independent.
And you messed up some things : you're dealing with Z1Z2, not Z1Z2^2. I corrected it.