1. ## gamma distribution 2

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!!!

2. W is the joint density of three i.i.d variates. What does this mean?

3. []EDIT[]Corrected post. Please refer to Matheagle and Moo's expert posts below.

4. are you asking for the mean and covariance matrix of a vector?

5. Thank you so much !!!!

Jacobian tranfomation? I am not familiar with it T-T

I need to find E(W) and Var(W)........that's what question was asking does it help?

6. I think you can use my method. Unless there are any objections?

Let me know if you get stuck.
Anonymous

7. $E(W) = \left(E(Z_1 ^2), E(Z_2 ^2), E(Z_1)E(Z_2)\right)=(2,2,1)$

The variance/covariance matrix has in the $ij^{th}$ 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.

8. Originally Posted by Anonymous1
Since they are i.i.d you can multiple the three densities $f_{Z}^2(z)$ where $f_{Z}(z)$ is the density of an $exponential(1)$ to find your joint distribution. See what you get. Does it look gamma? Then simply compute the first and second moments to find the variance of W.

$\int_0^{\infty} z\times f_{Z_1}^2(z)*f_{Z_2}^2(z)*f_{Z_3}^2(z)dz$

$\int_0^{\infty} z^2\times f_{Z_1}^2(z)*f_{Z_2}^2(z)*f_{Z_3}^2(z)dz$

wait a minute... You may have to use a Jacobian transform. Does anyone know?
What the... ?

The pdf of $Z^2$ when Z has the pdf $f_Z$ is certainly not $f_Z^2$

9. 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

10. Originally Posted by dymin3
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)
cov (Z2^2, Z1^2) var (Z2^2) cov (Z2^2, Z1Z2)
cov (Z1Z2, Z1^2) cov (Z1Z2, Z2^2) var (Z1Z2)

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.

11. Wow, thank you so much. I have one last question

Now i have Var(Z1Z2) = E[(Z1Z2)^2] - [E(Z1Z2)]^2 = ? - 1^2
How can i find E[(Z1Z2)^2]? I can't use the moments of an exp(1) for this, can I?

12. ... It's algebra : (Z1Z2)^2=Z1^2 Z2^2
And since Z1 and Z2 are independent, we'll have E[(Z1Z2)^2]=E[Z1^2]E[Z2^2]

13. Originally Posted by Moo
... It's algebra : (Z1Z2)^2=Z1^2 Z2^2
And since Z1 and Z2 are independent, we'll have E[(Z1Z2)^2]=E[Z1^2]E[Z2^2]
oops.... i kept forgetting about the meaning of the independent
Thanks!!

Now I have EW = (2,2,1) and VarW = (20, 20, 3)
Problem solved!!! YEY!!