Hi,

I want to find the variance of the product of two random variables normally distributed with common mean and variance.

X, Y normally distributed with common mean and variance Var(X) = Var(Y); X, Y are independent identical distributed.

what is variance(x * y)?

For the special case where the mean is 0 & non zero Cov(x, x)=0, after running some calculations on a list of random numbers it appears to follow the pattern:

Var(x*y) = Var(X)^2 = Var(Y)^2

For the special case where the mean is 0 and where x=x for each realization of x {non zero Cov(x, x)}, after running some calculations on a such list of random numbers it appears to follow the pattern:

Var(x*x) = 2*Var(X)^2 = 2*Var(Y)^2.

To make this case clear say the random number stream is

2.106, -1.392, 4.435, 1.940, -2.933, ...

then I am calculating the variance of: 2.106*2.106, (-1.392*-1.392), (4.435*4.435), ...

Confirmation of this apparent relationship would help, a general formula would be better, a proof or location of a proof would be much better

Thank you Peter

I want to find the variance of the product of two random variables normally distributed with common mean and variance.

X, Y normally distributed with common mean and variance Var(X) = Var(Y); X, Y are independent identical distributed.

what is variance(x * y)?

For the special case where the mean is 0 & non zero Cov(x, x)=0, after running some calculations on a list of random numbers it appears to follow the pattern:

Var(x*y) = Var(X)^2 = Var(Y)^2

For the special case where the mean is 0 and where x=x for each realization of x {non zero Cov(x, x)}, after running some calculations on a such list of random numbers it appears to follow the pattern:

Var(x*x) = 2*Var(X)^2 = 2*Var(Y)^2.

To make this case clear say the random number stream is

2.106, -1.392, 4.435, 1.940, -2.933, ...

then I am calculating the variance of: 2.106*2.106, (-1.392*-1.392), (4.435*4.435), ...

Confirmation of this apparent relationship would help, a general formula would be better, a proof or location of a proof would be much better

Thank you Peter

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