- Thread starter gigiola
- Start date

If you know the distribution then you can calculate.

var(Z) =E[ (Z^2-E[Z])^2 ]

var(Z^2) =E[ (Z^4-E[Z^2])^2 ]

var(Z) =E[ (Z^2-E[Z])^2 ]

var(Z^2) =E[ (Z^4-E[Z^2])^2 ]

\(Var(Z^2) = E[(Z^2 - E[Z^2])^2] = E[Z^4] - E[Z^2]^2 \)

gigiola,

if distribution is not specified,

May be you can assume normal distribution and come up an answer of Var(Z^2)

E[Z^4] = S^4 + 6

E[Z^2] =

Now you can calculate Var(Z^2)

http://en.wikipedia.org/wiki/Normal_distribution

on internet I found something like:

var(XY)=Var(X)var(Y)+var(X)E[Y]^2+var(Y)E[X]^2

but there is the assumption of Yand X independent, that is hardly my case.

Probably I am complicating too much the question, so I give you the equation I have to solve, because probably there is an easier solution:

E[-e^-alpha(Y+Z)^2] that is:

-e^-alpha[E[Y+Z]^2-alpha/2*var[(Y+Z)^2] with Y and Z distributed as normal

thank you

var(XY)=Var(X)var(Y)+var(X)E[Y]^2+var(Y)E[X]^2

but there is the assumption of Yand X independent, that is hardly my case.

Probably I am complicating too much the question, so I give you the equation I have to solve, because probably there is an easier solution:

E[-e^-alpha(Y+Z)^2] that is:

-e^-alpha[E[Y+Z]^2-alpha/2*var[(Y+Z)^2] with Y and Z distributed as normal

thank you

Last edited: