Originally Posted by

**ash89** I need to work out E((bb*)^2) for b~N(0,1/2)+iN(0,1/2) b* is complex conjugate.

Using prior working:

E[bb*]= E[ Re{b}^2 + Im{b}^2] = E[Re{b}^2] + E[Im{b}^2]

using the variance formula

E[Re{b}^2] = var[Re{b}] + E[Re{b}]^2 = 1/2

E[Im{b}^2] = var[Im{b}] + E[Im{b}]^2 = 1/2

so E[bb*] = 1

So far I have that:

E[(bb*)^2] = E[Re{b}^4 + Im{b}^4 + 2*Re{b}^2*Im{b}^2]

= E[Re{b}^4] + E[Im{b}^4] + 2*E[Re{b}^2]*E[Im{b}^2]

which tells me that 2*E[Re{b}^2]*E[Im{b}^2] = 2*1/2*1/2 = 1/2

But I am unsure of how to compute E[Re{b}^4] and E[Im{b}^4] easily as I do not know the variance of bb*

Your work seems correct so far. What you would need is $\displaystyle E[X^4]$ when $\displaystyle X\sim\mathcal{N}(0,\sigma^2)$, i.e. compute

$\displaystyle E[X^4]=\int x^4 e^{-x^2/(2\sigma^2)}\frac{dx}{\sqrt{2\pi\sigma^2}}$

You can integrate by parts by writing the integrand as $\displaystyle (x^3) (xe^{-x^2/(2\sigma^2)})$, deriving the first factor and integrating the second one. Thus you will resume to something like $\displaystyle C x^2 e^{-x^2/(2\sigma^2)}$ for some constant C, and you can relate this to

$\displaystyle \sigma^2=E[X^2]=\int x^2 e^{-x^2/(2\sigma^2)}\frac{dx}{\sqrt{2\pi\sigma^2}}$.

You should find $\displaystyle E[X^4]=3\sigma^4$, I think. Another way consists in differentiating the characteristic (or moment generating) function four times and evaluating it at 0, if you know that technique.