# Fourth moment of complex Gaussian r.v

Let $x \sim \mathcal{CN}(0,a)$ (a complex Gaussian random variable) and I know that the p.d.f. of $z$ is defined by:

$$p(z) = \frac{1}{\sqrt{2\pi a}} e^{-\frac{|z|^2}{2a}}$$.

Given that, how can I calculate

$$\mathbb{E}[|z|^4] = ?$$

Could someone show how to get $\mathbb{E}[|z|^4]$ step by step with the integrals, please?

• – whuber Feb 4 '18 at 15:07
• Thanks but I'm looking for the complex case. – Felipe Augusto de Figueiredo Feb 4 '18 at 17:46
• The complex case is a special case of the real distribution: complex numbers are just ordered pairs of real numbers and the complex norm is the same as the real norm. – whuber Feb 5 '18 at 14:19
• OK, but I still don get how to solve... sorry... – Felipe Augusto de Figueiredo Feb 5 '18 at 18:51
• Presumably $a$ refers to a positive real number, and it (rather than its square root) should appear in the denominator of the normalizing constant, for otherwise your notation needs explanation. In that case, writing $z=x+iy,$ $$\frac{|z|^2}{2a} = \frac{x^2}{2a} + \frac{y^2}{2a}$$ shows that $x$ and $y$ are independently Normal with mean $0$ and variance $a$. Moreover, expanding $$|z|^4 = (|z|^2)^2 = (x^2+y^2)^2 = x^4+2x^2y^2+y^4$$ reduces the question to finding the expectations $E[x^2]=E[y^2]$ and $E[x^4]=E[y^4]$. – whuber Feb 5 '18 at 19:44