# Mean and Variance of norm of multivariate normal

Let $$a\in\mathbb{C}^n$$ be some complex vector, and let $$Y \sim \mathbb{C}\mathcal{N}(0, \sigma^2I)$$, i want to find the mean and variance of this random variable: $$X = \left\lVert Y \right\rVert_2^2 - \left\lVert Y - a \right\rVert_2^2.$$

This is a random variable describing the distance of $$Y$$ to its mean versus another specific point. I was able to calculate the mean of $$X$$ as $$-\left\lVert a \right\rVert_2^2.$$ But the variance computations has proven to be quite complicated. Can someone help with this?

Edit: Work I've done so far To compute $$E[X^2]$$, we have that: $$X^2 = \left\lVert Y \right\rVert^4 + \left\lVert Y - a \right\rVert^4 - 2\left\lVert Y \right\rVert^2 \left\lVert Y - a \right\rVert^2.$$ Computing the terms one at a time, we start with: \begin{align} \left\lVert Y - a \right\rVert^4 = {} & ( \left\lVert Y\right\rVert^2 + \left\lVert a \right\rVert^2 - 2\Re \left\langle Y, a \right\rangle)^2 \\ = {} & \left\lVert Y\right\rVert^4 + \left\lVert a\right\rVert^4 + 4\left(\Re \langle Y,a\rangle\right)^2 \\ &+ 2\left\lVert Y\right\rVert^2\left\lVert a\right\rVert^2 \\ &- 4 \left\lVert Y\right\rVert^2 \Re\left\langle Y, a \right\rangle \\ &\underbrace{ {} - 4 \left\lVert g\right\rVert^2 \Re\left\langle Y, a \right\rangle}_{E[\cdot]= 0} \end{align} Moreover, we have that: \begin{align} \left\lVert Y\right\rVert^2 \left\lVert Y - a \right\rVert^2 &= \left\lVert Y\right\rVert^4 + \left\lVert Y \right\rVert^2\left\lVert a \right\rVert^2 - 2\left\lVert Y \right\rVert^2\Re \langle Y,a \rangle \\ \end{align} Subtracting $$E[X]^2 = \left\lVert a \right\rVert_2^4$$ from $$E[X^2]$$, and removing all the terms that cancel out: \begin{align} \operatorname{Var} &= E[X^2] - E[X]^2 \\ &= E\left[\left\lVert a\right\rVert_2^4 + 4(\Re\langle Y,a\rangle)^2\right] - \left\lVert a\right\rVert_2^4 \\ &= E\left[4(\Re \langle Y,a\rangle )^2\right] \\ &= E\left[\big(\langle Y,a \rangle + \langle a,Y\rangle\big)^2\right] \end{align} I'm stuck on what to do with the expression above.

• I've never worked with complex random variables, but wouldn't $\|Y\|_2^2$ and $\|Y-a\|_2^2$ each have a noncentral chi squared distribution, and $X$ have a generalized chi-squared distribution?
– Joe
Commented Jun 17, 2021 at 23:59
• Where did you get stuck with the variance? I agree that it might be quite painful, but ultimately this is a doable computation
– md5
Commented Jun 18, 2021 at 10:19
• @md5 I've arrived at the expression above that and i don't know how to proceed from there. Commented Jun 18, 2021 at 18:31

I will take $$\langle a,b\rangle$$ to mean $$a'b$$ where $$a'$$ is the conjugate-transpose of $$a$$ and $$a'b$$ is the product of $$a'$$ and $$b$$ as matrices. And $$\|a\|_2^2 = \langle a,a\rangle.$$
And if $$\mu=\operatorname E(W)\in\mathbb C^{n\times1}$$ then $$\operatorname{var}(W) = \operatorname E\big((W-\mu)(W-\mu)'\big) \in\mathbb C^{n\times n}.$$
Consequently $$\operatorname{var} \langle a,Y\rangle = a'\big(\operatorname{var}Y\big) a.$$
Since $$\langle Y,a\rangle = \langle a,Y\rangle'\in\mathbb C,$$ we have $$\operatorname{var}\langle Y,a\rangle = \big( \operatorname{var}\langle a,Y\rangle\big)'.$$
Here $$a'\in\mathbb C^{1\times n},$$ $$\operatorname{var}Y\in\mathbb C^{n\times n},$$ and $$a\in\mathbb C^{n\times1}.$$
\begin{align} \|Y-a\|_2^2 & = \|Y\|_2^2 - \langle Y,a\rangle - \langle a,Y\rangle + \|a\|_2^2. \\[8pt] \|Y\|_2^2 - \|Y-a\|_2^2& = \langle Y,a\rangle + \langle a,Y\rangle -\|a\|_2^2. \end{align} The variance of that last item is what you seek, and only the first two terms are random. So then we have \begin{align} \operatorname{var}\langle a,Y\rangle = a'\big(\operatorname{var}Y\big)a = a'\big(\sigma^2I_n\big)a = \sigma^2\|a\|_2^2. \end{align}