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.
 A: 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}
