How to link the variance of the distance between two vectors to the variance of their norms? I'm looking at a simplified case of this question where I have a random vector variable $X$ in dimension $k$. I know this about $X$: its mean $E[X]$, its covariance matrix, the mean of its $L^2$ norm $E[{||X||]}$, the variance of its $L^2$ norm $\operatorname{Var}(||X||)$.
I have another random vector variable $Y$ in the same space, and I know that $\overline{||X - Y||} = A$ where $A$ is some constant, $\operatorname{Var}(||Y||)$ = $\operatorname{Var}(||X||)$ and $X$ and $Y$ are independent.
I'm trying to figure out how $\operatorname{Var}(||X - Y||)$ relates to the properties I know of $X$ and $Y$, either through a strict equality or an upper bound. Here's my attempt at finding an upper bound:
$$
\begin{align}
\operatorname{Var}(||X - Y||) & = E[(||X - Y|| - E[||X - Y||])^2] \\
                 & = E[(||X - Y|| - A)^2] \\
                 & = E[||X - Y||^2] - 2 A \times E[||X - Y||] + A^2 \\
                 & = E[||X - Y||^2] - 2A^2 + A^2 \\
                 & = E[||X - Y||^2] - A^2 \\
                 & \leq E[(||X|| + ||Y||)^2] - A^2 \\
                 & \leq E[||X||^2] + E[||X||||Y||] + E[||Y||)^2] - A^2
\end{align}
$$
But I'm stopped at the second order moment. Plus, I think the $A$ constant should somehow disappear. I'm conjecturing that

*

*$\operatorname{Var}(||X - Y||) \leq \operatorname{Var}(||X||) + \operatorname{Var}(||Y||) = 2 \operatorname{Var}(||X||)$

*The equality is strict if $X$ and $Y$ are independent.

I'm also wondering if there is a formula in the more general case.
 A: Counter-example
Take $X, Y$ i.i.d. with $P[X=-1]=P[X=1]=1/2$.  Then $Var(|X|)+Var(|Y|)=0$ but $Var(|X-Y|)>0$.
A modified statement:
Let $X=(X_1, ..., X_n)$ and $Y=(Y_1,...,Y_n)$ be random vectors in $\mathbb{R}^n$.  Assume $E[X_i^2]$ and $E[Y_i^2]$ are finite for all $i \in \{1, ..., n\}$, and that $X_i$ and $Y_i$ are uncorrelated for each $i \in \{1, ..., n\}$.   Let $||x||= \sqrt{\sum_{i=1}^n x_i^2}$ denote the Euclidean norm. Then
\begin{align*}
 Var(||X-Y||) &\leq Var(||X||) + Var(||Y||) \\
& \quad + (E[||X||]^2-||E[X]||^2) + (E[||Y||]^2 - ||E[Y]||^2)
\end{align*}
Proof:
We have
\begin{align*}
&||X-Y||^2 = ||X||^2 + ||Y||^2 -2\sum_{i=1}^n X_iY_i\\
\implies & E[||X-Y||^2] = E[||X||^2] + E[||Y||^2] -2\sum_{i=1}^n E[X_i]E[Y_i] \quad (Eq. *)
\end{align*}
where we have used the fact that $X_i$ and $Y_i$ are uncorrelated for each $i\in \{1, \ldots, n\}$.
On the other hand, by Jensen's inequality with the convex function $||\cdot||$ we have
$$ E[||X-Y||] \geq ||E[X]-E[Y]|| \geq 0$$
and since $a\geq b \geq 0 \implies a^2 \geq b^2$ for any real numbers $a,b$, we have
\begin{align}
E[||X-Y||]^2 &\geq ||E[X]-E[Y]||^2 \\
&= ||E[X]||^2 + ||E[Y]||^2 - 2\sum_{i=1}^n E[X_i]E[Y_i] \quad (Eq. **)
\end{align}
Thus
\begin{align*}
Var(||X-Y||) &= E[||X-Y||^2] - E[||X-Y||]^2\\
&\leq  E[||X||^2] - ||E[X]||^2  + E[||Y||^2] - ||E[Y]||^2
\end{align*}
where the final inequality combines (Eq. *) and (Eq. **).
$\Box$
