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

1. $$\operatorname{Var}(||X - Y||) \leq \operatorname{Var}(||X||) + \operatorname{Var}(||Y||) = 2 \operatorname{Var}(||X||)$$
2. The equality is strict if $$X$$ and $$Y$$ are independent.

I'm also wondering if there is a formula in the more general case.

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. **). $$\Box$$

• Thank you! Are there any "common" classes for X where E[||X||]^2 = ||E[X]||^2? I'm wondering for example, if X is Gaussian this may be the case Commented Jul 2, 2020 at 20:14
• After joting down notes, E[||X||]^2 - ||E[X]||^2 = Sum_i(Var[X_i]). So they would be equal if every Var(X_i) = 0 ie X always takes the same value Commented Jul 2, 2020 at 20:28
• I cannot get your variance equality. However since $f:\mathbb{R}^n\rightarrow\mathbb{R}$ given by $f(x) = ||x||$ is a strictly convex function, it holds that $E[f(X)]\geq f(E[X])$ with equality if and only if $P[X=E[X]]=1$. Commented Jul 2, 2020 at 21:50