How to show that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for a multivariate distribution? I am trying to prove that $\left|\left|E\left[\frac{x-X}{||x-X||}\right]\right|\right|\xrightarrow{||x||\to\infty}1$ for any multivariate distribution.
My Progress: I have proved the result for univariate distributions as the variable $\frac{x-X}{|x-X|}$ takes the value $1$ with probability $F(x)$ and $-1$ with probability $1-F(x)$, so the expectation becomes $2F(x)-1$ with converges to $\pm1$ as $|x|\to\infty$. But I couldn't generalize this to higher dimensions.

Approach:

*

*First, we approach through a sequence $\{x_n\}_{n\in\mathbb{N}}$ such that $||x_n||\to\infty$ as $n \to \infty.$ We fix an $x_n$ first.

*We take an $L_\epsilon$ such that $P(||X||>L_\epsilon) <\epsilon$, i.e., we first take the n-ball where 'most of the $X$' lies.

*Then, we will show that for all such 'good' values of $X$'s, $\frac{x_n-X}{||x_n-X||}$ is close to 'something' whose norm is 1.

*Next, the rest of the $X$'s have a minuscule probability of contributing, and corresponding values of the random variable are bounded.


Image:
First, it would be easier to look at this scratch image:

$O$ is the origin and $U$ is the unit hypersphere(circle in 2D). Denote the large circle by $C_\epsilon$ where 'most of the X lies'. Choose one $X_1$ from there. $Y$ is the vector corresponding to the $\frac{x_n-X}{||x_n-X||}.$ Denote $P_n$ as the $\frac{x_n}{||x_n||},$ i.e., the value corresponding to the $X=0.$
Carefully note that any point $X$ from inside the $C_\epsilon$ would have corresponding $\frac{x_n-X}{||x_n-X||}$ value inside the arc $B_1  P_n B_2.$ Let the angle $B_1OP_n$ is denoted by $\theta$ (which depends on both $\epsilon$ and $n$). We will show that this length ($YP_n$) is smaller than $B_1P_n$, which can be 'very small'.
The images are in 2D, but I think they are valid for any fixed finite-dimension, $p\in\mathbb{N}$ also.

Proof:

*

*Fix $\epsilon>0$ and get $L_\epsilon$ such that $P(||X||>L_\epsilon) <\epsilon$ assuming $X$ is $\mathbb{R}^p$ valued for a fixed $p\in\mathbb{N}.$


*If $||X||\le L_\epsilon,$ then writing $\sin\theta_{n,\epsilon}=L_\epsilon/||x_n||$ and $P_n=\frac{x_n}{||x_n||}$
$$||\frac{x_n-X}{||x_n-X||} - P_n||\le 2\sin(\theta_{n,\epsilon}/2) = \sqrt{2}\sqrt{1-\cos\theta_{n,\epsilon}} = \sqrt{2}\sqrt{1-\sqrt{1-L_\epsilon^2/||x_n||^2}}$$


*Then
\begin{align}
&E\left(||\frac{x_n-X}{||x_n-X||} - P_n||\right) \\ 
&=E\left(||\frac{x_n-X}{||x_n-X||} - P_n||\cdot I_{(||X||\le L_\epsilon)}\right) + E\left(||\frac{x_n-X}{||x_n-X||} - P_n||\cdot I_{(||X||> L_\epsilon)}\right) \\ 
&\le 2\sin(\theta_{n,\epsilon}/2) \cdot P(||X||\le L_\epsilon) + 2 \cdot P(||X||> L_\epsilon) \\ 
&\le 2\sin(\theta_{n,\epsilon}/2) + 2 \epsilon
\end{align}


*We want to show that, $E\left(||\frac{x_n-X}{||x_n-X||} - P_n||\right) \to 0$ as $n\to \infty$ so that $||E \frac{x_n-X}{||x_n-X||} - E P_n||\to 0$ and hence $||E \frac{x_n-X}{||x_n-X||} || \to 1$ as $n\to\infty$
($0\le |\ ||EW|| -||EV||\ | \le ||E(W-V)|| \le E||W-V||$)


*Now, for any fixed $\epsilon>0,$ $\sin(\theta_{n, \epsilon})\to 0$ as $n\to \infty.$ This concludes $\lim_{n\to \infty} E\left(||\frac{x_n-X}{||x_n-X||} - P_n||\right) \le \epsilon$ from which you can conclude your result.
The (almost) proof I gave looks a little complex - however, there are possibly easier proofs. There should be some more arguments to handle the
$X=x$ case indefiniteness as $P(X=x)\to 0$ as $||x||\to\infty$ assuming $X\in\mathbb{R}^p.$ There might be some argument problem; please mention if you find any.
