# Expectation of Spherically Symmetric Random Vector

Suppose we are considering a vector valued random variable $$X \in \mathbb{R}^d$$ that is "spherically symmetric" in the sense that $$E \frac{X}{\|X\| } = 0.$$ Suppose $$a_n \in \mathbb{R}$$ is a deterministic sequence tending to zero. What I want to know is: what additional conditions (if any) are needed to imply that, for any fixed nonzero vector $$u$$, $$E \frac{X}{\|X - a_n u\| } \to 0 ?$$ And $$E \frac{\|X\|^2}{\|X - a_n u\|^2} \mbox{ is bounded}?$$ Intuitively it seems these should follow from some "dominated convergence" condition since as $$a_n \to 0$$, $$\frac{X}{\|X - a_n u\| } \to X/\|X\|$$ almost surely, and $$\| X/\|X\| \| =1$$. Any help is much appreciated!

• Does the law for the random vector have a density? (If so, there is hope) May 1, 2022 at 9:02

Pick $$X$$ such that for $$n \in \{1, 2, 3, \ldots \}$$ $$\mathbb P \left (X = \frac 1n + \frac1{n^4} \right) = \frac 1{2 \zeta (2) n^2}$$ $$\mathbb P \left (X = -\frac 1n -\frac1{n^4} \right) = \frac 1{2 \zeta (2) n^2}$$ It should be clear that $$\mathbb E \left[ \frac{X}{|X|} \right]= 0$$ Now pick $$a_n = \frac 1n$$ and $$u=1$$. And note that since $$X \ne a_n$$ almost surely then $$\left | \frac{X-a_n}{|X-a_n|} \right| = 1$$ and hence by the dominated convegence theorem $$\mathbb \lim_{n\to \infty} \mathbb E \left [ \frac{X-a_n}{|X-a_n|} \right] = \mathbb E \left [ \frac{X}{|X|} \right] = 0.$$ This means that if the limit exists
$$\mathbb \lim_{n\to \infty} \mathbb E \left [ \frac{X}{|X-a_n|} \right] = \mathbb \lim_{n\to \infty} \mathbb E \left [ \frac{X-a_n}{|X-a_n|} \right]+\mathbb E \left [\frac{a_n}{|X-a_n|} \right]=\lim_{n\to \infty} \mathbb E \left [ \frac{a_n}{|X-a_n|} \right]$$ However note that $$E \left [ \frac{a_n}{|X-a_n|} \right] \ge \frac{a_n}{|\frac1n + \frac 1{n^4}|}\frac 1{2 \zeta (2) n^2} = \frac{n}{2\zeta(2)}$$ So the limit doesn't exist. You need some additional criterion for convergence to be guaranteed. One is that $$X$$ is bounded below. Then if $$\left\lVert X \right\rVert \ge M$$ for some $$M>0$$, then we can pick $$N$$ such that $$\left \lVert a_nu\right \rVert \le \frac 12 M$$ for $$n \ge N$$ and so $$\left\lVert \frac{a_nu}{\left\lVert X-a_nu \right\rVert} \right\rVert \le \frac {2\left \lVert a_n u\right \rVert }{M}$$ and so $$\frac{a_nu}{\left\lVert X-a_nu \right\rVert} \to 0$$ almost surely, giving you that $$\lim_{n \to \infty}\mathbb E \left [\frac{X}{\left\lVert X-a_nu \right\rVert}\right] = \lim_{n \to \infty}\mathbb E \left [\frac{X-a_nu}{\left\lVert X-a_nu \right\rVert}\right]+\lim_{n \to \infty}\mathbb E \left [\frac{a_nu}{\left\lVert X-a_n u\right\rVert}\right] = \mathbb E \left [\frac{X}{\left\lVert X \right\rVert}\right] =0$$

Edit:

Inspired by H. H. Rugh's comment here is another criterion:

$$X$$ has a density function $$f$$ that is bounded in some neighbourhood of the origin.

So to see why this works note that like before, if we prove that $$\frac{a_n u}{\left \lVert X- a_nu \right \rVert} \to 0$$, then we are done. This is clearly true along the subsequence where $$a_n = 0$$. Let $$b_n$$ be the subsequence where $$a_n >0$$ and $$c_n$$ be the subsequence where $$a_n <0$$. If we prove that the limit is $$0$$ along both of theses subsequences then we are done. Let $$Y_n = \frac{a_n}{\left \lVert X- a_nu \right \rVert}$$ then for $$y>0$$ $$1-F_{Y_n}(y) = \mathbb P(Y_n >y ) = \int_{ \frac{a_n}{\left \lVert x- a_nu \right \rVert}>y} f(x) dx = \int_{\left \lVert \frac{x}{a_n}- u \right \rVert < \frac 1y} f(x) dx = \int_{\left \lVert z- u \right \rVert < \frac 1y} a_nf(a_nz) dz$$ This obviously converges to $$0$$ when $$f$$ is bounded in some neighbourhood of the origin. So for $$y > 0$$ $$\lim_{n \to \infty} F_{Y_n} (y) = 1$$

It is also clear that for $$y<0$$ $$\lim_{n \to \infty} F_{Y_n} (y) = 0$$

Hence $$F_{Y_n} \to F_{0}$$ in the continuity set of $$F_0$$ hence hence $$Y_n \to 0$$ in distribution and hence $$\frac{a_n u}{\left \lVert X- a_nu \right \rVert} \to 0$$ along the subsequence where $$a_n >0$$. The subsequence where $$a_n<0$$ is similar. Hence we can see $$\frac{a_n u}{\left \lVert X- a_nu \right \rVert} \to 0$$ in general and we have our result.