4
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ Does the law for the random vector have a density? (If so, there is hope) $\endgroup$
    – H. H. Rugh
    Commented May 1, 2022 at 9:02

1 Answer 1

4
+50
$\begingroup$

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .