0
$\begingroup$

Let $X_i \sim \mathcal N(0, I_{d(n)})$ be iid normal random vectors taking values in $\mathbb R^{d(n)}$ where ${d(n)}$ is a sequence (dependent on a positive integer $n$) that grows such that ${d(n)}/n \rightarrow \gamma >0$ as $n\rightarrow \infty$ for some positive constant $\gamma \in (0,1)$. Consider the following quantity:

$$Y_n = \left\|\frac{1}{\sqrt n} \sum_i^n X_i\right \|_2^2$$

Does $Y_n$ have a limiting distribution? If so, what is it?

What I've tried

Rewriting, we have:

$$Y_n = \left\|\sqrt n\,\frac{1}{n} \sum_i^n X_i\right \|_2^2 = \left\|\sqrt n\, \bar X\right \|_2^2$$

where $\bar X \sim \mathcal N\left(0, I_{d(n)}/n\right)$ so that $Y_n$ is clearly chi squared distributed with $d(n)$ degrees of freedom. The chi squared distribution does not have a limiting distribution as the number of degrees of freedom increases to $\infty$. This shows that $Y_n$ does not converge in distribution.

The homework problem suggests to apply the Paley-Zygmund inequality, but I don't understand where that might be relevant. Any hints would be appreciated.

Paley-Zygmund:

$$\mathbb P[X\ge\theta \mathbb EX]\ge (1-\theta)^2\frac{\mathbb E[X]^2}{\mathbb E[X^2]}$$

$\endgroup$
4
  • $\begingroup$ Hint: Have you computed the mean and variance of $Y_n$ and then plug into the bound? $\endgroup$
    – E-A
    Mar 9, 2021 at 21:22
  • $\begingroup$ @E-A $P(Y_n \geq \theta d) \geq d^2/(2d+d^2) \rightarrow 0$ but I don't see how this lower bound tells me anything about the convergence in distribution. $\endgroup$
    – dmh
    Mar 9, 2021 at 21:29
  • 1
    $\begingroup$ That limit -assuming your expression is right- is 1/2, not 1? The point is that you can show that this violates tightness, i.e. you have some part of your measure escaping to infinity. $\endgroup$
    – E-A
    Mar 9, 2021 at 22:32
  • $\begingroup$ Oh that's great, I messed up that limit calculation. I now see how it violates tightness. $\endgroup$
    – dmh
    Mar 9, 2021 at 23:00

1 Answer 1

0
$\begingroup$

As pointed out in the comments, the lower bound from the Paley-Zygmund inequality allows us to show that the limit distribution is not tight, and so is not actually a probability measure.

Evaluating the terms in the expression:

$$\mathbb P[Y_n\ge\theta d]\ge (1-\theta)^2\frac{d^2}{2d + d^2} \rightarrow 1/2$$

and so the limit distribution of $Y_n$ is not tight and so does not exist.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.