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.


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

  • $\begingroup$ 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


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.


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