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Setup

Let random variables $X_1,\ldots, X_n$ be i.i.d. $N(0, \sigma^2)$. Suppose that the observed value $x^2_i$ of $X^2_i$ is given.

What I think

If $$ \frac{\max_{i=1,\ldots, n}{x^2_i}}{\sum_{i=1}^n x^2_i} \to 0, $$ then $$ \frac{\sqrt{n}\bar{X}}{\sqrt{\frac{1}{n}\sum_{i=1}^n x^2_i}}= \frac{\sum_{i=1}^n |x_i| Y_i}{\sqrt{\sum_{i=1}^n x^2_i}} \overset{L}{\to} N(0, 1) $$ holds where $Y_i = \pm 1$ with probability 1/2 each.

Problem

What is the difference between $$ \frac{\max_{i=1,\ldots, n}{X^2_i}}{\sum_{i=1}^n X^2_i} \overset{P}{\to} 0 $$ and $$ \frac{\max_{i=1,\ldots, n}{x^2_i}}{\sum_{i=1}^n x^2_i} \to 0 $$

Background

I could show convergence in probability, but I really want to show convergence below. I hope these are equivalent, but I don't understand the relationship between the two limits. can I just replace $X^2_i$ and $x^2_i$ in a straightforward way?

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    $\begingroup$ The usual setup of probability theory/statistics doesn't usually have the notion of "observed values," insofar as random variables remain random variables (i.e. maps from a probability space into $\mathbb R$) and never get transmuted into constants (i.e. values of $\mathbb R$). Can you say exactly how $x_i$ and $X_i$ are related? Are $x_i$ just some sequence of real numbers? $\endgroup$
    – daisies
    Feb 26 at 3:36
  • $\begingroup$ en.m.wikipedia.org/wiki/Realization_(probability) $\endgroup$
    – ytnb
    Feb 26 at 3:46
  • $\begingroup$ We now consider $X^2_i= x^2_i$ are given. $\endgroup$
    – ytnb
    Feb 26 at 3:47
  • $\begingroup$ I see; if you mean $x_i = X_i(\omega)$ for some specific $\omega$ in the probability space, then you can't relate any properties of $X_i$ to $x_i$, since the $x_i$'s only give you information on a set of probability zero. In particular $x_i$ doesn't convey any meaningful information about $X_i$. $\endgroup$
    – daisies
    Feb 26 at 4:09
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    $\begingroup$ If all the $x_i^2$ are given then the question of whether $\lim\limits_{n \to \infty}\frac{\max_{i=1,\ldots, n}{x^2_i}}{\sum_{i=1}^n x^2_i}$ exists and what it is, is a fact depending on the $x_i$s rather than the $X_i$s, though you do need to check them all (countably infinitely many of them) as for any $n$ there was a positive probability that $X_n$ was big enough to substantially shift the quotient and so $x_n$ may do. That is hard when you do not have an expression for the $x_i$s but just a list of values. $\endgroup$
    – Henry
    Feb 26 at 13:28

1 Answer 1

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I asked a similar question a while ago on MO: https://mathoverflow.net/questions/459782/definition-of-weak-conditional-convergence-of-random-variables


We want to show that $$Z_n := \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i Y_i}{\sqrt{\frac 1n\sum_{i=1}^n X_i^2}}$$ converges in distribution, conditional on $X_1, X_2, ..., X_n$, to random variable with distribution $\mathcal N(0, 1)$ as $n\rightarrow\infty$.

Let $(\Omega,\mathcal F, P)$ denote the underyling probability space, and assume that

  • $(X_n)_{n\in\mathbb N}$ is a sequence of iid random variables with finite second moments
  • $(Y_n)_{n\in\mathbb N}$ is a sequence of iid random variables with mean zero and unit variance.

Let $F_Y\big(\cdot\mid X_1(\omega), X_2(\omega), \dots, X_n(\omega)\big)$ denote the regular conditional cumulative distribution function of (each) $Y_i$ given that $x_1 := X_1(\omega), x_2 := X_2(\omega), \dots, x_n := X_n(\omega)$; note that $x_1,x_2,\dots,x_n\in\mathbb R$. Furthermore, let $\Phi$ denote the cumulative distribution function of a random variable with probability distribution $\mathcal N(0, 1)$.


We say that the sequence of random variables $(Z_n)_{n\in\mathbb N}$ converges in distribution to a random variable with distribution $\mathcal N(0, 1)$ iff

$$P\left(\omega\in\Omega : \lim_{n\rightarrow\infty}\sup_{z\in\mathbb R}\bigg\vert \Phi(z) - F_Y\big(z\mid X_1(\omega), X_2(\omega), \dots, X_n(\omega)\big)\bigg\vert = 0\right) = 1,$$

i.e., for almost all $\omega\in\Omega$ it holds that

$$\frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i(\omega) Y_i}{\sqrt{\frac 1n\sum_{i=1}^n X_i(\omega)^2}} = \frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n x_i Y_i}{\sqrt{\frac 1n\sum_{i=1}^n x_i^2}}$$ converges in distribution to a random variable with distribution $\mathcal N(0,1)$ as $n\rightarrow\infty$.


Let's fix some $\omega\in\Omega$ and assume that $$\lim_{n\rightarrow\infty}\max_{k=1}^n\frac{X_k(\omega)^2}{\sum_{i=1}^n X_i(\omega)^2} = 0.$$ Then, by Lindeberg Lévy CLT, it follows that $$\frac{\frac{1}{\sqrt{n}}\sum_{i=1}^n X_i(\omega) Y_i}{\sqrt{\frac 1n\sum_{i=1}^n X_i(\omega)^2}}$$ converges in distribution to a random variable with distribution $\mathcal N(0,1)$ as $n\rightarrow\infty$.


So, in order to have conditional convergence, it remains to show that the condition $$\lim_{n\rightarrow\infty}\max_{k=1}^n\frac{X_k(\omega)^2}{\sum_{i=1}^n X_i(\omega)^2} = 0$$ holds for almost all $\omega\in\Omega$, i.e., $$P\left(\omega\in\Omega : \lim_{n\rightarrow\infty}\max_{k=1}^n\frac{X_k(\omega)^2}{\sum_{i=1}^n X_i(\omega)^2} = 0\right) = 1.$$ If you wish, you could write $x_i$ instead of $X_i(\omega)$, but I believe emphasizing the dependence on $\omega$ is important and makes it more clear what is going on.


Finally, note that this condition is stronger than the condition you state: $$\lim_{n\rightarrow\infty}P\left(\omega\in\Omega : \max_{k=1}^n\frac{X_k(\omega)^2}{\sum_{i=1}^n X_i(\omega)^2} \geq \epsilon\right) = 0$$ for all $\epsilon>0$. I believe that this is not enough as we need the limit inside the probability.

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