# The difference between convergence in probability and convergence in observed value

### 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?

• 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? Feb 26 at 3:36
• en.m.wikipedia.org/wiki/Realization_(probability)
– ytnb
Feb 26 at 3:46
• We now consider $X^2_i= x^2_i$ are given.
– ytnb
Feb 26 at 3:47
• 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 $X_i$. Feb 26 at 4:09
• 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. Feb 26 at 13:28

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.