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Let's say a random variable is supported on a semi-infinite interval (say $(0, \infty)$ or all real numbers). We take a finite interval within the support. We then consider the distribution of this random variable conditional on it lying within the finite interval. Without loss of generality, we can even require that the finite interval be of size $1$.

It seems clear that as we increase the variance of the random variable, the part of the distribution within the finite interval will "flatten out" and in the limit, it should approach a uniform distribution over said interval. Is there any situation where this conjecture might be violated? And if not, is there a way to prove this in general (for any random variable supported on all real numbers or a semi-infinite interval)?

Context: this will help prove the conjecture in this question: Going "well into the lifetime" of a renewal process means the time until the next event will be uniform conditional on inter-arrival?

Rohit Pandey
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  • "Increase the variance" is a bit vague, since there are many ways to do this. If the mechanism is specifically $\alpha X$ as $\alpha \to \infty$, then I think your claim holds. But as crankk points out, there are many other sequences of random variables with increasing variance, and some do not satisfy your claim. You need to be more specific about how exactly you want to "increase the variance." – angryavian May 31 '21 at 06:53
  • Got it. In the specific instance I had in mind, more and more i.i.d random variables are added to increase the variance. So, I think this corresponds to your $\alpha X$ example. Would really appreciate a proof or outline of one for your claim. – Rohit Pandey May 31 '21 at 06:56
  • I modified the example below such that it is of the form $\alpha X$. – crankk May 31 '21 at 07:02
  • @crankk - what if I have i.i.d random variables, $\eta_i$ with mean $0$ and some finite variance. Now, I sum them $\sum\limits_{i=1}^{n} \eta_i$. I increase the variance by summing more and more of them (let $n \to \infty$). Can you think of a counter-example for this case? – Rohit Pandey May 31 '21 at 07:04

2 Answers2

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What means 'increasing the variance'? If there are no restrictions to that, than there are plenty of counter examples, e.g. let $f(x) =\left(1-\sin\left(\frac{1}{x}\right)\right)1_{[0,1)}(x)+\frac{1}{x^2}\left(1-\sin\left(\frac{1}{x}\right)\right)1_{[1,\infty)}(x)$ and $g= \frac{f}{\int f}$. Let $X$ be a random variable with density $g$. We consider $\alpha X$. Sending $\alpha\to \infty$ increases the variance, however, it will not approach a uniform distribution on $[0,1]$.

crankk
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  • Haven't completely understood this, but as you increase $\alpha$, the interval size increases with it, right? I need the interval size to stay constant. – Rohit Pandey May 31 '21 at 06:47
  • @RohitPandey The different values of $\alpha$ correspond to different random variables $X_\alpha$. As $\alpha \to \infty$, the variance of $X_\alpha$ increases. But on any interval, the conditional distribution stays as a fixed sine function for large enough $\alpha$. – angryavian May 31 '21 at 06:52
  • I moidified it such that the density is non zero on all the positive numbers for any $\alpha$. – crankk May 31 '21 at 06:53
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    @crankk I think OP wants to consider $f\mathbf{1}_{[a,b]} / \int_a^b f$ for some fixed interval $[a,b]$. – angryavian May 31 '21 at 07:14
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[Partial answer]

Let $S_n = \sum_{i=1}^n \eta_i$ where $\eta_i$ are i.i.d. with mean zero and finite variance. WLOG I will assume the variance is $1$.

By the CLT, $n^{-1/2} S_n$ converges in distribution to $N(0, 1)$.

Given some interval $[a,b]$, you are asking for $P(S_n \le t \mid S_n \in [a,b]) = \frac{P(S_n \in [a,t])}{P(S_n \in [a,b])}$ for $t \in [a,b]$.

When $n$ is large enough, $P(S_n \in [u,v]) \approx \Phi(v/\sqrt{n}) - \Phi(u/\sqrt{n})$ where $\Phi$ is the CDF of the standard normal distribution. So the desired conditional probability is closed to $$\frac{\Phi(t/\sqrt{n}) - \Phi(a/\sqrt{n})}{\Phi(b/\sqrt{n}) - \Phi(a/\sqrt{n})}.$$

This is the ratio of two areas under the normal PDF over intervals that are shrinking to zero. My intuition is that as the length of these intervals shrink to zero, the PDF gets flatter and flatter to the point that the ratio is approximately $\frac{(t-a)/\sqrt{n}}{(b-a)/\sqrt{n}} = \frac{t-a}{b-a}$ which would yield the CDF of the uniform distribution on $[a,b]$. I'm not sure how to formalize this intuition at the moment.

angryavian
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  • My intuition is that as the length of these intervals shrink to zero, the PDF gets flatter and flatter....

    Isn't this true for just about any curve (not just Gaussian CDF)? If you zoom in enough, you get a line.

     – Rohit Pandey May 31 '21 at 07:48
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    @RohitPandey For a differentiable function, that is true, but being locally linear is not enough here; you also need the slope to be zero, which is true of the normal PDF at zero. – angryavian May 31 '21 at 08:25
  • No matter how far $a$ and $b$ are from zero, as the variance grows, they're not that far from zero after all. – Rohit Pandey May 31 '21 at 16:04
  • @RohitPandey Yes, the two intervals $[a/\sqrt{n}, t/\sqrt{n}]$ $[a/\sqrt{n}, b/\sqrt{n}]$ get close to zero as $n \to \infty$, so the PDF should be approximately flat. But I did not explicitly prove this intuition [that the ratio of the two integrals is $\frac{t-a}{b-a}$] rigorously. – angryavian May 31 '21 at 17:29