As the variance of a random variable grows, the conditional distribution of it residing in an interval of length $1$ becomes uniform 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?
 A: 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 $[0,1]$.
A: [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.
