Normal with infinite variance I have a normal distribution with support $[a,b]$ $(a < 0 < b)$ with mean 0 and variance $\sigma^2$. I am wondering if $\sigma^2 \rightarrow \infty$, the normal distribution tends to a uniform $U(a,b)$. 
Any idea? 
I guess I can't suppose this support instead of $\mathbb{R}$. But graphically, as variance increases, the pdf turns flat. So, if the support is $\mathbb{R}$, I guess that pdf is zero for all $x \in \mathbb{R}$. That is why I think that with support $[a,b]$, maybe it converges to a uniform.   
 A: You can start by taking the ratio of the value at zero (the max of the distribution, if it's a truncated mean-zero Gaussian) to that of one of the ends, say $b$:
$$
\frac{f(0)}{f(b)} = \exp\left(b^2 / 2 \sigma^2 \right),
$$
where $f(x)$ is the truncated Normal distribution in question. In the limit, as $\sigma \rightarrow \infty$ we get,
$$
\lim_{\sigma \rightarrow \infty} \frac{f(0)}{f(b)} = 1.
$$
That is, the ratio of the peak of the distribution to the far-right end approaches $1.$ You can show this similarly for the ratio of the peak to $a$. If $f(a) = f(0) = f(b),$ and $f$ is non-increasing in either direction from $0$ (which is true if it's a mean-$0$ truncated Normal distribution with $a < 0 < b$), then $f$ is the same everywhere in $[a,b].$
If we assume, as well, that $f$ remains a probability distribution as $\sigma$ increases, which means that we properly truncate the denominator by the proper normalization factor that depends on $\sigma$, then, clearly, the distribution approaches $U(a,b).$
