# Does a bounded section of the normal distribution converge to the uniform distribution?

It was once asked on CrossValidated whether the normal distribution converges to a uniform distribution when the standard deviation grows to infinity. (The answer was no.) I am curious about a related yet slightly different question. Suppose I have an arbitrarily chosen fixed interval $$[A, B)$$. One can consider the uniform distribution on that interval, with density $$1/(B-A)$$ for $$A \le x < B$$. One can also consider the centered normal distribution $$\mathcal{N}(0, \sigma^2)$$, with pdf $$f(x)$$. But we are instead interested in a distribution defined by truncating this distribution to $$[A, B)$$, with pdf $$g(x) \propto f(x)$$ for $$A \le x < B$$ and 0 elsewhere. As $$\sigma \rightarrow \infty$$, does $$g(x)$$ converge to the uniform distribution with density $$1/(B-A)$$?

• Are you aware of the DCT? Note that by necessity $g_{\sigma}(x)=\frac{e^{\frac{-1}{2}(\frac{x}{\sigma})^{2}}}{\int\limits_{A}^{B}e^{\frac{-1}{2}(\frac{x}{\sigma})^{2}}}$ on the interval, so you just need an argument for interchanging the limit with the integral.
– FZan
Sep 17, 2023 at 2:21

Yes, even uniformly. Nothing fancy is needed, we just need to formalize the intuitive idea that as $$\sigma \to \infty$$ the Gaussian density on any fixed interval becomes closer and closer to constant.
On the interval $$[A, B]$$ the (unnormalized) Gaussian density $$f(x) = \exp \left( - \frac{x^2}{\sigma^2} \right)$$ is bounded from above by $$1$$ and bounded from below by $$\exp \left( - \frac{\text{max}(A^2, B^2)}{\sigma^2} \right)$$ (this slightly awkward expression is needed to handle the case that $$A$$ is negative and $$B$$ is positive), and as $$\sigma \to \infty$$ the lower bound converges to $$1$$. This gives that the normalized density is bounded from above and below by
$$\frac{\exp \left( - \frac{\text{max}(A^2, B^2)}{\sigma^2} \right)}{B - A} \le g(x) = \frac{\exp \left( - \frac{x^2}{\sigma^2} \right)}{\int_A^B \exp \left( - \frac{x^2}{\sigma^2} \right) } \le \frac{1}{(B - A) \exp \left( - \frac{\text{max}(A^2, B^2)}{\sigma^2} \right)}$$
so as $$\sigma \to \infty$$ we see that $$g(x)$$ converges uniformly to $$\frac{1}{B - A}$$ as desired.
This argument shows that $$A$$ and $$B$$ don't even need to be fixed and can grow slowly (sublinearly) with $$\sigma$$, e.g. we could have $$A = -B, B = O(\sqrt{\sigma})$$.