Anirban DasGupta writes:

Well done to Yixin Wang (Columbia University) and Vivian Meng (McGill University), who provided correct solutions to the various parts of the problem.

The problem asked was the following: suppose $X_1,X_2,\cdots ,X_n$ are iid $N(\mu ,{\sigma }^2)$ where the mean $\mu$ is some unknown positive integer, and $\sigma$ is a completely unknown standard deviation. We are to find the unique MLE of $\mu$ and ${\sigma }^2$, and show that the MLE of $\mu$ converges to the true $\mu$ exponentially fast and that the MLE of ${\sigma }^2$ is consistent.

Denote the likelihood function by $l(\mu ,\sigma )$. Then, directly, whenever $\overline{x}\ge \frac{1}{2}$, $$l(\mu +1,\sigma )\ge l(\mu ,\sigma )\iff \mu \le [\overline{x}–\frac{1}{2}],$$ where $[z]$ stands for the integer part of $z$. Thus, for $\overline{x}\ge \frac{1}{2}$, the unique MLE of $\mu$ is $1+[\overline{x}–\frac{1}{2}]=[\overline{x}+\frac{1}{2}]$; this is the same as the integer closest to the unrestricted MLE $\overline{X}$, a very intuitive result. If $\overline{X}<\frac{1}{2}$, $l(\mu ,\sigma )$ is monotone decreasing in $\mu$ over the set of positive integers, and in that case the MLE of $\mu$ is $1$. By a standard calculus argument, once we have found the MLE $\hat{\mu }$ of $\mu$, the MLE of ${\sigma }^2$ is $\widehat{{\sigma }^2}=\frac{1}{n}\sum _{i=1}^n(X_i-\hat{\mu }{)}^2$. To complete the solution, if the true $\mu >1$, $$P_{\mu }(\hat{\mu }=\mu )=P_{\mu }(\mu –\frac{1}{2}\le \overline{X}\le \mu +\frac{1}{2}).$$

The complementary probability, $P_{\mu }(\hat{\mu }\ne \mu )$ therefore is $2[1-\mathrm{\Phi }(\frac{\sqrt{n}}{2\sigma })]$, which is of the order of $\frac{e^{-\frac{n}{8{\sigma }^2}}}{\sqrt{n}}$.

This is a (somewhat faster than) exponential rate. The argument for the case $\mu =1$ is exactly similar; the expression just needs a very small modification. Obviously, therefore, by the Borel-Cantelli lemma, with probability one, for all large $n,\hat{\mu }=\mu$. This means, the MLE of ${\sigma }^2$ is in fact
even strongly consistent, because $\frac{1}{n}\sum _{i=1}^n(X_i–\mu {)}^2$ converges a.s. to ${\sigma }^2$ by the usual SLLN.