# Solution to Student Puzzle Corner 8

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.

Yixin Wang | Vivian Meng |

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 $\bar{x} \geq \frac{1}{2}$, \[ l(\mu + 1, \sigma ) \geq l(\mu , \sigma ) \Leftrightarrow \mu \leq [\bar{x} – \frac{1}{2}], \] where $[z]$ stands for the integer part of $z$. Thus, for $\bar{x} \geq \frac{1}{2}$, the unique MLE of $\mu $ is $1+[\bar{x} – \frac{1}{2}] = [\bar{x} + \frac{1}{2}]$; this is the same as the integer closest to the unrestricted MLE $\bar{X}$, a very intuitive result. If $\bar{X} < \frac{1}{2}$, $l(\mu , \sigma )$ 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 $\hat{\sigma ^2} = \frac{1}{n}\sum_{i = 1}^n (X_i - \hat{\mu })^2$. To complete the solution, if the true $\mu > 1$, \[ P_\mu (\hat{\mu } = \mu ) = P_\mu (\mu – \frac{1}{2} \leq \bar{X} \leq \mu + \frac{1}{2}).\]

The complementary probability, $P_\mu (\hat{\mu } \neq \mu )$ therefore is $2[1-\Phi (\frac{\sqrt{n}}{2\sigma})]$, 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.

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