Mar 31, 2015

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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