**Bulletin Editor Anirban DasGupta sets this problem. Student members of the IMS are invited to submit solutions (to bulletin@imstat.org with subject “Student Puzzle Corner”). The deadline is January 25, 2019.**

Anirban DasGupta says:

*The previous problem on inference based on the distribution of a nonsufficient statistic required the use of both Markov Chain theory and statistical inference [see solution below]. It was a problem on probability and statistics simultaneously. This month we pose a rather simple problem which should be fun to think about, and has many possible answers! So, hopefully, many of you will think of one of the correct answers.*

Let $C(\mu , 1)$ denote the Cauchy distribution on the real line with location parameter $\mu $ and scale parameter equal to one. Suppose $\mu $ belongs to $\mathcal{R}$ (the parameter space) and that we wish to estimate it under squared error loss function. Let $X_1, X_2, \cdots $ be an iid $C(\mu , 1)$ sequence. Assume that $n > 7$. Give, with proof, a sequence of estimators $T_n(X_1, X_2, \cdots , X_n)$ of $\mu $, such that:

(a) For every $n$, $T_n$ is inadmissible;

(b) For no $n$, $T_n$ is minimax;

(c) For every $n$, $T_n$ is unbiased;

(d) The sequence of estimators ${T_n}$ is asymptotically efficient.

(e) Compute the numerical value of the estimator you have proposed for the following data values:

0.1, 2.9, −0.6, 3.1, 3.6, −6.5, 0.2, 1.0, 2.4, −15.9.

## Solution to Puzzle 22

Embed the longest run problem into a stationary Markov chain with the following transition matrix. Denote the observed longest head run in $n$ tosses of a $p$-coin by $L_n$ and suppose we wish to find $P(L_n \geq m), m$ a general nonnegative integer. You go to state zero from state $i$ with probability $1-p$ and go to state $i+1$ from state $i$ with probability $p$, with state $m$ as an absorbing state. Denote this $(m+1) \times (m+1)$ matrix by $P_m$ and let $Q[n,m]$ denote its $n$th power.

Then $P(L_n \geq m)$ is the last element in the zero-th row of $Q[n,m]$.

By evaluating $P(L_n \geq 1) – P(L_n \geq 2)$ with $n = 10$, one gets

\[ P(L_n = 1) = 10p-54p^2+128p^3-189p^4+216p^5-205p^6+144p^7-63p^8+14p^9-p^{10}. \]

It is uniquely maximized at $p \approx .1616$, which is the value of the MLE of $p$ based on $L_n$ alone.

A moment estimate is easily found by inverting the expectation formula

\[ E(L_n) \approx \frac{\log n}{\log \frac{1}{p}} – \frac{\log (1-p)}{\log p}. \]

An approximate solution is

\[ \hat{p} = n^{-1/L_n}. \]

This estimate will have a fairly serious bias problem. However, with work, we can derive (a high order) asymptotic expansion for the bias of $\hat{p}$. Hence, we can correct $\hat{p}$ for its bias, at least to the first order. These are very classic ideas in large sample theory of inference.

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