Dec 14, 2018

Student Puzzle Corner 23

Bulletin Editor Anirban DasGupta sets this problem. Student members of the IMS are invited to submit solutions (to 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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