*Student members: submit your solution by March 15, 2022. *

**Anirban DasGupta poses a statistical puzzle, looking at an easily understood Bayes problem that appears paradoxical at first glance, and it is hard to find a non-mathematical purely intuitive explanation for it. All PhD students in statistics most likely have seen parts of this specific problem in a standard course.**

(a) Suppose $X \sim \mbox{Bin}(n,p), 0 < p < 1$. Show that the estimator $\frac{X+1}{n+2}$ is the posterior mean of $p$ with respect to a unique Beta prior on $p$.

(b) Show that $\frac{X+1}{n+2}$ is the posterior mean of $p$ with respect to a class of infinitely many different prior distributions on $p.$ For the special case $n = 2$, exhibit two such priors different from a Beta prior.

(c) Let $X_1, \cdots , X_n$ be iid $N(\mu , 1), -\infty < \mu < \infty $. Suppose $\delta (X_1, \cdots , X_n)$ is a statistic. Prove that $\delta (X_1, \cdots , X_n)$ is either not a posterior mean of $\mu $ with respect to any prior distribution on $\mu $, or it is so with respect to one and only one prior distribution on $\mu $.

## Solution to previous puzzle

Anirban DasGupta explains:

A reminder of the puzzle is here. **Ye Tian** and **Harry Xi** (both of whom are PhD students in statistics at Columbia University) have provided good solutions to the previous puzzle. Congratulations to both of them. They prove that they both have power!

Columbia University |
Columbia University |

Take the case of $X \sim N(0,1)$ and denote the number of calculator hits required by $N$. We get $ P(N > n) = P(|X| > (1.0001)^{2^n})$. We then have

\[ E(N) = P(N > 0) + \sum_{n = 1} ^ \infty \,[P(|X| > (1.0001)^{2^n})] \]

\[ = 0.317262 + 3.44934 = 3.7666. \]

Similarly, for the standard Cauchy case, $E(N) = 6.6548$. $E(N)$ need not exist. Indeed, $E(N)$ does not exist if the tail of the distribution of $X$ is so heavy that $E[\log \log |X|]$ does not exist.

This problem could have been asked in a more general manner. There is nothing special about taking repeated square roots; we can ask the same question about taking, say, repeated cube roots. Have you ever played with pressing the cosine key repeatedly on your pocket calculator? Regardless of what number you start with, you always end up with the same number eventually. Why? You can formulate and solve a problem similar to the previous puzzle’s square root problem in this case too. And, there is nothing special about the cosine key; you can generalize that too. Every door in the palace of mathematics has treasures behind it. Just open the door…