**Guest puzzler Stanislav Volkov, Centre for Mathematical Sciences at Lund University, sets a puzzle about a speeding random walk:**

For *n* = 1, 2, … let *X _{n}* = ±

*n*with equal probabilities and assume that

*X*s are independent. Define a “speeding” random walk by

_{n}*S*

_{0}= 0 and

*S*=

_{n}*X*

_{1}+

*X*

_{2}+ … +

*X*(note that if

_{n}*X*were ±1 then it would be the ordinary, simple, symmetric random walk). A possible path for this random walk, starting from 0, would be 0, 1, 3, 0, 4, 9, 15, 8, ….

_{n}**The Question:**

Is the random walk *S _{n}* recurrent or transient (in the usual sense)?

*Send us your solution by March 15, 2022.*

## Solution to Puzzle 36

Problem Corner Editor Anirban DasGupta writes on the previous puzzle:

For part (a), trivially, $\frac{X+1}{n+2}$ is the posterior mean of $p$ against the $U[0,1]$ prior, and this is the only Beta prior for which $\frac{X+1}{n+2}$ is the posterior mean.

Recall now that for a general prior distribution $G$, the posterior mean of $p$ equals

\[ \delta (x) = \frac{\int p^{x+1}\,(1-p)^{n-x}\,dG(p)}{

\int p^{x}\,(1-p)^{n-x}\,dG(p)}, \]

and hence depends only on the first $n+1$ moments of $G$. Hence, for any prior distribution $G$ whose first three moments are equal to those of the $U[0,1]$ distribution produces the posterior mean $\frac{X+1}{n+2}$ for $p$. Simple examples are finitely supported priors $G$, with support on four suitable points. This handles part (b).

For part (c), first note that if we denote the marginal density of $\bar{X}$ with respect to a given prior $G$ on $\mu $ as $m(\bar{X}) = m_G(\bar{X})$, then the posterior mean admits the formula

\[ \delta (\bar{x}) = \bar{x} + \frac{1}{n}\, \frac{m'(\bar{x})}{m(\bar{x})}. \]

Thus, if a given estimator $\delta(\bar{x})$ is the posterior mean against two different priors $G, H$, then one gets that

\[ \frac{m_G'(\bar{x})}{m_G(\bar{x})} = \frac{m_H'(\bar{x})}{m_H(\bar{x})}, \]

which would give that for some positive constant $c, m_H(\bar{x}) =c\,m_G(\bar{x})$, and $c$ must be equal to $1$. Now, note that since $m_G, m_H$ are Gaussian convolution densities, if $m_H = m_G$, then $G = H$ as measures. This handles part (c).