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

For n = 1, 2, … let Xn = ±n with equal probabilities and assume that Xns are independent. Define a “speeding” random walk by S0 = 0 and Sn = X1 + X2 + … + Xn (note that if Xn 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, ….

The Question:
Is the random walk Sn 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).