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_ns are independent. Define a “speeding” random walk by S₀ = 0 and S_n = X₁ + X₂ + … + X_n (note that if X_n 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 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 $\overline{X}$ with respect to a given prior $G$ on $\mu$ as $m(\overline{X})=m_G(\overline{X})$, then the posterior mean admits the formula
$$\delta (\overline{x})=\overline{x}+\frac{1}{n}\frac{m^{\prime }(\overline{x})}{m(\overline{x})}.$$
Thus, if a given estimator $\delta (\overline{x})$ is the posterior mean against two different priors $G,H$, then one gets that
$$\frac{m_G^{\prime }(\overline{x})}{m_G(\overline{x})}=\frac{m_H^{\prime }(\overline{x})}{m_H(\overline{x})},$$
which would give that for some positive constant $c,m_H(\overline{x})=c{m}_G(\overline{x})$, 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).