Anirban DasGupta says, “We are going to continue with our contest model introduced in the previous puzzle. Each correct answer receives 3 points, each incorrect answer receives -2 points, and each item left unanswered receives -1 point. The top three scorers will be recognized. You can answer just one of the two problems, 50.1 and 50.2, although answering both will be really nice.”

Puzzle 50.1 Suppose you have n i.i.d. observations from a one-dimensional normal distribution with mean μ, −∞ < μ < ∞, and a known variance. Let G be any proper prior on μ and δ_G(X¯ ) the posterior mean of μ. Prove, or disprove by providing a counterexample, that there exists a value of X¯ such that δ_G(X¯ ) < −10⁶ X¯.

Puzzle 50.2 For our contest problem, answer True or False, without the need to provide a proof. But reasoned answers are especially welcome. Here are the items.

(a) The sample median is an inadmissible estimator of a one-dimensional normal mean μ if the loss function is (μ−a)⁴.

(b) If X₁,X₂, …,X_n are i.i.d. standard exponential, then lim sup [E(max(X₁,X₂, …,X_n)) − log n] < .50.

(c) If F(x) is the CDF of an absolutely continuous distribution on the real line with a completely unknown density f(x), then it is possible to construct an unbiased estimate of f provided the sample size is 5 or more.

(d) If Z has a standard normal distribution, then one can find independent random variables X and Y, each possessing a density, such that Z = X Y in law.

(e) If F(x) is the CDF of an absolutely continuous distribution on the real line with a continuously differentiable density f(x), and |f′(x)| ≤ 100 for all x, then f is bounded above by 20.

(f) The Jeffrey prior on a multinomial probability vector (p₁, …,p_k), where p_k = 1 −∑^k−1_i=1 p_i, is improper if k ≥ 3.

(g) If X has a Poisson distribution with mean one, then E(X^X) < ∞.

(h) Suppose X, Y each has a finite variance, that E(X |Y ) = Y and E(Y |X) = 0. Then Y = 0 with probability one.

(i) Pick a permutation of {1, 2, …, n} at random. Then the probability that the cycle containing 1 has length k is the same for all k.

Solution to Puzzle 49

Well done Debanjan Bhattacharjee (ISI Delhi), and Bilol Banerjee (ISI Kolkata) for sending in correct solutions!

Puzzle 49.1

Denote the normal observations as $X_1, \cdots , X_n$ and the Cauchy observations as $Y_1, \cdots , Y_n$. Then, denoting the minimum and the maximum of $X_1, \cdots , X_n$ as $X_{(1)}, X_{(n)}$ respectively, $P(X_{(1)} \leq Y_1 \leq X_{(n)}) = E[\arctan (X_{(n)}) – \arctan (X_{(1)})]$ \[ = 2\,E[\arctan (X_{(n)})] = 2\,n\,\int \arctan(x)\, \Phi(x)^{n-1}\,\phi(x)\,dx = \theta_n, \] and $\mu_n = n\,\theta_n$. In particular, $\mu_2 = .493, \mu_5 = 2.46, \mu_{10} = 6.031, \mu_{30} = 20.951, \mu_{100} = 75.307$, etc. $\mu_n$ of course converges to $\infty $ as $n \to \infty$. A better question is how many of the Cauchy values do not fall within the convex hull of the normal values. The asymptotic behavior of $n-\mu_n$ can be found by an asymptotic expansion of the arctan function “at infinity”. You will get $n-\mu_n \, \sim \, 2\,n/(\pi \,\sqrt{2\,\log n})$. For $n = 500$, the true value of $n-\mu_n$ is 102, and the theoretical order above predicts it to be a little less than 91.

Puzzle 49.2

We can safely classify $-50.64, -6.41, 6.65, 12.42$ as not coming from a standard normal distribution. So we have to identify the remaining three Cauchy values among
$-1.39, -0.72, -0.70, -0.16, -0.11, 0.24, 0.92, 1.01, 1.17, 1.75$.
The three largest likelihood ratios in favor of Cauchy for these observations are for $1.75, -0.16, -0.11$. Quite interestingly, $-0.16, -0.11$ are indeed Cauchy observations in our simulated set, though the other Cauchy observation is $-1.39$, not $1.75$. One can also compute the likelihood ratio for each of the three-element subsets of the above 10 values, and select that subset with the largest likelihood ratio. There are many other options as well.