Anirban DasGupta says this problem seems impossible at first sight, but thinks you will probably arrive at a solution quickly and it might surprise some of you that this is possible. This is the kind of thing statistics students used to learn routinely forty or fifty years ago. The problem is extremely easy to state:

(a) X is a single observation from N(μ, σ²), where μ, σ are both completely unknown, i.e., −∞ < μ < ∞, σ > 0. It is emphasized that the sample size in our experiment is n = 1. Explicitly find and plot a joint confidence region for (μ, σ) that has a coverage probability constantly equal to 0.95.

(b) Now suppose you have a sample of size n = 2. Derive and plot the corresponding joint confidence region for (μ, σ) that has a coverage probability constantly equal to 0.95, and find its expected area.

Solution to Puzzle 33

Student Puzzle Editor Anirban DasGupta explains the solution to the last puzzle:

Uttaran Chatterjee

Casey Bradshaw

One first proves that the number of n-digit palindromes is $9\,\times \, 10^{[(n-1)/2]}$ and the sum of all $n$-digit palindromes is $99/2 \, \times \,
10^{[3 (n-1)/2]}$, where $[x]$ denotes the integer part of a nonnegative number $x$. On division, one gets $E(X_n)$ for general $n$.

By summing over $n$ the above expressions separately and then dividing, one gets $E(Y_n)$. Careful summation of finite geometric series is required. For example, one will require the sums $\sum_{n = 2}^{2k}\, 10^{[(n-1)/2]} = \frac{2\,\times 10^k-11}{9}, $
$ \sum_{n = 2}^{2k}\, 10^{[3(n-1)/2]}
= \frac{11\,\times \,1000^k-1010}{999}, $
and the corresponding sums when the range of $n$ ends at $2k+1$.

It then turns out that the expected value of a random palindrome less than or equal to $10^8$ is $\frac{30280280280}{1111}$
and of a random palindrome less than or equal to $10^{12}$ is $\frac{30280280280280280}{111111}$.

The pattern that emerges is clearly striking; but note that this specific pattern is for palindromes less than or equal to $10^n$ for an even $n$.