**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*, *σ*^{2}), 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: **

Well done to student members Casey Bradshaw of Columbia University, who sent a detailed, complete and correct solution; Uttaran Chatterjee of the University of Calcutta also made significant advances.

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 $n$.