Bulletin Editor Anirban DasGupta writes the solution to the previous puzzle.

The exact problem was this:

Fix $\epsilon > 0$. Give examples of two absolutely continuous distributions with densities $f$ and $g$ such that $|f(x)-g(x)| \leq \epsilon $ for all $x$, and one of the two distributions is infinitely divisible but the other is not. Such an example can be constructed using numerous different sufficient conditions for noninfinite divisibility. One can use the no zero property of the characteristic function, or property about tails, and if we don’t mind it if one has a bounded support, then by also using the bounded support condition. For example, given $\epsilon > 0$, take $f$ to be the density of the $N(0, \frac{2}{\pi \,\epsilon ^2})$ distribution, and $g$ to be the density of the uniform distribution on $[-\,\frac{1}{\epsilon }, \frac{1}{\epsilon }]$.

Then, $\sup _{x \in \mathcal{R}}\,|f(x)-g(x)| \leq \epsilon $, and the $N(0, \frac{2}{\pi \,\epsilon ^2})$ distribution is infinitely divisible, while the uniform distribution on $[-\,\frac{1}{\epsilon }, \frac{1} {\epsilon }]$ is not, because it has a bounded support. Examples can be constructed easily even if both $f, g$ are supported on the whole real line: just perturb a standard normal density slightly.

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