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This is problem 10 of the seventh Kolmogorov Student Olympiad in Probability Theory as translated by Jonathan Christensen in this thread.

Given a sample of size one from the random variable $\xi \sim N(\mu,\sigma^2)$, both of whose parameters are unknown, give a confidence interval for $\sigma^2$ with confidence level at least $99\%$.

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You may also be interested in the discussion of this question over at stat.SE. –  Jonathan Christensen Dec 14 '12 at 2:49
    
@JonathanChristensen Thanks, it didn't occur to me to search stat.SE, too, before posting. (It's telling that we chose the same question among them all to post, though :-) ) –  Phira Dec 14 '12 at 6:11
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up vote 4 down vote accepted

The density for $\xi$ is bounded from above by its maximum $(\sqrt{2\pi}\sigma)^{-1}$. Thus, if we ensure that the range of $\xi$ in which we miss $\sigma^2$ has width at most $\sqrt{2\pi}\sigma/100$, this ensures that the probability of missing $\sigma^2$ is at most $1\%$. Thus a suitable confidence interval is $\sigma^2\in[0,20000\xi^2/\pi]$, since the range of $\xi$ values that causes us to miss $\sigma^2$ is $-\sqrt{2\pi}\sigma/200\lt\xi\lt\sqrt{2\pi}\sigma/200$, with width exactly $\sqrt{2\pi}\sigma/100$. Of course we could use $\xi-\xi_0$ with arbitrary $\xi_0\in\mathbb R$ instead.

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+1. Nice answer to a nice question :) –  Stefan Hansen Dec 13 '12 at 14:02
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