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A favorite example in theoretical statistics is this:

A sample of individuals are drawn independently from a distribution with density $f(x)$, where $f(x)$ is unknown, but is known to be symmetric about an unknown center $\mu$. Estimate $\mu$.

To explain what I mean by "a favorite example", I'll list a few papers that provide procedures for this problem. Huber's original paper on robust estimators discusses this problem at length. In Diaconis and Freedman's paper about inconsistent Bayes estimates, one of the results is about a Bayesian prior over possible symmetric density functions $f$. The literature on the empirical characteristic function includes applications to this problem, here and here.

However, I can't think of situations where this happens in practice. I know, of course, that people often make normal assumptions, and the normal distribution is symmetric. However, if you're looking for estimators like these that don't assume specific distributions, presumably you think there are deviations from a normal assumption, or other common distributional assumptions. But I can't think of a reason one would assume that these deviations happen equally on both sides.

So, my question is: what situation justifies the assumption of symmetry, but not the assumption of a specific form of a distribution?

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  • $\begingroup$ Uniform distribution? Or if you want infinite support: what about the example in your very first reference? "Contamination model <...> observations are assumed to be normal with variance 1, but a fraction $\epsilon$ of them is affected by gross errors." As long as the errors are symmetric and not normal (e.g. uniform), the sum distribution is not normal. $\endgroup$
    – juod
    Sep 5 '18 at 2:03
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One common situation where it's implied by the assumptions people often make (because they think they're reasonable) is when there's paired observations which are assumed to have the same distribution apart from a possible location shift, and that between-individual differences are the source of pair-dependence

(e.g. where some treatment is presumed to simply move the distribution of values without changing the spread or shape)

The pair differences would then be symmetric.

This isn't the only situation you'd see symmetry, though.

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  • $\begingroup$ $Y-X = (\mu_Y+\epsilon_Y)-(\mu_X+\epsilon_X) = (\epsilon_Y - \epsilon_X) + (\mu_Y - \mu_X)$. Since $\epsilon_Y$ has the same distribution as $\epsilon_X$, then $\epsilon_Y - \epsilon_X$ has the same distribution as $\epsilon_X - \epsilon_Y = -(\epsilon_Y - \epsilon_X)$. So $\epsilon_Y - \epsilon_X$ is symmetric about zero, and $Y-X$ is symmetric about $\mu_Y-\mu_X$. OK, got it. $\endgroup$
    – user54038
    Sep 6 '18 at 14:46

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