Does a proof exist that the mean of two symmetric, independent random variables is also symmetric? Or is the conjecture false?
I would be interested to learn about references to this question, or perhaps a shown proof.
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@Xi'an Why would having the same center matter? if you had one centered at $a$ and another at $b$, they could be given the same center by a simple location shift (followed by a shift back). E.g. $X$ has center $a$, $Y$ has center $b$, then $Y_1=Y-b+a$ has center $a$, but $X+Y$ is just $X+Y_1$ shifted by $b-a$ -- its symmetry property cannot be altered by that mere shift (only where it occurs). Why would we worry about specifying $a=b$? Or do you just mean it's easier to prove that special case and then generalize from there? – Glen_b Apr 08 '20 at 07:14
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@StéphaneLaurent If weighting each variable by half does not matter, then I suppose it answers my question. Although I would prefer to see why it does not matter. – user120911 Apr 08 '20 at 15:16