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Jun 10, 2010 at 16:25 comment added James Martin Hi Wadim, what would they complain about? (Maybe the fact that it's written out in terms of random variables rather than integrals; just because I'd seen Robin's answer and I was feeling lazy and it seemed easier to write it out that way... no doubt my answer could be shortened.) I think your response was fine, no need to delete it. The answer I gave was somewhat more general since it covers any case where the distribution has mean 0, not just those which are symmetric around 0.
Jun 9, 2010 at 23:29 comment added Wadim Zudilin James, I can't get the voting logic: if I give my students your solution, they would clearly complain. Anyway, seeing a very strong preference of yours by the MO community, I have to delete mine.
Jun 9, 2010 at 10:43 comment added Wadim Zudilin James, I vote for your scepticism! The condition $\int_{\mathbb R}xf(x)\ dx=0$ is to replace the parity of $f(x)$, so your proof is essentially the same. And I now show why this is not true in general.
Jun 9, 2010 at 9:23 history answered James Martin CC BY-SA 2.5