Is it true that in probability theory the third central moment of a sum of two independent random variables is equal to the sum of the third central moments of the two separate variables?
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1$\begingroup$ Suppose that the random variables both have zero means. Can you write equations instead of words to express the main point of your question? Can you figure out the answer from the alleged equation? Finally, if you got the answer Yes after all this, read this article on Wikipedia. $\endgroup$– Dilip SarwateCommented Sep 29, 2023 at 13:57
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1$\begingroup$ You may find it helpful to investigate cumulants en.wikipedia.org/wiki/Cumulant $\endgroup$– Glen_bCommented Sep 29, 2023 at 15:59
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As Dilip Sarwate pointed out in a comment, this is similar to looking at $(X+Y)^3=X^3+Y^3$ but just with expectations and the centering. Maybe I'm ridiculous but I tried but here's my progress after a few expansions and a helpful comment:
\begin{align*} E\left\{[(X+Y)-E(X+Y)]^3\right\}&=E(X^3)+2 (EX)^3 -3 E(X^2)(EX) -3 E(Y^2)(EY) +2 (EY)^3 +E(Y^3) \end{align*}
Then as commented below by whuber, we may assume without loss of generality that $E(X)=E(Y)=0$ when dealing with centered moments.
\begin{align*} E\left\{[(X+Y)-E(X+Y)]^3\right\}&=E(X^3)+2 (0)^3 -3 E(X^2)(0) -3 E(Y^2)(0) +2 (0)^3 +E(Y^3)\\ &=E(X^3)+E(Y^3)\\&=E [(X-E(X))^3]+E[(Y-E(Y))^3] \end{align*}
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2$\begingroup$ Without any loss of generality, you may assume $E[X]=E[Y]=0$ when computing central moments, because they (obviously) are not changed when you add any constant to a random variable. What happens to the right hand side of your equation then? ;-) $\endgroup$– whuber ♦Commented Sep 29, 2023 at 16:38
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$\begingroup$ @whuber Ohhh I didn't know that about central moments yet. Thank you very much for your comment. $\endgroup$– DerfCommented Sep 29, 2023 at 16:59
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$\begingroup$ So is the answer to my question yes? $\endgroup$– AdVenCommented Sep 29, 2023 at 17:37