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Suppose $X, Y$ and $Z$ are three non-degenerate random variables. Suppose that $X$ and $Z$ are independent and that $Y$ and $Z$ are not independent. Is it possible for $X + Y$ to be independent of $Z$? Can you provide an example?

I am also interested in strengthening the assumptions about the relationship between $Y$ and $Z$. Suppose that $Y$ is discretely distributed with support $\{y_{1},y_{2},y_{3}\}$ and $Z$ is discretely distributed with support $\{z_{1}, z_{2}\}$. Suppose that $P[Y = y_{j} \vert Z = z_{0}] \neq P[Y = y_{j} \vert Z = z_{1}]$ for each $j = 1,2,3$. As before, assume that $X$ and $Z$ are independent. Now is it possible for $X + Y$ to be independent of $Z$?

Does the answer change if $X$ is continuously distributed?

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  • $\begingroup$ Just an idea, no proof, In non degenerate cases, $X+Y$ is always dependent to $Y$ and as a result to $Z$. $\endgroup$ – hhsaffar Dec 6 '13 at 21:05
  • $\begingroup$ If $X+Y$ is independent of $Z$ and $X$ is independent of $Z$, then $(X+Y)-X = Y$ must also be independent of $Z$. $\endgroup$ – whuber Dec 6 '13 at 21:51
  • $\begingroup$ @whuber No. A classic mistake. $\endgroup$ – Did Dec 6 '13 at 22:13
  • $\begingroup$ @Did Thank you for pointing that out! $\endgroup$ – whuber Dec 6 '13 at 22:14
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Assume that $(X,Z)$ is uniform on $\{-1,1\}^2$ and consider $Y=XZ-X$. Then $(X,Z)$ is independent. Furthermore, $X+Y=XZ$ and $(XZ,X)$ is uniform on $\{-1,1\}^2$ hence $(X+Y,Z)$ is independent. Finally, $[Z=1]=[Y=0]$ and $[Z=-1]=[|Y|=2]$ hence $(Y,Z)$ is not independent.

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  • $\begingroup$ This is very nice, would you please tell me how you came up with that? $\endgroup$ – hhsaffar Dec 6 '13 at 22:22
  • $\begingroup$ @hhsaffar Just a variation on a classical example of a non independent (n+1)-tuple such that every n-tuple is independent. $\endgroup$ – Did Dec 6 '13 at 22:29
  • $\begingroup$ Thank you this is very helpful. As for the other two parts of the question, do you think that they will change the conclusion? I understand coming up with a concrete example may be hard (after all that's why I'm asking this question to begin with), but do you have any suggestions on how to proceed? $\endgroup$ – evencoil Dec 6 '13 at 23:19

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