# If $X$ is independent of $Z$ and $Y$ is dependent with $Z$ is it possible for $X + Y$ to be independent of $Z$?

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?

• Just an idea, no proof, In non degenerate cases, $X+Y$ is always dependent to $Y$ and as a result to $Z$. – hhsaffar Dec 6 '13 at 21:05
• If $X+Y$ is independent of $Z$ and $X$ is independent of $Z$, then $(X+Y)-X = Y$ must also be independent of $Z$. – whuber Dec 6 '13 at 21:51
• @whuber No. A classic mistake. – Did Dec 6 '13 at 22:13
• @Did Thank you for pointing that out! – whuber Dec 6 '13 at 22:14

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.