4

Is it possible to prove that if $X, Y, Z$ independent, then $X + Y$ and $Z$ are independent? Is it even true? Intuitively it seems true but I am unsure.

dan
  • 85

1 Answers1

4

By definition $X,Y,Z$ independent means $$P(X \in A, Y \in B, Z \in C) = P(X \in A)P(Y \in B)P(Z \in C).$$ In particular the pair $(X,Y)$ is independent of $Z$, $$ P((X,Y) \in A \times B, Z \in C) = P(X \in A)P(Y \in B)P(Z \in C) = P((X,Y) \in A \times B) P(Z \in C). $$ Then we note that if $U,V$ are independent, then $f(U), g(V)$ are independent for any borel $f,g$., see e.g. If $X$ and $Y$ are independent then $f(X)$ and $g(Y)$ are also independent.

Apply this to $U=(X,Y)$ and $V=Z$ with $f((X,Y)) = X+Y$ and $g(Z) = Z$ to find that $X+Y$ is independent of $Z$.

nullUser
  • 27,877
  • A priori, the fact that "the pair $(X,Y)$ is independent of $Z$" requires more than $P((X,Y) \in A \times B, Z \in C)$ being what you say--in other words, you should explain why what you say is enough for the pair $(X,Y)$ to be independent of $Z$. – Did Mar 05 '16 at 17:04
  • @nullUser Could you explain a little bit about why $(X,Y)$ is independent of $Z$, considering the comment given by Did? – Kenneth.K May 09 '17 at 21:25
  • The definition of independence of $(X,Y)$ from $Z$ is that $P((X,Y) \in E, Z \in C) = P((X,Y) \in E)P(Z \in C)$ for all measurable $E \subseteq \mathbb{R}^2$ and $C \subseteq \mathbb{R}$. The triple $X,Y,Z$ being an independent family by definition gives this when $E = A \times B$. The step I left out was that sets of the form $A\times B$ form a $\pi$-system generating the product $\sigma$-algebra, so if the independence formula holds for $E = A\times B$, then it will hold for all product measurable $E$. – nullUser May 10 '17 at 12:58