# $X$,$Y$,$Z$ mutually independent implies $X+Y$ independent of $Z$

Supposing $X$, $Y$ and $Z$ and mutually independent real random variables, how can we prove that $X+Y$ and $Z$ are independent from the definition? If not from the definition, using $\sigma$-algebras?

I know that if $X$ and $Y$ are not independent this doesn't work so I know I have to make use of their independence in my proof but I can't figure out how. And it means that if I write the following, I'm making a mistake, and I can't find it:

$$\mathbb{P}(X+Y \leq k,Z \leq l)=\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x, Z \leq l) dx=\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x) \mathbb{P}(Z \leq l) dx = \mathbb{P}(Z \leq l)\int_{\mathbb{R}} \mathbb{P}(X=x,Y \leq k-x)dx= \mathbb{P}(Z \leq l) \mathbb{P}(X+Y \leq k)$$

PS: I have seen the proof here but it much more advanced stuff, kind of a sledgehammer... X,Y,Z are mutually independent random variables. Is X and Y+Z independent?

• Have you tried drawing a picture, like a Venn diagram? – RustyStatistician Nov 25 '15 at 18:57
• I couldn't do anything sensible with pictures, but I have tried... – Kika Nov 26 '15 at 11:00

Let $g:\mathbb R^2\to\mathbb R$ be the map $(a,b)\mapsto a+b$. Clearly for any $t\in\mathbb R$, $$g^{-1}((-\infty,t])=\{(a,b)\in\mathbb R^{2}:a+b\leqslant t\}$$ is a measurable set. Therefore $g$ is a measurable map, so $X+Y=g(X,Y)$ and $\sigma(g(X,Y))\subset \sigma(X,Y)$. It follows that if $E\in\sigma(g(X,Y))$, $F\in\sigma(Z)$, then $E\in\sigma(X,Y)$ so that $$\mathbb P(E\cap F)=\mathbb P(E)\mathbb P(F),$$ from which we conclude.
• Where have you used the independence of $X$ and $Y$? – Kika Nov 26 '15 at 11:00
• @Kika That isn't actually necessary for $X+Y$ to be independent of $Z$. We just need that $(X,Y)$ and $Z$ are independent (which is implied by $X, Y, Z$ being mutually independent). – Math1000 Nov 27 '15 at 5:15