Fundamental question on stochastic independence

Question

Let $X, Y, Z$ be random variables. If $(X,Y)$ is independent of $Z$, is then also $X$ independent of $Z$?

Answer 1

Intuitively, if $(X,Y)$ is independent of $Z$, then $Z$ provides no information about both $X$ and $Y$, so $X$ and $Z$ are independent.

Formally (for discrete case): $$p_{X,Y,Z}(x,y,z)=p_{X,Y}(x,y)p_Z(z)$$ \begin{align} p_{X,Z}(x,z)=&\sum_y p_{X,Y,Z}(x,y,z)\\ =&\sum_y p_{X,Y}(x,y)p_Z(z)\\ =&p_{X}(x)p_Z(z) \end{align}

Answer 2

I thinks the answer is yes.

method 1)

$$P(X\in A , Z\in B)=P(X\in A, Y\in (-\infty, \infty), Z\in B)$$ $$=P(X\in A, Y\in (-\infty, \infty) )P(Z\in B)$$ $$=P(X\in A )P(Z\in B)$$

Method2) since

$$\sigma(X) \subset \sigma(X,Y)$$ so

when $Z$ is independent of $(X,Y)$ then any events in $\sigma(Z)$ are independent of $\sigma(X,Y)$. IndependentStochasticProcesses

In hence any events in $\sigma(Z)$ are independent of $\sigma(X)\subset \sigma(X,Y)$.

So $Z$ and $X$ are independent.

On the other hand if $Z$ is independent of $(X,Y)$ so for any $A\in \sigma(Z)$ and $B\in \sigma(X,Y)$ $$P(Z\in A ,(X,Y)\in B)= P(Z\in A ) P((X,Y)\in B) $$ .

so if $D\in \sigma(X)$

$$P(Z\in A ,X\in D)= P(Z\in A ) P(X\in D) $$ since $D\in \sigma(X)\subset \sigma(X,Y)$