# If random variables $X,Y, Z$ are pairwise independent, why are $X+Y$ and $Z$ independent as well & Generalization to infinite sums

probabilityrandom variablesstochastic-calculus

Let $$X_{i}, Y_{i}: \Omega \to \mathbb{R} \cup \{+ \infty, – \infty \} , i \in \mathbb{N}$$ two families of random variables, which are all independent to each other, ie $$X_i$$ and $$Y_i$$ are pairwise independent, but for eny $$i, j$$ the $$X_i$$ and $$Y_j$$ are independent as well.

Consider two new random variables, two infinite sums $$S_X:=\sum_{i =1}^{\infty}X_i, S_Y:=\sum_{i =1}^{\infty}Y_i$$.

Question: How to show that $$S_X$$ and $$S_Y$$ are independent as well, ie $$P_{S_X, S_Y}= P_{S_X} \cdot P_{S_Y}$$?

Firstly, the finite case: If random variables $$X, Y, Z$$ are
independent, are then $$X+Y$$ and $$Z$$ are also independent?

(almost) idea: For the distribution of $$X+Y$$
have the convolution $$P_{X+Y}= P_{X*Y} = P_X * P_Y$$ and therefore

$$P_{X+Y}(a)= \int_{\mathbb{R}} P_X(x) \cdot P_Y(a-x) d \lambda(x)$$

How can we deal with $$P_{X+Y, Z}= P_{Y*Y, Z}$$? Can it somehow be
pulled into the integral? If we can do it, the independence is
reduced to independence of $$X, Y, Z$$ and we are done. But I not see
any reason why we can do something like this:

$$P(a \in X+Y, b \in Z)= P_{X+Y, Z}(a,b)= \int_{\mathbb{R}} P_{X,Z}(x, b) \cdot P_{Y,Z}(a-x, b) d \lambda(x)$$

If we can show that $$X+Y$$ and $$Z$$ are independent, we can proceed inductively,
and such that finite sums $$S^n_X:=\sum_{i =1}^{n}X_i, S^m_Y:=\sum_{i =1}^{m}Y_i$$ are independent. Can we then use a limit
argument to pass to $$S_X$$ and $$S_Y$$?

In summary:
Problem 1: If RV $$X, Y, Z$$ are
independent, why are then $$X+Y$$ and $$Z$$also independent?
Problem 2: If $$S_X^n, S_Y^m$$ are inddpendent for all $$n,m$$, why are $$S_X$$ and $$S_Y$$ also
independent?

If $$X, Y, Z$$ are independent, then vector $$(X,Y)$$ and $$Z$$ are independent and hence $$f((X, Y))$$ is independent of $$g(Z)$$ for any Borel functions $$f$$ and $$g$$. It follows from definition immediately. Now put $$f(a,b) = a+b$$ and $$g(a)=a$$.
$$S_X = \lim_{n \to \infty} S_X^n$$, but what type of limit do you imply? A.s. convergence? Anyway for sums of independent r.v. we have that convergence a.s. is equivalent to convergence in probability and to convergence in distribution, so I will suppose that you imply convergence a.s. Hence $$(S_X, S_Y)= \lim_{n \to \infty} (S_X^n, S_Y^n)$$ a.s. and thus almost surely, thus a characteristic function of vector $$S_X, S_Y$$ is a limit of characteristic functions of $$(S_X^n, S_Y^n)$$: $$Ee^{i (uS_X + vS_Y)} = \lim_{n \to \infty}Ee^{i (uS_X^n + vS_Y^n)}$$ By independence $$Ee^{i (uS_X^n + vS_Y^n)} = Ee^{i uS_X^n} Ee^{i ( vS_Y^n)}$$. As $$S_X = \lim_{n \to \infty} S_X^n$$ we have $$Ee^{i uS_X} = \lim_{n \to \infty}Ee^{i uS_X^n}$$. So, $$Ee^{i (uS_X + vS_Y)} = \lim_{n \to \infty}Ee^{i (uS_X^n + vS_Y^n)} = \lim_{n \to \infty} Ee^{i uS_X^n} Ee^{i ( vS_Y^n)} =$$ $$= \lim_{n \to \infty} Ee^{i uS_X^n} \lim_{n \to \infty} Ee^{i ( vS_Y^n)} =Ee^{i uS_X} Ee^{i vS_Y}.$$ Independence of $$S_X$$ and $$S_Y$$ follows from equality $$Ee^{i (uS_X + vS_Y)} =Ee^{i uS_X} Ee^{i vS_Y}.$$