Prove that $X,Y$ are independent iff the characteristic function of $(X,Y)$ equals the product of the characteristic functions of $X$ and $Y$

Solution 1:

You're saying that the pair $(X,Y)$ has the same distribution as the pair $(\bar X,\bar Y)$ and $\bar X,\bar Y$ are independent and you want to prove $X,Y$ are independent. \begin{align} & \Pr(X\in A\ \&\ Y\in B) \\[10pt] = {} & \Pr((X,Y)\in A\times B) \\[10pt] = {} & \Pr((\bar X,\bar Y)\in A\times B) & & \text{(since the joint distributions are the same)} \\[10pt] = {} & \Pr(\bar X\in A)\Pr(\bar Y\in B) & & \text{(since $\bar X,\bar Y$ are independent)} \\[10pt] = {} & \Pr(X\in A)\Pr(Y\in B) & & \text{(since $X\sim\bar X$ and $Y\sim\bar Y$)}. \end{align} Hence $X,Y$ are independent.

The part of this that took some work to prove is that the joint distributions are the same, and you seem to have done that part already.

Solution 2:

Denote by $\mathbb{P}_X$ the distribution of a random variable $X$ and by "$\stackrel{d}{=}$" equality in distribution.

Since $\tilde{X}$ and $\tilde{Y}$ are independent, the distribution of $\tilde{Z}$ equals

$$\mathbb{P}_{\bar{Z}} = \mathbb{P}_{\tilde{X}} \otimes \mathbb{P}_{\tilde{Y}}.$$

Moreover, $\tilde{X} \stackrel{d}{=}X$ and $\tilde{Y} \stackrel{d}{=}Y$ and therefore

$$\mathbb{P}_{\bar{Z}} = \mathbb{P}_X \otimes \mathbb{P}_Y.$$

Finally, since $\tilde{Z} \stackrel{d}{=} Z$, we get

$$\mathbb{P}_Z = \mathbb{P}_X \otimes \mathbb{P}_Y. \tag{1}$$

Hence,

$$\begin{align*} \mathbb{P}(X \in A, Y \in B) &= \mathbb{P}_Z(A \times B) \\ &\stackrel{(1)}{=} (\mathbb{P}_X \otimes \mathbb{P}_Y)(A \times B) \\ &=\mathbb{P}_X(A) \mathbb{P}_Y(B) \\ &= \mathbb{P}(X \in A) \mathbb{P}(Y \in B) \end{align*}$$

for any two Borel sets $A,B$. This shows that $X$ and $Y$ are independent.