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Let

  • $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $X$ and $Y$ be random variables on $(\Omega,\mathcal A,\operatorname P)$ with values in $\mathbb{R}^m$ and $\mathbb{R}^n$, respectively
  • $\varphi_Z$ denote the characteristic function of a random variable $Z$

Claim: $\;$ $X$ and $Y$ are independent iff $$\varphi_{(X,Y)}(s,t)=\varphi_X(s)\varphi_Y(t)\;\;\;\text{for all }s\in\mathbb{R}^m\;\text{and}\;t\in\mathbb{R}^n\tag{1}$$

Proof: $\;$ "$\Rightarrow$":

  • Let $Z:=(X,Y)$ and $u:=(s,t)\in\mathbb{R}^m\times\mathbb{R}^n$
  • $X$ and $Y$ are independent $\Rightarrow$ $e^{i\langle s,\;\cdot\;\rangle}\circ X$ and $e^{i\langle t,\;\cdot\;\rangle}\circ Y$ are independent $\Rightarrow $

\begin{equation} \begin{split} \varphi_Z(u)&\stackrel{\text{def}}{=}\operatorname E\left[e^{i\langle u,Z\rangle}\right]\\ &=\operatorname E\left[e^{i\langle s,X\rangle+i\langle t,Y\rangle}\right]\\ &=\operatorname E\left[e^{i\langle s,X\rangle}e^{i\langle t,Y\rangle}\right]\\ &=\operatorname E\left[e^{i\langle s,X\rangle}\right]\operatorname E\left[e^{i\langle t,Y\rangle}\right]\\ &\stackrel{\text{def}}{=}\varphi_X(s)\varphi_Y(t) \end{split} \end{equation}

"$\Leftarrow$":

  • Let $\tilde X\sim X$ and $\tilde Y\sim Y$ be independent
  • Since a finite measure on $\mathbb{R}^d$ is uniquely determined by its characteristic function, $$\varphi_X=\varphi_{\tilde X}\;\;\;\text{and}\;\;\;\varphi_Y=\varphi_{\tilde Y}\tag{2}$$
  • Thus, \begin{equation} \begin{split} \varphi_{(X,Y)}(s,t)&\stackrel{(1)}{=}\varphi_X(s)\varphi_Y(t)\\ &\stackrel{(2)}{=}\varphi_{\tilde X}(s)\varphi_{\tilde Y}(t)\\ &=\varphi_{(\tilde X,\tilde Y)}(s,t) \end{split} \end{equation} by "$\Rightarrow$"
  • Again, since the distribution of $(X,Y)$ is uniquely determined by $\varphi_{(X,Y)}$, we've got $$(X,Y)\sim (\tilde X,\tilde Y)$$
  • Especially, $Z:=(X,Y)$ and $\tilde Z:=(\tilde X,\tilde Y)$ have the same distribution function $F$

Now, I got stuck. From the definition of $F$ and the definition of independence, it seems to be obvious, that we can conclude the independence of $X$ and $Y$. However, how do we need to argue in detail?

saz
  • 120,083
0xbadf00d
  • 13,422
  • This question is highly motivated by the answer of @saz to the following question: http://math.stackexchange.com/questions/287138/moment-generating-functions-characteristic-functions-of-x-y-factor-implies-x – 0xbadf00d Jul 05 '15 at 17:26

2 Answers2

4

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.

user103828
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Michael Hardy
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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.

saz
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  • $\tilde Z$ is a random variable with values in $(\mathbb{R}^{m+n},\mathcal{B}(\mathbb{R}^{m+n})$. Now, $$\mathcal D:=\left{A\times B:A\in\mathcal{B}(\mathbb{R}^m),B\in\mathcal{B}(\mathbb{R}^n)\right}$$ is a $\cap$-stable generator of $$\mathcal{B}(\mathbb{R}^m)\otimes \mathcal{B}(\mathbb{R}^n)=\mathcal{B}(\mathbb{R}^{m+n})$$ – 0xbadf00d Jul 05 '15 at 18:07
  • Since a finite measure on a $\sigma$-algebra $\mathcal S$ is uniquely determined by it's values on any $\cap$-stable generator of $\mathcal S$, we've got $$\operatorname P_{\tilde Z}=\operatorname P_{\tilde X}\otimes\operatorname P_{\tilde Y}\tag{3};,$$ since this relation holds on $\mathcal D$ by independecne. This is the argumentation you've used for $(3)$, right? – 0xbadf00d Jul 05 '15 at 18:07
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    @0xbadf00d No, not at all. Note that we do already know that $\tilde{X}$ and $\tilde{Y}$ are independent and therefore $$\mathbb{P}(\tilde{Z} \in A \times B) = \mathbb{P}(\tilde{X} \in A, \tilde{Y} \in B) = \mathbb{P}(\tilde{X} \in A) \mathbb{P}(\tilde{Y} \in B) = (\mathbb{P}{\tilde{X}} \otimes \mathbb{P}{\tilde{Y}})(A \times B).$$ – saz Jul 05 '15 at 19:27