• $(\Omega,\mathcal A,\operatorname P)$ 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}


  • 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?


2 Answers 2


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.


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

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


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

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

  • $\begingroup$ $\tilde Z$ 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 $$\mathcal{B}(\mathbb{R}^m)\otimes \mathcal{B}(\mathbb{R}^n)=\mathcal{B}(\mathbb{R}^{m+n})$$ $\endgroup$
    – 0xbadf00d
    Jul 5, 2015 at 18:07
  • $\begingroup$ 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? $\endgroup$
    – 0xbadf00d
    Jul 5, 2015 at 18:07
  • 1
    $\begingroup$ @0xbadf00d No, not at all. Note that we do already know that $\tilde{X}$ and $\tilde{Y}$ are independent and therefore $$\mathbb{P}(\tilde{Z} \in A \times B) = \mathbb{P}(\tilde{X} \in A, \tilde{Y} \in B) = \mathbb{P}(\tilde{X} \in A) \mathbb{P}(\tilde{Y} \in B) = (\mathbb{P}_{\tilde{X}} \otimes \mathbb{P}_{\tilde{Y}})(A \times B).$$ $\endgroup$
    – saz
    Jul 5, 2015 at 19:27

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