# Moment generating functions/ Characteristic functions of $X,Y$ factor implies $X,Y$ independent.

This is solely a reference request. I have heard a few versions of the following theorem:

If the joint moment generating function $\mathbb{E}[e^{uX+vY}] = \mathbb{E}[e^{uX}]\mathbb{E}[e^{vY}]$ whenever the expectations are finite, then $X,Y$ are independent.

And there is a similar version for characteristic functions. Could anyone provide me a serious reference which proves one or both of these theorems?

• I'm not at all sure that this is true. (Certainly it's true when "if" and "then" are interchanged.) – Michael Hardy Jan 26 '13 at 6:18
• If this were true, every random variables with subexponential tails on both sides would be independent. Please reach a plausible statement. – Did Jan 26 '13 at 10:57
• In this post a theorem with a more general result was presented: math.stackexchange.com/questions/1802289/… – Carlos Mendoza May 29 '16 at 13:55

## 1 Answer

Theorem (Kac's theorem) Let $X,Y$ be $\mathbb{R}^d$-valued random variables. Then the following statements are equivalent.

1. $X,Y$ are independent
2. $\forall \eta,\xi \in \mathbb{R}^d: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot \mathbb{E}e^{\imath \, Y \cdot \eta}$

Proof:

• $(1) \Rightarrow (2)$: Straightforward, use $\mathbb{E}(f(X) \cdot g(Y)) = \mathbb{E}(f(X)) \cdot \mathbb{E}(g(Y))$
• $(2) \Rightarrow (1)$: Let $(\tilde{X},\tilde{Y})$ be such that $\tilde{X}$, $\tilde{Y}$ are independent, $\tilde{X} \sim X$, $\tilde{Y} \sim Y$. Then $$\mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} \stackrel{(2)}{=} \mathbb{E}e^{\imath \, X \cdot \xi} \cdot \mathbb{E}e^{\imath \, Y \cdot \eta} = \mathbb{E}e^{\imath \tilde{X} \cdot \xi} \cdot \mathbb{E}e^{\imath \tilde{Y} \cdot \eta} = \mathbb{E}e^{\imath (\tilde{X},\tilde{Y}) \cdot (\xi,\eta)}$$ i.e. the characteristic functions of $(X,Y)$ and $(\tilde{X},\tilde{Y})$ coincide. From the uniqueness of the Fourier transform we conclude $(X,Y) \sim (\tilde{X},\tilde{Y})$. Consequently, $X$ and $Y$ are independent.

Reference (not for the given proof, but the result):David Applebaum, B.V. Rajarama Bhat, Johan Kustermans, J. Martin Lindsay, Michael Schuermann, Uwe Franz: Quantum Independent Increment Processes I: From Classical Probability to Quantum Stochastic Calculus (Theorem 2.1).

• Do we need $X,Y \in L^1$? I don't see it being used in the proof. – nomadicmathematician Jun 13 '16 at 21:04
• @takecare You are right; it's not needed for the proof. – saz Jun 14 '16 at 5:38
• How do you know you can find such a pair of independent variables $(\tilde{X},\tilde{Y})$ – Ansel B Feb 3 '17 at 14:03
• @AnselB If $\mathbb{E}e^{i (X \xi + Y \eta)} = \mathbb{E}e^{i \xi X} \mathbb{E}e^{iY \eta}$ for all $\xi$, $\eta$, then $X$ and $Y$ are independent; that's exactly what the proof shows. If you prefer, think about like this: Denote by $\mathbb{P}_X$ and $\mathbb{P}_Y$ the distribution of $X$ and $Y$, respectively. Then the characteristic function of the product measure $\mu = \mathbb{P}_X \times \mathbb{P}_Y$ is given by $$\hat{\mu}(\xi,\eta) = \mathbb{E}e^{i \xi X} \mathbb{E}e^{i \eta Y}$$ which is, by assumption, also the characteristic function of $(X,Y)$. – saz Feb 4 '17 at 6:29
• @takecare Yes... if $$\mathbb{E}\exp \left( i \sum_{j=1}^n \xi_j X_j \right) = \prod_{j=1}^n \mathbb{E}\exp(i \xi_j X_j)$$ for any $\xi_1,\ldots,\xi_n$, then the random variables are independent. – saz Jun 4 '17 at 4:45