Let $X$ and $Y$ be real-valued random variables defined on the same space. Let's use $\phi_X$ to denote the characteristic function of $X$. If $\phi_{X+Y}=\phi_X\phi_Y$ then must $X$ and $Y$ be independent?

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    $\begingroup$ No, as the case $X=Y=$ standard Cauchy shows. $\endgroup$ – Did May 1 '13 at 7:58
  • $\begingroup$ @Did why this is a counterexample? $\endgroup$ – Connor Dec 12 '16 at 13:19
  • $\begingroup$ @Connor Did you try to compute the characteristic function of a standard Cauchy random variable? If you did, this should be clear. If you did not, what are you asking? $\endgroup$ – Did Dec 12 '16 at 13:43
  • $\begingroup$ I flagged Connor's question as a duplicate but ended up expanding your comment to an answer, @Did $\endgroup$ – Therkel Dec 12 '16 at 14:02

No. The property in your post is called subindependence, and it is strictly weaker than independence. (Note that some people use the term "subindependent" as a synonym for "uncorrelated".) In addition to the references given in Wikipedia, you can find an example in this short note. Unfortunately it's behind a paywall. The example consists of two random variables with joint pdf $$h(x,y)=f(x)f(y)(1+\cos x\cos 3y)$$ where $$f(x)=C\left(\int_0^{1/2} \exp(1/(4s^2-1))\cos (sx)\,ds\right)^2$$ with appropriate normalizing constant $C$.


As user75064 already pointed out, the answer is "no". However, there is the following result:

Let $X,Y$ be $\mathbb{R}^d$-valued random variables. Then the following statements are equivalent.

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

i.e. if the characteristic function of the random vector $(X,Y)$ equals the product of the characteristic function of $X$ and $Y$, then $X$ and $Y$ are independent (proof).

  • $\begingroup$ the small letter that you have in the exponent is the imaginary number $i$, right? $\endgroup$ – Carlos Mendoza May 27 '16 at 20:25
  • $\begingroup$ @CarlosMendoza Yes.... $\endgroup$ – saz May 28 '16 at 5:33

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