# Does $\varphi_{X+Y}(t)=\varphi_X(t)\varphi_Y(t)$ imply independence of $X$ and $Y$? [duplicate]

It shouldn't, but I am blanking on a counterexample.

ETA: Note that the $t$ is shared on both sides - which differentiates this from this question. Similarly $F_{X,Y}(x,y)=F_X(x)F_Y(y)$ implies independence, but $F_{X,Y}(t,t)=F_X(t)F_Y(t)$ doesn't.

## marked as duplicate by user228113, Shailesh, user91500, user296602, Claude LeiboviciJan 17 '16 at 7:20

• Interesting. – Em. Jan 17 '16 at 0:37
• Could you please specify what $\varphi_{X}(t)$ denotes. Is it the characteristic function of random variable $X$? – Anders Muszta Jan 17 '16 at 0:38
$X=Y$ Cauchy is a counter-example I was looking for.
• your characteristic function relation is equivalent to $f_{X+Y}(x) = \int_{-\infty}^{\infty} f_X(u) f_Y(x-u) du =\int_{-\infty}^{\infty} f_{X,Y}(u,x-u) du$ but this doesn't imply $f_X(u) f_Y(x-u) = f_{X,Y}(u,x-u)$ almost everywhere : your are viewing $f_{X,Y}(u,x-u)$ a $\mathbb{R}^2 \to \mathbb{R}$ function only by summing it on parallel lines, but to prove it is really the product of two $\mathbb{R} \to \mathbb{R}$ functions you would need more information on it. how many more ? that is the question. – reuns Jan 17 '16 at 1:38