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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.

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Interesting.$ $ $\endgroup$ – Em. Jan 17 '16 at 0:37
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    $\begingroup$ Could you please specify what $\varphi_{X}(t)$ denotes. Is it the characteristic function of random variable $X$? $\endgroup$ – Anders Muszta Jan 17 '16 at 0:38
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    $\begingroup$ @AndersMuszta Judging by the tag, it is. $\endgroup$ – Clarinetist Jan 17 '16 at 0:38
  • $\begingroup$ Close vote retracted. $\endgroup$ – Clarinetist Jan 17 '16 at 1:06
  • $\begingroup$ Related, may be helpful for people who see this question in the future. Note that the claim in the linked question is still true (the answerer to that question also responded to the related question I have here). $\endgroup$ – Clarinetist Jan 17 '16 at 1:17
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$X=Y$ Cauchy is a counter-example I was looking for.

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  • $\begingroup$ 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. $\endgroup$ – reuns Jan 17 '16 at 1:38

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