Is a sequence of r.v. are independent if and only if their characteristic factors as a product?

It is obvious that if $\vec{X} = (X_1,X_2\cdots,X_n)^T$ are independent random variables, and their marginal characteristic function and joint characteristic function exists, then they are related by $$Ee^{i\sum_{j=1}^n t_ix_i} = \prod_{j = 1}^n E e^{i t_i x_i},$$ but is the converse true? That is, if there exists such a factorization, then the variables are independent? My professors says so, but I cannot find proof of it, nor in my literature or online. is it true? Is it true for normal r.v.? If it is true, can you provide a reference or proof?

• – kjetil b halvorsen May 21 '18 at 20:43
• @kjetilbhalvorsen Marius is not looking at the sum, but rather the joint distribution... – Lorenzo Najt May 21 '18 at 20:55
• @kjetilbhalvorsen Oh that's interesting. Apparently this property of MGFs doesn't apply to CFs, I guess. – BCLC May 26 '18 at 17:02

Try computing what the characteristic function of the product of independent marginals would be (that is, if we treat the $x_i$ in your formula as independent random variables). You will get that it splits up as a product in the same way. Then use the theorem that the characteristic function identifies the density.
• @MariusJonsson as a start , can you show that if X and Y are independent then the $\phi_(X,Y)(a,b)= \phi_X(a) \phi_Y(b)$? Here phi denotes the characteristic function. – Lorenzo Najt May 21 '18 at 21:36
• Yes, I call this obvious in my question: Since $X,Y$ independent and exp is continuous, $\phi(X,Y)(a,b)=E [e^{i(Xa + Yb)}] = E [e^{iXa}e^{iYb}] = E [e^{iXa}]E[e^{iYb}] = \phi(X)(a)\phi(Y)(b)$ – Mikkel Rev May 22 '18 at 15:38