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

Question

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?

Answer 1

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.

(The characteristic function is the same as the Fourier transform, so you may find it easier to find a statement like: if two probability distributions have the same Fourier transform, they are equal. This is also proven somewhere in Durret.)