Let $X$ and $Y$ be random variables such that $\mathbb{E} e^{it(X+Y)} = \mathbb{E} e^{itX} \mathbb{E} e^{itY}$ and covariance exists.
I want to show that they are uncorrelated, i.e., $\mathbb{E}(X -\mathbb{E}X) (Y - \mathbb{E}Y) = 0$. According to this paper https://www.tandfonline.com/doi/pdf/10.1198/tast.2009.09051?needAccess=true, this is true. I can prove it when both have finite second moments, but I do not know how I should proceed for general general random variables. Any hints or reference are appreciated.
Differentiating $\varphi_{X+Y}(t) = \varphi_X(t) \varphi_Y(t) $ twice and letting $t=0$, \begin{equation} \varphi_X ''(0) \varphi_Y(0) + 2 \varphi_X'(0) \varphi_Y'(0) + \varphi_X(0) \varphi_Y''(0) = \varphi_{X+Y}''(0) \end{equation} Since \begin{equation} \varphi_X^{(k)}(t) = \mathbb{E} (( iX)^k e^{itX})\end{equation} holds for $X \in L^k$, \begin{equation} \mathbb{E} ( iX)^2 + 2 \mathbb{E} ( iX) \mathbb{E} ( iY)+\mathbb{E} ( iY)^2 = \mathbb{E} (i^2(X+Y)^2) \end{equation}