Suppose I have a joint moment generating function $M_{X,Y}(s,t)$ for a joint distribution with CDF $F_{X,Y}(x,y)$. Is $M_{X,Y}(s,t)=M_{X,Y}(s,0)⋅M_{X,Y}(0,t)$ both a necessary and sufficient condition for independence of $X$ and $Y$? I checked a couple of textbooks, which only mentioned necessity:
$$F_{X,Y}(x,y)=F_X(x)\cdot F_Y(y) \implies M_{X,Y}(s,t)=M_X(s) \cdot M_Y(t)$$
That result is clear as independence implies $M_{X,Y}(s,t)=\mathbb{E}(e^{sX+tY})=\mathbb{E}(e^{sX}) \mathbb{E}(e^{tY})$. Since the MGFs of the marginals are determined by the joint MGF we have:
$$X,Y\text{ independent} \implies M_{X,Y}(s,t)=M_{X,Y}(s,0)⋅M_{X,Y}(0,t)$$
But after searching online I found only a fleeting reference, without proof, to the converse. Is the following sketch proof workable?
Given a joint MGF $M_{X,Y}(s,t)$, this uniquely determines the marginal distributions of $X$ and $Y$ and their MGFs, $M_X(s)=M_{X,Y}(s,0)$ and $M_Y(t)=M_{X,Y}(0,t)$. The marginals alone are compatible with many other possible joint distributions, and uniquely determine a joint distribution in which $X$ and $Y$ are independent, with CDF $F_{X,Y}^{\text{ind}}(x,y)=F_X(x) \cdot F_Y(y)$ and MGF:
$$M_{X,Y}^{\text{ind}}(s,t) = M_X(s) \cdot M_Y(t) = M_{X,Y}(s,0)⋅M_{X,Y}(0,t)$$
So if we are given, for our original MGF, that $M_{X,Y}(s,t) = M_{X,Y}(s,0)⋅M_{X,Y}(0,t)$, this is sufficient to show $M_{X,Y}(s,t) = M_{X,Y}^{\text{ind}}(s,t)$. Then by the uniqeness of MGFs, our original joint distribution has $F_{X,Y}(x,y) = F_{X,Y}^{\text{ind}}(x,y) = F_X(x) \cdot F_Y(y)$ and $X$ and $Y$ are independent.
