TITLE: Joint moment generating function
QUESTION [0 upvotes]: How to prove that if joint moment generating functions of $n$  random variables is equal to the product of each of their moment generating functions then the random variables are independent.

REPLY [4 votes]: Let $(X,Y)$ be random variables such that
$$
M_{(X,Y)}(t,s)=M_X(t)\cdot M_Y(s),\quad s,t\in\mathbb{R}.
$$
Let $(\tilde{X},\tilde{Y})$ be an independent copy of $(X,Y)$, i.e. $(X,Y)\sim (\tilde{X},\tilde{Y})$ and $(X,Y)\perp\!\!\perp (\tilde{X},\tilde{Y})$. In particular we have that

$X\perp\!\!\perp \tilde{Y}$
$Y\sim\tilde{Y}$

Then
$$
M_{(X,\tilde{Y})}(t,s)=M_X(t)\cdot M_{\tilde{Y}}(s)=M_X(t)\cdot M_Y(s)=M_{(X,Y)}(t,s),
$$
where we have used that $X\perp\!\!\perp \tilde{Y}$. Since the moment-generating function determines the distribution, we have that $(X,\tilde{Y})\sim (X,Y)$ and hence
$$
\begin{align}
P(X\in A,Y\in B)&=P((X,Y)\in A\times B)=P((X,\tilde{Y})\in A\times B)\\
&=P(X\in A,\tilde{Y}\in B)=P(X\in A)P(\tilde{Y}\in B)\\
&=P(X\in A,Y\in B),
\end{align}
$$
for $A,B\in\mathcal{B}(\mathbb{R})$. Thus $X\perp\!\!\perp Y$. Now generalize this to $n$ random variables.
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