Does $M_{X_1+...+X_n}(t)= M_{X_1}(t)×...×M_{X_n}(t)$ for all real $t$ conclude indepedence? Let $X_1,X_2,...,X_n$ be $n$ random variables. Let $M_X(t)$ be the moment generating function of $X$.
It is given that $$M_{X_1+...+X_n}(t)= M_{X_1}(t)×...×M_{X_n}(t)$$ for all real $t$. 
Can we conclude from here that the $n$ random variables are independent?
I know,that if the random variables are independent then the above equation holds true but I cannot prove or disprove the converse.
Any help would be appreciated. Thanks in advance!
 A: No. The fact that $M_{X_1+\ldots+X_n}(t)=M_{X_1}(t)×...×M_{X_n}(t)$ for all $t$ does not imply independence of $X_1,\ldots,X_n$. 
Let us take $n=2$. Consider random variables $X$ and $Y$ with the following joint distribution: 
\begin{array}{r|c|c|c}
 & Y=0 & Y=1 & Y=2 \\ \hline
X=0   & 1/9     & 0   & 2/9\\  \hline
X=1    &2/9 &1/9 &0 \\  \hline
X=2    &0 & 2/9 &1/9
\end{array}
The marginal distributions of $X$ and $Y$ are the same and 
$$
\mathbb P(X=0)=\mathbb P(X=1)=\mathbb P(X=2)=\frac13.
$$
The distribution of $X+Y$ is 
$$
\mathbb P(X+Y=0)=\mathbb P(X+Y=4)=\frac19, 
$$
$$
\mathbb P(X+Y=1)=\mathbb P(X+Y=3)=\frac29, 
$$
$$
\mathbb P(X+Y=2)=\frac39.
$$
MGF for $X+Y$ is 
$$
M_{X+Y}(t) = \frac19+\frac29e^t+\frac39 e^{2t}+\frac29 e^{3t}+\frac19 e^{4t}.
$$
And MGF of $X$ or $Y$ is
$$
M_X(t)=M_Y(t)=\frac13+\frac13e^t+\frac13e^{2t},
$$
so 
$$
M_X(t)\times M_Y(t)=\left(\frac13+\frac13e^t+\frac13e^{2t}\right)^2 = M_{X+Y}(t).
$$
And $X$, $Y$ are dependent since for example
$$\mathbb P(X=0,Y=1)=0\neq \mathbb P(X=0)\times\mathbb P(Y=1)=\frac13\times\frac13.$$
