Given MGF of X, find MGF of $ Y=X_1\dot\ X_2 \dot\ X_3$ Let $X_1$, $X_2$, $X_3$ be a random sample from a discrete distribution with probability funciton 
$p(0)= 1/3$
$p(1) = 2/3$
Calculate moment generating function, $M(t)$, of $Y=$$X_1$$X_2$$X_3$
My Work 
$M_x(t) = \frac{1}{3} + \frac{2}{3}e^t$
then $E[e^{t(X_1X_2X_3)}]$ 
$=E[e^{tX_1}+e^{tX_2}+e^{tX_3}]$
$= E[e^{tX_1}]+E[e^{tX_2}]+E[e^{tX_3}]$
$=3(\frac{1}{3} + \frac{2}{3}e^t)$
$=1+2e^t$
However, $M_x(0)=3\neq1$, so this must be wrong, but why? 
 A: $X_1 X_2 X_3 = 1$ if and only if $X_1 = X_2 = X_3 = 1$.  If $X_i$ and $X_j$ are independent whenever $i \ne j$, then what is $\Pr[Y = 1]$?  Then what is $$M_Y(t) = \operatorname{E}[e^{tY}] = e^t \Pr[Y = 1] + e^0 \Pr[Y = 0]?$$
A: If $X_1\,,\, X_2$ and $X_3$ be independent , then
$$P(Y=0)=\left( \begin{matrix}
   3  \\
   1  \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{1}}{{\left( \frac{2}{3} \right)}^{2}}+\left( \begin{matrix}
   3  \\
   2  \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{2}}{{\left( \frac{2}{3} \right)}^{1}}+\left( \begin{matrix}
   3  \\
   3  \\
\end{matrix} \right){{\left( \frac{1}{3} \right)}^{3}}
$$
$$P(Y=0)=\frac{12}{27}+\frac{6}{27}+\frac{1}{27}=\frac{19}{27}$$
$$P(Y=1)=\left( \begin{matrix}
   3  \\
   3  \\
\end{matrix} \right){{\left( \frac{2}{3} \right)}^{3}}=\frac8{27}$$
$$M_Y(t) = \operatorname{E}[e^{tY}] = \frac8{27}e^t  +  \frac{19}{27}$$
A: A standard identity in algebra says $A^{x+y+z} = A^x A^y A^z$.
You seem to be trying to rely on a similar identity that would say $A^{t(xyz)} = A^{tx} + A^{ty} + A^{tz}$.
The problem is that the latter proposed identity is not true.
