Moment Generating Function of the Product of Three I.I.D Bernoulli random variables

Question

Let X1, X2, X3 be i.i.d. random variables with distribution $P(X_1 =0)=\frac13, P(X_1 =1)= \frac {2}{3}$ Calculate the moment generating function of $Y = X_1X_2X_3.$

My work:

$M_x(t)= E(e^{tx}) = \sum e^{tx} p(x) = (1-p) + e^{t}p = \frac {1}{3} + \frac{2e^t}{3}$

My question is this statement true? $M_y(t)= \sum e^{tx_1x_2x_3}p_{x_1}(x)p_{x_2}(x)p_{x_3}(x) \ne [M_x(t)]^3$

change made in above equation to correspond with given answer [not equal] added

Answer 1

The moment generating function is given by $$ M_Y(t)=\operatorname Ee^{tY}=(1-(2/3)^3)e^{t\cdot0}+(2/3)^3e^{t\cdot 1}=1-(2/3)^3+(2/3)^3e^t $$ and it is not equal to $M_{X_1}^3(t)$.

Answer 2

The simplest thing to do is to observe that the product $Y = X_1 X_2 X_3$ is itself Bernoulli, and that the probability that $Y = 1$ is equal to the probability that each of $X_1$, $X_2$, and $X_3$ is equal to $1$, since that is the only way that the product is $1$.