Finding MGF of Multiple RV

Question

I am working on a problem and am a bit stuck.

The problem: Let X1, X2, X3 be i.i.d random variables with distribution

P(X1=0) = $1\over3$

P(X1=1) = $2\over3$

Calculate the MGF of Y = X1X2X3

Not sure exactly how to approach this problem.

Answer 1

Since $P(Y=1)=\frac{8}{27}$, $P(Y=0)=\frac{19}{27}$. The MGF of $Y$ is $\dfrac{19+8e^t}{27}$.

Answer 2

$Ee^{tX_i}=\frac 1 3 +e^{t} \frac 2 3$. So $E(e^{tX_1X_2X_3}|X_2X_3)=\frac 1 3 +\frac 2 3e^{tX_2X_3}$ and $Ee^{tX_1X_2X_3}=E(\frac 1 3 +\frac 2 3 e^{tX_2X_3})$. Now $E(\frac 1 3 +\frac 2 3e^{tX_2X_3}|X_3) =\frac 1 3+\frac 2 3 (\frac 1 3+\frac 2 3 e^{tX_3})$. Now take expectation again. Can you now complete the computation?

Answer 3

The definition of the MGF is $$ M_Y(t)=E(e^{tY}),$$ so you need to compute this expectation value. Notice that $Y$ can be either zero or one, so the expectation value can be written $$ E(e^{tY}) = P(Y=0)e^{t\cdot 0}+P(Y=1)e^{t\cdot 1} = P(Y=0)+P(Y=1)e^t.$$

So it's just a matter of computing $P(Y=0)$ and $P(Y=1).$ Notice if you compute one of these the other follows from the fact that they need to add up to one. $P(Y=1)$ is slightly easier to compute cause we can write $$ P(Y=1) = P(X_1=1,X_2=1,X_3=1).$$ I'll let you take it from there.

Answer 4

Note that $Y=0$ or $1$ with probability $1$ whence the mgf of Y is given by $$ Ee^{tY}=e^{t\cdot0}P(Y=0)+e^tP(Y=1). $$ You only need to find $P(Y=1)$. But $Y=1$ iff $X_{i}=1$ for all $i$, so...