Weak convergence of moment generating function I have the following sequence of rvs
$$Z_1 = X_0*Y_0$$
$$Z_{n+1} = Z_n /2 + X_n*Y_n$$
Where $X_n$ and $Y_n$ are independent, with $X_n$ having Bernoulli distribution with p=1/2 and $Y_n$ having exponential distribution with parameter λ
I'm trying to find the moment generating function for $Z_n$ for any n>0 and show that $Z_n$ converges in distribution to Z, where Z is exponentially distributed.
I found that the CDF for $Z_1$ is $1-0.5e^{-λz_1}$ by considering conditional probability for when $X_0 = 0$ vs $X_0 = 1$, and used it to find $M_{x1}(t) = λ/[2(λ-t)]$. I know that since the rvs are independent, I should probably use something like $M_{u+v}(t) = M_u(t)M_v(t)$. For example, $M_{X_2}(t) = M_1(t/2)M_1(t)$. But I am not sure if this is correct or how to proceed from here.
 A: Most of what you did is correct, but there is one critical error.  So let's back up and start over.
First, since $\lambda$ is a positive scale factor, by choosing suitable units of measurement you may assume with no loss of generality that it equals $1.$  This simplifies the work a little.
Second, you have noticed that the sequence involves two operations at each stage: scale $Z_n$ by $1/2$ and add the random variable $X_nY_n.$  You know what each of these does to the mgf: the first replaces its argument $t$ by $t/2$ while the second multiplies the mgf by that of $X_nY_n.$
Let's compute that mgf.  One can just look at this problem and write it down directly, because the formula $X_nY_n$ is merely a fancy way of stating that $X_nY_n$ is a mixture of a constant $0$ and an exponential variable, with equal weights.  Since the mgf of $0$ is the constant function $1$ and the mgf of an exponential variable is $t\to 1/(1-t),$ we conclude the mgf is the mixture of the mgfs of these components,
$$\phi(t) = \frac{1}{2}(1) + \frac{1}{2}\left(\frac{1}{1-t}\right) = \frac{1-t/2}{1-t}.$$
But if you're unconvinced, you may start with the definition of the mgf and compute it using an iterated expectation, thus:
$$\begin{aligned}
\phi(t) &= E\left[\exp(tX_nY_n)\right]\\
& =  E\left[E\left[\exp(tX_nY_n)\mid Y_n\right]\right] \\
& = \frac{1}{2}E\left[\exp(tX_n(0))\right] + \frac{1}{2}E\left[\exp(tX_n(1)\right) \\
&= \frac{1}{2} + \frac{1}{2} \frac{1}{1-t}\\
&= \frac{1-t/2}{1-t}.
\end{aligned}$$
Adding distributions corresponds to multiplying their mgfs.  Thus, at each stage, to compute the next mgf $\phi_{n+1},$ you will (1) replace $t$ by $t/2$ in the current mgf $\phi_n$ and then (2) multiply that result by $\phi(t).$  In mathematical terms this is
$$\phi_{n+1}(t) = \phi_n\left(\frac{t}{2}\right)\,\phi(t).$$
Let's start this sequence to see what happens:
$$\begin{aligned}
\phi_1(t) &= \phi(t) &=\frac{1-t/2}{1-t};\\
\phi_2(t) &= \phi_1\left(\frac{t}{2}\right)\,\phi(t) = \frac{1-(t/2)/2}{1-t/2}\frac{1-t/2}{1-t} &= \frac{1-t/4}{1-t};\\
\phi_3(t) &= \phi_2\left(\frac{t}{2}\right)\,\phi(t) = \frac{1-(t/2)/4}{1-t/2}\frac{1-t/2}{1-t} &= \frac{1-t/8}{1-t};
\end{aligned}$$
and so on.  The pattern is evident.  The rest of the solution involves an easy proof (by induction) of this pattern followed by finding the limit of this sequence of functions.
