Moment generating function of the average of two variables I am given two variables $Y_1$ and $Y_2$ obeying an exponential distribution with mean $\beta= 1$
We are asked what the distribution of their average is and the solution must be found using moment generating functions.
The solution to this exercise says:

Well first, the solution already has a typo since it should be $E[e^{t((1/2)Y_1+(1/2)Y_2)}]$ but moreover, 
I don't agree that $E[e^{t((1/2)Y_1+(1/2)Y_2)}]=M_{Y_1}(t)M_{Y_2}(t)$
Since $M_{Y_1}(t)M_{Y_2}(t)=E[e^{t(Y_1+Y_2)}]$, from my understanding.
So what is the actual solution to this problem?
 A: I think your doubt is right. We have $U=\frac12 Y_1+\frac12 Y_2$. Let´s see how $\frac12 Y_1$ is distributed.
$P(Z_1\leq z_1)=P(\frac12 Y_1\leq z_1)=P(Y_1\leq 2z_1)$ 
Thus $\frac12 Y_1\sim Exp(2)$. And the moment generating function of $\frac12 Y_1$ is
$M_{\color{blue}{\frac12 Y_1}}(t)=(1-\frac{t}2)^{-1}$. The subscripts are not right, since $M_{\frac12 Y_1}(t)\cdot M_{\frac12 Y_2}(t)=(1-\frac{t}2)^{-2}$
A: Yes, clearly, too many typos were made when type setting the solution.   The solution is otherwise okay.
$$\begin{align}\mathsf M_U(t) &= \mathsf E(\mathsf e^{t U})&&\text{Definition of MGF: note the }t \\ &= \mathsf E(\mathsf e^{t(Y_1+Y_2)/2})&&\text{Definition of }U\\ &=\mathsf E(\mathsf e^{(t/2)Y_1}\mathsf e^{(t/2)Y_2}) &&\text{Exponent Algebra} \\ &= \mathsf E(\mathsf e^{(t/2)Y_1})~\mathsf E(\mathsf e^{(t/2)Y_2})&&\text{Independence}\\ &= \mathsf M_{Y_1}(t/2)~\mathsf M_{Y_2}(t/2) &&\text{Definion of MGF: note }t/2\\ &= (1-t/2)^{-1}~(1-t/2)^{-1} &&\text{MGF of Exponential Distribution}\\ &= (1-t/2)^{-2} &&\text{Exponent Algebra} \end{align}$$

Notes:
Because $\mathsf M_{Y_k}(t)=(1-t)^{-1}$ when $Y_k\sim\mathcal{Exp}(1)$
Therefore $\mathsf M_{Y_k}(t/2)=(1-t/2)^{-1}$ when $Y_k\sim\mathcal{Exp}(1)$
