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Suppose that $X$ and $Y$ have moment generating functions $M_X(t),M_Y(t)$ respectively and $U$ has uniform distribution on $[0,1]$.

What is a random variable that is a function of either X,Y, and or Z such that it's moment generating function is 1) $\int_0^1 M_X(tu) \, du$ and 2) $\frac{M_X(t)+M_Y(t)}2$?

I know the linear properties of moment generating functions, such as $M_U(t) = \int_0^1 e^{ut}f(u)\,du$ and the properties $M_{X+Y}(t) = M_X(t)M_Y(t)$ as well as $M_{aX+b}(t) = e^{bt} M_X(at)$ but don't know how other properties would apply. Would the product of UV yield a moment generating function corresponding to $\int_0^1 M_X(tu) \, du$ Any help would be really appreciated!

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    $M_{X+Y}(t) = M_X(t)M_Y(t)$ if $X,Y$ are independent, but not always if they're not. But the question refers to $(M_X(t) + M_Y(t))/2,$ and which function that is does not depend on whether $X,Y$ are independent. – 2017-02-01
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    @MichaelHardy yes, are there any properties for products of random variables that relate to either $\int_0^1 M_X(tu) \, du$ or $\frac{M_X(t)+M_Y(t)}2$ though? – 2017-02-01

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Suppose $W=\begin{cases} 1 & \text{with probability } 1/2, \\ 0 & \text{with probability } 1/2, \end{cases}$
$\vphantom{\dfrac11}$and $Z=WX + (1-W)Y,$ so that $Z=X$ if $W=1$ and $Z=Y$ if $W=0.$ Then $$ M_Z(t) = \operatorname{E}(e^{tZ}) = \operatorname{E}(\operatorname{E}( e^{tZ}\mid W)). $$ Oberserve that $$ M_{Z\,\mid\, W=1}(t) = \operatorname{E}(e^{tZ}\mid W=1) = \operatorname{E}(e^{tX}) \\ M_{X\,\mid\,W=0}(t) = \operatorname{E}(e^{tZ}\mid W=0) = \operatorname{E}(e^{tY}) $$ so $$ \operatorname{E}(e^{tZ}\mid W) = \begin{cases} \operatorname{E}(e^{tX}) & \text{with probability } 1/2, \\ \operatorname{E}(e^{tY}) & \text{with probability } 1/2. \end{cases} $$ Therefore $$ \operatorname{E}(\operatorname{E}(e^{tZ}\mid W)) = \frac 1 2 \operatorname{E}(e^{tX}) + \frac 1 2 \operatorname{E}(e^{tY}) = \frac{M_X(t) + M_Y(t)} 2, $$ so $$ M_Z(t) = \frac{M_X(t) + M_Y(t)} 2. $$

The question involving $\displaystyle \int_0^1 M_X(tu)\,du$ is a bit subtler. Note that $$ M_X(tu)=\operatorname{E}(e^{tXu}) = \operatorname{E}(e^{tXU} \mid U=u), $$ so $$ M_X(tU) = \operatorname{E}(e^{tXU} \mid U). $$ It follows that \begin{align} \int_0^1 M_X(tu) \,du = \operatorname{E}(M_X(tU)) = \operatorname{E}(\operatorname{E}(e^{tXU} \mid U)) = \operatorname{E}(e^{tXU}) = M_{XU}(t). \end{align}