Determining the distribution of a variable using moments Is this proof right? My only problem in my solution is that my "proof" would be true for the general case and not only when $U ∼ U(0, 1)$ . 
Prompt
Is it possible for $X, Y, Z$ to have the same distribution and satisfy $X =
U(Y +Z),$ where $U ∼ U(0, 1)$, and Y, Z are independent of $U$ and each
other?
Proof
$$M_x (t) = E(e^{tx})$$
$$M_y (t) = E(e^{ty})$$ 
$$M_z (t) = E(e^{tz})$$ 
$Z + Y ∼ M_z(t) + M_y(t) => E(e^{tz}) + E(e^{ty}) = E(e^{ty} + e^{tz})$ . Now we have $U ∼ U(0,1)$ and its moment generating function is denoted as $M_u (t) = E(e^{ut})$. Then $$U(Y + Z) ∼ M_u(t) * (M_y(t) + M_z(t)) = E(e^{tu})E(e^{yt} + e^{zt}) = E(e^{tu} (e^{yt} + e^{zt}))$$ 
However this form is not the same as that of $M_x(t) = E(e^{tx})$, so it is not possible. 
 A: Let us assume that the variables $U\sim \mathcal{U}(0,1)$, and $Y$, $Z$ are i.i.d. continuous random variables independent from $U$. Therefore, the joint probability distribution function (PDF) of the random variable $(U,Y,Z)$ is the product of the PDFs of its components. We consider the function $g: (u,y,z) \mapsto \left(u(y+z),y,z\right)$, and want to determine the PDF $f_{(X,Y,Z)}$ of the variable $(X,Y,Z) = g(U,Y,Z)$. To do so, one can use the change of variable method. Thus, we consider a measurable bounded function $h$, and compute the expected value $\mathbb{E}h(X,Y,Z)$:
\begin{aligned}
\mathbb{E}(h\circ g)(U,Y,Z) &= \int_{\mathbb{R}^3} (h\circ g)(u,y,z)\, f_U(u) f_Y(y) f_Z(z) \,\mathrm{d}u\,\mathrm{d}y\,\mathrm{d}z \, ,\\
&= \int_{\mathbb{R}^3} h\!\left(x,y,z\right) f_U\!\left(\frac{x}{y+z}\right) f_Y(y) f_Z(z) \frac{1}{|y+z|}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \, ,\\
&= \int_{\mathbb{R}^3} h\!\left(x,y,z\right) \underbrace{\mathbf{1}_{0\leq x/(y+z)\leq 1}  \frac{f_Y(y) f_Z(z)}{|y+z|}}_{f_{(X,Y,Z)}(x,y,z)}\,\mathrm{d}x\,\mathrm{d}y\,\mathrm{d}z \, .
\end{aligned}
The condition for $X$, $Y$ and $Z$ to be identically distributed writes $f_X(x) = f_Y(x) = f_Z(x)$ for all $x$, where $f_X$ denotes the marginal PDF of $X$. If $X$, $Y$ and $Z$ are positive, it amounts to
$$
f_X(x) = \int_{\mathbb{R}^2} \mathbf{1}_{0\leq x\leq y+z}  \frac{f_X(y) f_X(z)}{y+z}\,\mathrm{d}y\,\mathrm{d}z \, .
$$
Maybe somebody can deduce something out of it.
