Let $W_1, W_2, W_3\sim Exp(s)$ independent RV's. Define $S=W_1+W_2+W_3$. Let $Z_1,Z_2$ be independent RV, not neceserrly with the same distribution, such that $Z_1,Z_2\mid S \sim Uniform([0,S])$. I need to show that $\min(Z_1,Z_2)\sim Exp(s) \sim W_1$, and that $\max(Z_1,Z_2)\sim W_1 +W_2$. I tried to use analytic approach by using $f_{x\mid Y=y}=\frac{f_{x,y}}{f_y}$, and use the law of total probability, but did not succeeded. Any ideas?
Following Graham Kemp approach:
$ f_{\min\left\{ Z_{1},Z_{2}\right\} }\left(z\right) =\int_{z}^{\infty}f_{S}\left(s\right)\cdot2\cdot f_{Z_{1}\mid S}\left(z\mid s\right)\cdot\left(1-F_{Z_{1}\mid S}\left(z\mid s\right)\right)ds= \int_{z}^{\infty}f_{S}\left(s\right)\cdot2\cdot\frac{1}{s}\cdot\left[\left(1-\frac{z}{s}\right)\cdot1_{z\leq s}\right]ds= \int_{z}^{\infty}\frac{\lambda^{3}s^{3-1}\cdot e^{-\lambda s}}{\Gamma\left(3\right)}\cdot2\cdot\frac{1}{s}\cdot\left[\left(1-\frac{z}{s}\right)\cdot1_{z\leq s}\right]ds= \int_{z}^{\infty}\frac{\lambda^{3}s^{3-1}\cdot e^{-\lambda s}}{\Gamma\left(3\right)}\cdot2\cdot\frac{1}{s}\cdot\left(\frac{s-z}{s}\right)ds= \int_{z}^{\infty}\frac{\lambda^{3}\cdot e^{-\lambda s}}{\Gamma\left(3\right)=3!}\cdot2\cdot\left(s-z\right)ds= \frac{\lambda^{3}}{3}\int_{z}^{\infty}e^{-\lambda s}\cdot\left(s-z\right)ds= \frac{\lambda^{3}}{3}\left[\int_{z}^{\infty}se^{-\lambda s}ds-z\int_{z}^{\infty}e^{-\lambda s}ds\right]= \frac{\lambda^{3}}{3}\cdot-1\cdot\left[\frac{e^{-\lambda z}\cdot\left(-\lambda z+\lambda z-1\right)}{\lambda^{2}}\right]=\frac{\lambda}{3}\cdot e^{-\lambda z} $
Any ideas where the $3$ is coming from?
The key is that the two $Z_\star$ distributions should be conditionally independent given $S$.
So, first use the Law of Total Probability, then use the rules for order statistics.
$$\begin{align} f_{\min\{Z_1,Z_2\}}(z) & =\int_z^\infty f_S(s)\cdot f_{\min\{Z_1,Z_2\}\mid S}(z)~\mathrm d s \\[1ex] &= \int_z^\infty f_S(s)\cdot 2!~f_{Z_\star\mid S}(z\mid s)~\big(1-F_{Z_\star\mid S}(z\mid s)\big)~\mathrm d s \end{align}$$
Likewise
$$\begin{align} f_{\max\{Z_1,Z_2\}}(z) & =\int_z^\infty f_S(s)\cdot f_{\max\{Z_1,Z_2\}\mid S}(z)~\mathrm d s \\[1ex] &= \int_z^\infty f_S(s)\cdot 2!~f_{Z_\star\mid S}(z\mid s)~F_{Z_\star\mid S}(z\mid s)~\mathrm d s \end{align}$$
The rest is left to you.