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.

$$\begin{align} f_{\min\{Z_1,Z_2\}}(z) & =\int_z^\infty f_S(s)\cdot f_{\min\{Z_1,Z_2\}\mid S}(z)~\mathrm d s \\[1ex] &= \int_z^\infty f_S(s)\cdot 2!~f_{Z_\star\mid S}(z\mid s)~\big(1-F_{Z_\star\mid S}(z\mid s)\big)~\mathrm d s \end{align}$$


$$\begin{align} f_{\max\{Z_1,Z_2\}}(z) & =\int_z^\infty f_S(s)\cdot f_{\max\{Z_1,Z_2\}\mid S}(z)~\mathrm d s \\[1ex] &= \int_z^\infty f_S(s)\cdot 2!~f_{Z_\star\mid S}(z\mid s)~F_{Z_\star\mid S}(z\mid s)~\mathrm d s \end{align}$$

The rest is left to you.

  • $\begingroup$ PS: The sum of $n$ identical and independent Exponential Distributed random variables follows what kind of distribution ? $\endgroup$ – Graham Kemp Jul 23 '19 at 2:19
  • $\begingroup$ It has Gamma distribution. I don't familier with rules for order statistics, so I think the solution can't use them. I don't see how the second equality is justified (for both the minimum and maximum). $\endgroup$ – user3708158 Jul 23 '19 at 6:30
  • $\begingroup$ The minimum of $Z_1,Z_2$ is equal to $z$ iif one of them equals $z$ and the other is at least $z$. $$\{\omega: \min\{Z_1(\omega),Z_2(\omega)\}=z\}=\{\omega: (Z_1(\omega)=z\land Z_2\geqslant z)\lor(Z_1>z\land Z_2=z)\}$$ and since they are conditionally independent with respect to the sum $S$, then$$f_{\min\{Z_1,Z_2\}\mid S}(z\mid s)=f_{Z_1\mid S}(z\mid s)\mathsf P(Z_2\geqslant z\mid S{=}s)+\mathsf P(Z_1\gt z\mid S{=}s)f_{Z_2\mid S}(z\mid s)$$Finally, since they are identically distributed:$$f_{\min\{Z_1,Z_2\}\mid S}(z\mid s)=2~f_{Z_\star\mid S}(z\mid s)~\big(1-F_{Z_\star\mid S}(z\mid s)\big)$$ $\endgroup$ – Graham Kemp Jul 23 '19 at 9:13
  • $\begingroup$ By similar argument. $$f_{\max\{Z_1,Z_2\}\mid S}(z\mid s)~=~2~f_{Z_\star\mid S}(z\mid s)~F_{Z_\star\mid S}(z\mid s)$$ $\endgroup$ – Graham Kemp Jul 23 '19 at 9:18
  • $\begingroup$ Thanks. Can you help me figure out where the $3$ is coming from? See edited question. $\endgroup$ – user3708158 Jul 23 '19 at 12:19

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