# Conditional probability of sum of exponential random variables

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?

• – Graham Kemp Jul 23 '19 at 22:12

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.

• PS: The sum of $n$ identical and independent Exponential Distributed random variables follows what kind of distribution ? – Graham Kemp Jul 23 '19 at 2:19
• 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). – user3708158 Jul 23 '19 at 6:30
• 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:$$f_{\min\{Z_1,Z_2\}\mid S}(z\mid s)=2~f_{Z_\star\mid S}(z\mid s)~\big(1-F_{Z_\star\mid S}(z\mid s)\big)$$ – Graham Kemp Jul 23 '19 at 9:13
• By similar argument. $$f_{\max\{Z_1,Z_2\}\mid S}(z\mid s)~=~2~f_{Z_\star\mid S}(z\mid s)~F_{Z_\star\mid S}(z\mid s)$$ – Graham Kemp Jul 23 '19 at 9:18
• Thanks. Can you help me figure out where the $3$ is coming from? See edited question. – user3708158 Jul 23 '19 at 12:19