I need help or some advice to solve this exercise about functions random variables.

Question

I've thought about it for a while using the cumulative distribution, but I've not concluded.

Three people $A$, $B$ and $C$ arrive at the same time at a telephone booth that has two telephone sets. Both devices are used immediately by A and B. Person C replaces the first person to end their call and each person leaves the telephone booth after their call is finished. Let $X_1$, $X_2$ and $X_3$ be the connection times of $A$, $B$ and $C$, respectively. Suppose that $X_1$, $X_2$ and $X_3$ are independent and identically distributed random variables with exponential distribution with parameter $\lambda$. Determine the probability density function of $Z = max \{X_1, X_2\} - min \{X_1, X_2\}$ and calculate $P(Z<X_3)$.

Answer 1

I will try to give you a start: $X_1, X_2 \stackrel{iid}{\sim}\mathsf{Exp}(\lambda).$ Two 'parameterizations' of exponential distributions are in common use: rate and mean. I will assume $\lambda$ is the rate (so that the mean is $1/\lambda.$

Then let $W = \max(X_1,X_2).$ Then $$F_W(t) = P(W \le t) = P(X_1\le t,\,X_2\le t) = P(X_1\le t)P(\,X_2\le t) =(1 - e^{-\lambda t})^2,$$ for $t > 0,$ from which you can find $F_W(t)$ the density function $f_w(t)$. and $E(W).$

Somewhat similarly, for $V = \min(X_1,X_2),$ you have $$1 - F_V(t) = P(V > t) = P(X_1 > t, X_2 > t),$$ from which you can deduce that $V \sim \mathsf{Exp}(2\lambda).$

Here is a simulation for $\lambda = 2,$ which gives good approximations of the distributions of $W,V,$ and $Z.$ You can check your exact analytic (and perhaps intuitive) answers against these approximate ones. [With a million iterations, it is reasonable to expect about three-place accuracy.]

set.seed(111)
x1 = rexp(10^6, 2);  x2 = rexp(10^6, 2)
v = pmin(x1, x2);  w = pmax(x1, x2)
z = w - v

summary(v); sd(v)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.000001 0.071874 0.173258 0.250102 0.346770 3.383253 
[1] 0.2505        # both mean and sd exactly 1/4
summary(w); sd(w)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.000295 0.346604 0.613705 0.750089 1.004757 7.450897 
[1] 0.5595383     # not exponential, mean not = sd
summary(z); sd(z)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
 0.0000  0.1435  0.3463  0.5000  0.6927  7.0444 
[1] 0.5001754

par(mfrow=c(1,3))
 hist(v, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 4), add=T, col="red", lwd=2)
 hist(w, prob=T, br=40, col="skyblue2")
 hist(z, prob=T, br=40, col="skyblue2")
  curve(dexp(x, 2), add=T, col="red", lwd=2)
par(mfrow=c(1,1))

Finally, what is the (non-exponential) distribution of the sum of two exponential distributions with different rates? Consider these additional simulation results:

x3 = v + rexp(10^6, 2)
summary(x3); sd(x3)
    Min.  1st Qu.   Median     Mean  3rd Qu.     Max. 
0.000316 0.346711 0.614066 0.750727 1.006107 7.288232 
[1] 0.5598703
mean(z < x3)
[1] 0.66665   # aprx P(Z < X3)
mean(w > x3)
[1] 0.500151  # aprx P(W > X3)