Joint density of minimum and maximum - minimum.

Question

Let X = min(S, T), Y = max(S, T) for independent exponential ($\lambda$) variables S, T. Let Z = Y - X.

Find the joint density of X and Z. Are they independent? Identify the marginal distributions of X and Z.

Answer 1

Here is an exploration of your problem using simulation in R statistical software, with some additional hints. I used $\lambda = 0.1$, so $E(S) = SD(S) = 10$ and $E(T) = SD(T) = 10.$ For a million iterations of your experiment, here are some findings (correct to about three significant digits).

 s = rexp(10^6, .1);  t = rexp(10^6, .1)
 x = pmin(s,t);  y = pmax(s,t)
 z = y - x
 mean(x);  sd(x)
 ## 5.004163  # aprx E(X)
 ## 5.01231   # aprx SD(X)
 mean(y);  sd(y)
 ## 15.0305   # aprx E(Y)
 ## 11.19881  # aprx SD(Y)
 mean(z);  sd(z)
 ## 10.02633  # aprx E(Z) 
 ## 10.01379  # aprx SD(Z)

Here are histograms of the simulated distributions of $X, Y,$ and $Z$. The density curve of $Exp(2\lambda)$ is a good fit to the histogram of $X.$ The distribution of $Y$ is clearly not exponential [notice the shape of the histogram and the fact that $E(Y) \ne SD(Y)$.] The density curve of $Exp(\lambda)$ is a good fit to the histogram of $Z$.

To derive the distribution of $X,$ notice that its CDF is $$ F_X(u) = 1 - P(X > u) = 1- P(S > u)P(T > u) = 1- e^{-\lambda u}e^{-\lambda u} = 1- e^{-2\lambda u},$$ and differentiate the CDF to find the PDF.

To derive the distribution of $Y,$ notice that its CDF is $$ F_Y(u) = P(Y \le u) = P(S \le u)P(T \le u) = \cdots ,$$ and differentiate the CDF to find the (nonexponential) PDF. As a check, you can see that $E(Y) = \frac{1}{2\lambda} + \frac{1}{\lambda}.$ Roughly, the rationale is that the mean is the sum of the waiting time for the minimum [at which time we start the clock again because of the 'no-memory' property] and we wait another $1/\lambda$ unit of time for the second exponential event to occur.

To find the distribution of $Z$ ponder the intuitive rationale above for $E(Y)$ and @Henry's Hint.

This does not answer everything you asked, but I hope it gives you enough hints and intuition to finish the rest for yourself.