Variance of max - min of 2 exponential random variables

Question

Suppose we have 2 random variables, $S\sim Exp(\lambda)$ and $T\sim Exp(\mu)$. Let $U=\min(S,T)$ and $V=\max(S,T)$.

What is the variance of $W=V-U$?

I calculated:

$Var(U) = \frac{1}{(\lambda+\mu)^2}$
$Var(V) = \frac{1}{\lambda^2} + \frac{1}{\mu^2} + \frac{1}{(\lambda+\mu)^2}$

But I don't know how to proceed further.

Answer 1

Note that $$W = V - U = |S - T|.$$ Then note $$\operatorname{Var}[W] = \operatorname{E}[|S-T|^2] - \operatorname{E}[|S-T|]^2.$$ if $S$ and $T$ are independent, then $$\operatorname{E}[|S-T|] = \int_{s=0}^\infty \int_{t=0}^s (s-t) f_S(s) f_T(t) \, dt \, ds + \int_{s=0}^\infty \int_{t=s}^\infty (t-s) f_S(s) f_T(t) \, dt \, ds.$$ This turns out to be, for scale parameters $\operatorname{E}[S] = \lambda$, $\operatorname{E}[T] = \mu$, $$\operatorname{E}[|S-T|] = \frac{\lambda^2 + \mu^2}{\lambda + \mu}.$$ Then the rest is trivial, since $|S-T|^2 = (S-T)^2 = S^2 - 2ST + T^2$: $$\operatorname{Var}[W] = 2\lambda^2 - 2 \lambda \mu + 2 \mu^2 - \left(\frac{\lambda^2 + \mu^2}{\lambda+\mu}\right)^2 .$$

Answer 2

Not to discourage posting an analytic solution, but here's a simulation with rates $\lambda = 3,\; \mu = 5$, which confirms one of your results and casts doubt on the other.

 lam = 3;  mu = 5 # rates, not means
 s = rexp(10^6, lam);  t = rexp(10^6, mu)
 mean(s);  sd(s)
 ## 0.3332149     # approx E(S) to 3 places
 ## 0.3326488     # approx SD(S)
 mean(t);  sd(t)
 ## 0.2000782
 ## 0.2001233
 u = pmin(s,t);  v = pmax(s,t);  w = v-u 
 mean(u);  sd(u)
 ## 0.1248017     # U ~ Exp(rate=8);  E(U) = 1/8
 ## 0.1247771     # SD(U) = E(U)
 mean(v);  sd(v)  # maximum V not exponential
 ## 0.4082914   
 ## 0.3221837     # far from sqrt(1/9 + 1/25 + 1/64) = .4082
 mean(w);  sd(w)  # difference W not exponential
 ## 0.2835815
 ## 0.2976735

In the histograms below the simulated values of $U$ and $W$ approximate their distributions; vertical dotted line indicates $E(U).$