Solved – How to evaluate R = max(X1,X2) – min(X1,X2)

How would you evaluate this if X1, X2 are independent exponentials?

I know what $min(X1,X2)$ is and what $max(X1,X2)$ is. However, while $min(X1,X2)$ is exponential with parameter (lambda1 + lambda2) — the max is not.

Is the only way to proceed is to say:

1) Suppose $X_1 > X_2$. Then I find $Pr(R <= r) = Pr(X_1 – X_2 <= r)$ by doing a double integral.
2) Suppose $X_2 <X_1$. Then I find $Pr(R <= r) = Pr(X_2 – X_1 <= r)$ by doing a double integral.
3) Calculate the CDF as:

$Pr(X_1 – X_2 <= r) Pr(X_2 <= X_1) + Pr(X_2 – X_1 <= r)*Pr(X_1 <= X_2)$.

That should give me $Pr(R leq r)$. I then would then differentiate with respect to r to get $F_R(r)$.

Is there perhaps a more clever way to do it?


For any $alpha > 0$, $P{X > alpha} = e^{-lambda alpha}, P{Y > alpha} = e^{-mu alpha}$ and since $X$ and $Y$ are independent, these are also the values of the conditional probabilities that $X$ and $Y$ exceed $alpha$ given the value of the other random variable. Consequently, $$begin{align} P{R > z}&= P{max(X,Y) – min(X,Y) > z}\ &= P{X > Y+z} + P{Y > X+z}\ &= int_0^infty P{X > y+z mid Y = y}f_Y(y),mathrm dy + int_0^infty P{Y > x+z mid X = x}f_X(x),mathrm dx\ &= int_0^infty e^{-lambda(y+z)}cdotmu e^{-mu y},mathrm dy + int_0^infty e^{-mu(x+z)}cdotlambda e^{-lambda x},mathrm dx\ &= e^{-lambda z}int_0^infty mucdot e^{-(lambda+mu)y},mathrm dy + e^{-mu z}int_0^infty lambda cdot e^{-(lambda + mu)x},mathrm dx\ end{align}$$ Can you take it from here to get $P{R > z} = 1-F_R(z)$ and thus determine the density of $R$? You will get what is called a mixture of two exponential densities.

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