1

Let X, Y be independent r.v.s each with exponential distribution of parameter $1$. Let $U=\min(X, Y)$ and $V=\max(X, Y)$. Show that $V-U$ is independent of $U$. Hence, show that $V$ has the same distribution as $X+\frac{1}{2}Y$.

It is easy to see that U is exponential with parameter $2$. The first part is answered here:

Let $U=\operatorname{min}\{X,Y\}$ and $V=\operatorname{max}\{X,Y\}$. Show that $V-U$ is independent of $U$.

Not too sure how 'hence' comes into play for the second part (or how to do it without the 'hence').

user112358
  • 149
  • 1
    cf. https://math.stackexchange.com/a/147391/75808 (their $U$ is the same as yours, but their $V$ is your $V-U$). Then use the characteristic function to show that, for all $t$ (with your notation), $$\phi_V(t) = \phi_{V-U}(t)\phi_U(t) = \phi_{Y/2}(t)\phi_X(t) = \phi_{Y/2+X}(t)$$ – Clement C. Apr 08 '19 at 15:43

1 Answers1

2

You've shown $U$ is exponential with parameter $2$. Show that this means $U$ is equal in distribution to $\frac12 Y$.

Then, show that $V-U$ is exponential with parameter $1$, so $V-U$ is equal to $X$ in distribution. Conclude by noting that $V=U+(V-U)$, and $U$ is independent of $V-U$, just like $X$ is independent of $\frac12 Y$.

Mike Earnest
  • 75,930