Let $X \sim \text{exp}(\lambda)$, $Y \sim \text{exp}(\mu)$ and $Z \sim \text{exp}(\gamma)$ be independent r.v.'s where $\lambda$ and $\mu$ are the rates. Let $U = \min(X, Y)$ and $V = \min(Y, Z)$.
(a) Find $P(U > u, V > v)$ for $u, v > 0$ by considering the cases $u > v$ and $ u < v$ separately.
(b) Differentiate what you get in (a) for a "joint density"
(c) Show the double integral over all $(u, v)$ is $(\lambda + \gamma)$/$(\lambda + \mu + \gamma)$. Explain why it doesn't equal one. Explain why.
(d) Find P(U = V) and say how this explains (c).
(e) Find Cov(U, V)
(a) I got this one (Distribution with three exponential random variables). It's:
$P(U > u, V > v) = e^{-\lambda u - \mu v - \gamma v}$ if $u < v$ and $e^{-\lambda u - \mu v - \gamma v}$ if $v < u$.
(b)
I did two derivative with u and v and got $\lambda(\mu + \gamma) e^{-\lambda u - \mu v - \gamma v}$ if $u < v$ and also $\gamma(\lambda + \mu)e^{-\lambda u - \mu u - \gamma v}$ if $v > u$.
Are these ok ?
(c)
I think I need to do this:
$$\int_{0}^{\infty} \int_{0}^{v} \lambda(\mu + \gamma) e^{-\lambda u - \mu v - \gamma v} dudv + \int_{0}^{\infty} \int_0^u \gamma(\lambda + \mu)e^{-\lambda u - \mu u - \gamma v} dvdu$$
Is this right?
(d) I don't know how to find P(U = V). But I think it explain (c) because we do not have P(U = V) in our density. So my guess is P(U = V) + the integral gotten in c = 1.
(e) Find Cov(U, V). How I can get E(UV) for this one?
Thanks
Your computations for (a), (b), and (c) are correct. For (d), we have $$ \mathbb P(U=V) = \mathbb P(X\wedge Y=Y\wedge Z) = \mathbb P(Y < X\wedge Z) = \frac\mu{\lambda+\mu+\gamma}. $$ Your guess is correct; this quantity plus the integral over the "density" is equal to one.
I am not sure how to compute $\mathbb E[UV]$; when I tried I got a really nasty expression and my intuition leads me to believe that it should be "nice."