Let $X$ and $Y$ be independent exponential random variables of parameters $\lambda$ and $\mu$ respectively. In other words, their density functions are: $$ f_X(x) = \lambda e^{-\lambda x} \\ f_Y(y) = \lambda e^{-\lambda y} $$ Then we define $$ U = min (X, Y) \quad V = 1_{U=X} $$ I want to find the joint law of (U, V). This is what I have got so far. I want to find $F(u, v) = p(U\leq u, V \leq v)$
First Case: $v < 0 \implies p(V \leq v) = 0 \implies F(u, v) = 0$
Second Case: $0 \leq v < 1 \implies p (V \leq v) = p (V = 0) = p(X>Y)$ $$ p(X>Y) = \int_0^ \infty \int_0^x \lambda \mu e^{-\lambda x} e^{-\mu y} dy dx = \int_0^\infty \lambda e^{-\lambda x} (1 - e^{-\mu x})dx = \int _0^\infty \lambda e^{-\lambda x} - \lambda e^{-(\lambda + \mu)x} = 1 - \frac{\lambda}{\lambda + \mu} = \frac{\mu}{\lambda + \mu} $$ Then, using that $p(A \cap B) = p(A|B)p(B)$: $$ p(U\leq u, V \leq v) = p(U\leq u|X>Y)p(X>Y) = \frac{\mu}{\lambda + \mu} p(Y\leq u) = \frac{\mu}{\lambda + \mu} \int_0^u \mu e^{-\mu y} dy = \frac{\mu}{\lambda + \mu} (1-e^{-u\mu}) $$ Third case: $v \geq 1 \implies p(V \leq v) = 1$. In this case the V random variable is not adding any information $$ p(U\leq u, V \leq v) = p(U \leq u) = p(X \leq u \cup Y \leq u) = \\ p(X\leq u) + p(Y \leq u) - p(X\leq u \cap Y \leq u) = \\ =(1-e^{-u\mu}) + (1-e^{-u\lambda}) - (1-e^{-u\mu})(1-e^{-u\lambda}) = \\ =2 - e^{-u\mu} - e^{-u\lambda} - 1 +e^{-u\lambda}+e^{-u\mu} - e^{-u\mu}e^{-u\lambda} = \\ = 1- e^{-u\mu}e^{-u\lambda} $$
I do not know if any of this is right. But then I am asked if $U$ and $V$ are independent. I do not know if I am just stuck or I have done something wrong in the previous steps. Any help would be appreciated. Thanks in advance : )