I'm learning probability and need help with the following problem :
Let $X_1, X_2$ be independent and identically distributed random variables with probability density function
$$f(x_i) = \begin{cases} \lambda_i e^{\lambda_i x_i}, & x_i > 0 \;\; (i = 1, 2) \\ 0 & \text{elsewhere}.\end{cases}$$
$(1)$ Find the distribution of the random variable $V = \min(X_1, X_2)$.
$(2)$ Show that the random variables $Z = X_1/(X_1+X_2)$ and $X_1+X_2$ are independent.
Since I'm having difficulties for $(2)$, I'm going to share my work for $(1)$.
$(1)$ If two variables are iid, then they must have the same distribution. Which means they have the same parameters. So I assumed $\lambda_1 = \lambda_2$, i.e.
$$X_1, X_2 \overset{idd}\sim exp(rate = \lambda).$$
To find the distribution of the minimum, we look at cdfs. The cdf for the individual $X_i$’s is
$$F(x) = \int_{-\infty}^{x} f(u) \, du = \int_{0}^{x} f(u) \, du = \int_{0}^{x} \lambda e^{-\lambda u} \, du = 1 - e^{-\lambda x}.$$
The cdf for $V$, the minimum, is :
\begin{align*} F_V(v) = P(V \leq v) &= P(\min(X_1, X_2) \leq v) \\ \\ &= 1 - P(\min(X_1, X_2) > v) \\ \\ &\overset{indep}= 1 - P(X_1 > v) \cdot P(X_2 > v) \\ \\ &\overset{ident}= 1 - [P(X_1 > v)]^n. \end{align*}
From are computed cdf, we have that
$$P(X_1 > v) = 1 - P(X_1 \leq v) = 1 - (1 - e^{-\lambda v}) = e^{-\lambda v}.$$
So, the cdf for the minimum $V$ is
$$F_V(v) = 1 - [e^{-\lambda v}]^2 = 1 - e^{-\lambda 2 v}.$$
The pdf for the minimum is
$$f_V(v) = F'_V(v) = 2\lambda e^{-2\lambda v}.$$
So we proved that $$V \sim exp(rate = 2\lambda).$$
Is my work correct for $(1)$? Unfortunately I have no idea how to solve $(2)$. Any help would be appreciated.