If $B\sim Beta(j,k)$, why is $-logB \sim X_{(k)}$ where $X_{(k)}$ is the $k$-th order statistic of $X_1, \ldots, X_n$ are iid Expo rv's?

Question

Suppose that $B\sim Beta(j,k)$ with $j,k$ positive integers. I read inside a book that we can represent $B \sim U_{(j)}$, where $U_1, \ldots, U_n$ are iid $Unif(0,1)$ and $n = j+k-1$.

Then I am told that (what I do not understand at all):

$$ -logB \sim X_{(k)} $$

where $X_1, \ldots, X_n$ are iid Expo(1).

The book states that this is "because $-log$ is a decreasing function, so that $U_{(1)}$ corresponds to $X_{(n)}$, $U_{(2)}$ corresponds to $X_{(n-1)}$, etc"

Could anyone shed light on why this is true? Thanks!

Answer 1

First, notice that the negative log of a uniform random variable is an exponential random variable: $$P(U\le u) = u$$ $$P(X \ge x) = P(-\log(U) \ge x) = P(U \le e^{-x}) = e^{-x}$$

where $U$ is uniform and $X = -\log(U).$

Now, imagine transforming a finite sequence of independent uniforms $U_i$ to a sequence of independent exponentials $X_i = -\log(U_i).$ Negative log is a decreasing function, so the largest of the $U_i$ (i.e. $U_{(n)}$) is going to turn into the smallest of the exponentials ($X_{(1)})$. And the smallest $U_i$ is going to turn into the largest of the exponentials. The exponentials will have the reverse order in size from the parent uniforms.