Statistics random variables

Question

I'm having trouble with the following example:

If $a, b$ are functions of $\theta$ and $X\sim U(0,\theta)$, show that $Y=-\log(X)\sim E(a,b)$.

Answer 1

HINT

Note that $$ F_Y(y) = \mathbb{P}[Y \le y] = \mathbb{P}[-\ln X \le y] = \mathbb{P}[X \ge e^{-y}] = 1 - F_X(e^y). $$ Hence $$ f_Y(y) = F'_Y(y) = -f_X(e^y) e^y. $$ Can you substitute $f_X(y)$ and recognize in the final form what $a$ and $b$ have to become to make this work?

Answer 2

Alternative hint: This is just the inverse transformation method on $F(y) = \exp(-y)$:

\begin{align} P(Y \le y) &= F(y) = \exp(-y). \end{align}

Since the cumulative distribution of $Y$ is exponential, by Portemanteau Theorem, the density function of $Y$ is exponential.

Question: Why is $F(y) = \exp(-y)$ chosen?
Answre: Observe that the inverse function $F^{-1}$ encloses r.v. $X \sim \operatorname{unif}(0,\theta)$ in the proof of correctness of the linked wiki page. Therefore, $F^{-1}(X) = -\log(X) = Y$, so $F(Y) = \exp(-Y)$.