Show that a $-\log$ transformation of a Pareto distribution is exponentially distributed

Question

Question:

Given that $y$ is distributed as:

$$ f(y; \theta) = \theta y^{(\theta-1)} $$

$$0<y<1 , \theta>0$$

If $Z = -\log(Y)$, show that $Z$ has an exponential distribution (ie $E(Z) = 1/\theta$).

My Working:

$Y = e^{-z}$

$f(z; \theta) = \theta e^{-z(\theta - 1)}$

However I can't seem to get that into the standard exponential form of: $$\lambda e^{-z\lambda}$$

The question states that the fact for the gamma random variable $X$, the following may be useful:

$$E\left(\frac{1}{X}\right) = \frac{1}{\beta ( \alpha - 1 )}$$

My other avenue of thought was that to find the expected value of a continuous variable, the following is used:

$$E(Z) = \int z f(z) dz$$

When I use that on the function I derived ($\theta e^{-z(\theta - 1)}$) using the support $-\log(0)$ to $-\log(1)$, ie $0$ to infinity, i don't get the correct answer.

Do I need to make some sort of transformation of my function to get it into the standard exponential form?

Answer 1

For a one-to-one transformation $Z = g(Y)$ of a univariate distribution, the relevant formula is $$f_Z(z) = f_Y(g^{-1}(z)) \left| \frac{dg^{-1}}{dz} \right|.$$ Therefore, if $g(x) = - \log x$, the density of $Z$ easily follows.

Answer 2

$$f(y; \theta) = \theta y^{(\theta-1)}$$

Calling $Z=-\log Y$, we aim to show that $Z$ is exponentially distributed.

We recall that $Z$ is exponential with mean $\mu$ iff $\mathbb{E}\exp(tZ)=\frac{1}{1-\mu t}$.

Now:

$$\mathbb{E}\exp(tZ)=\mathbb{E}\exp(t(-\log Y))=\mathbb{E}Y^{-t}$$

$$=\int_0^1 \theta y^{(\theta-1)}y^{-t}\,dy=\int_0^1 \theta y^{\theta-t-1}\,dy=\frac{\theta}{\theta-t} \text{ for }(\theta-t-1)>-1$$

$$=\frac{1}{1-\frac{1}{\theta}t}$$

as sought.