Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.


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 cant 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:

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

My other avenue of thought was that to find the expected value of a continious 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 dont get the correct answer.

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

share|improve this question
Check change of variable and Jacobian in your lecture notes, to see where you misapplied these. –  Did May 26 at 7:29
Thanks. So: $$F_z(z) = P(Z <= z) = P(-log(Y) <= z)$$ $$F_z(z) = P(Y <= e^{-z})$$ $$F_z(z) = F_y(e^{-z})$$ Therefore $$f_Z(z) = \frac{d}{dz} F_Y(e^{-z})$$ $$f_Z(z) = -e^{-z} f_Y(e^{-z}) $$ $$f_Z(z) = -e^{-z} \theta (e^{-z})^{\theta - 1}$$ $$f_Z(z) = \theta e^{-z \theta} $$ Thus now in exponential form! –  James May 26 at 7:54

1 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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.