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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?

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Check change of variable and Jacobian in your lecture notes, to see where you misapplied these. – Did May 26 '14 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 '14 at 7:54

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.

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