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If $X$ has the density function

$$ f_\vartheta (x) = \Big \{ \begin{array}{cc} (\vartheta - 1)x^{-\vartheta} & x \geq 1\\ 0 & otherwise \end{array}$$

How can I see that $\log X \sim Exp(\vartheta - 1)$? I had the idea to look at $f(\log x)$ but I think that's not right. Many thanks for your help!

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    Yeah, cave paintings. I'm pretty bad at those, I know... Sorry about that.2012-07-13
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    Sorry that I missed you in chat, and I didn't even really say hi to you. Hope you're okay. Best wishes, Theo2012-07-16
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    @t.b. There is a smoke signal for you. And a cave painting. Not sure you get pinged if I send you a comment but I'm going to try.2014-06-18

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One way is to use the cdf technique. Let $Y = \log X$. Then $$P(Y \leq y) = P(\log X \leq y) = P(X \leq e^y) = \int_1^{e^y} (\vartheta - 1)x^{-\vartheta} dx = \left.-x^{-\vartheta+1}\right|_1^{e^y} = 1-e^{-(\vartheta-1)y}.$$ Differentiating with respect to $y$ yields $(\vartheta-1) e^{-(\vartheta-1)y}$ as the pdf of $Y$, which is also the pdf of an $Exp(\vartheta - 1)$ random variable.

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    Don't you prefer the change of variable method? You know, the one where one computes $E(\varphi(Y))$ for every (bounded measurable) test function $\varphi$. Using cumulative distribution functions where one is given a probability density function seems like a kind of *détour*, going from the PDF of $X$ to the CDF of $X$ to the CDF of $Y$ to the PDF of $Y$, instead of going directly from the PDF of $X$ to the PDF of $Y$. Well, I guess everything works here...2011-02-20
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    @Didier: You're right, of course, that the change of variable method requires fewer steps on this problem. When I was a student I used that method almost exclusively. I've found, though, that the cdf technique seems to make more sense to my probability students, especially when the function defining the transformation isn't one-to-one. So it looks like my emphasizing the cdf technique in the classroom has rubbed off on my practice. Funny how teaching will do that to you. :)2011-02-20
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    Thanks for your very interesting answer. Students did not manage to have this effect on me. Yet... :-)2011-02-20