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Let $$f_X (x, \theta) = \frac{1}{\theta} x^{\frac{1}{\theta} - 1}, \; x \in (0, 1)$$ find the distribution of: $$Y = - \frac{1}{\theta} \ln X$$ [Solution provided: $Y \sim \mathcal{E}(1)$ ]

I did: $$P(Y = y) = P(- \frac{1}{\theta} \ln X = y) = P( X = e^{-\theta y}),\; y \in (0, +\infty)$$ and so: $$f_Y(y, \theta) = f_X(e^{-\theta y}, \theta) = \frac{1}{\theta} e^{-y(1-\theta)}$$

Am I missing something?

Is this distrubition a $\mathcal{E}(1)$?If so why?

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2 Answers

up vote 4 down vote accepted

If I may quote another answer:

The simplest and surest way to compute the distribution density or probability of a random variable is often to compute the means of functions of this random variable. In the case at hand, one wants to write $\mathrm E(u(Y))$ as $$ \color{blue}{\mathrm E(u(Y))=\int u(y)g(y)\mathrm{d}y}, $$ for every bounded measurable function $u$. Then one can be sure that $g$ is the density of the distribution of $Y$. So, in a way, the functions $u$ play the role of a dummy variable and one wants the equality above to hold for every $u$.

The rest is easy: introducing $a=1/\theta$ for notational convenience and using the fact that for every bounded measurable function $v$, by definition of the distribution of $X$, $$ \mathrm E(v(X))=\int_0^1 v(x)ax^{a-1}\mathrm dx, $$ one gets $$ \mathrm E(u(Y))=\mathrm E(u(-a\log X))=\int_0^1 u(-a\log x)ax^{a-1}\mathrm dx=\int_0^1 u(-a\log x)x^{a}\frac{a\mathrm dx}x. $$ The change of variable $y=-a\log(x)$ yields $y\gt0$, $x^a=\mathrm e^{-y}$ and $\mathrm dy=a\mathrm dx/x$ hence $$ \mathrm E(u(Y))=\int_0^{+\infty} u(y)\mathrm e^{-y}\mathrm dy. $$ This proves that the density $g$ of $Y$ is defined by $g(y)=\mathrm e^{-y}$ if $y\gt0$ and $g(y)=0$ otherwise. In other words, $Y$ is a standard exponential random variable.

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The value of $P(Y=y)$ is irrelevant (for continuous random variables such as $X$ and $Y$, this is identically zero). –  Did May 27 '12 at 10:44
    
Why bring expectations into the picture? You can do this without bringing u(Y) in. The change of variables and the jacobian can be brought in without the function U. I think it is unnecessary complication. The OPs mistake was treating a density like a probability and not accounting for the jacobian of the transformation. –  Michael Chernick May 27 '12 at 18:53
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@Michael Why use expectations? Amongst other reasons, for pedagogical ones: the technique advocated here, because almost entirely automatized, makes more difficult to err, than the various more or less ad hoc approaches which people on this site are too often encumbered with. Sure, other methods do exist--but I have never seen anyone using this one and forgetting the jacobian, for example. Somehow, one is forced by the notations to use the right path. This is nice, don't you think? –  Did May 27 '12 at 19:15
    
Okay. Maybe that is a good reason. Whether or not it teaches better or causes confusion is open for debate. What I really want to know is why so many people are downvoting my response? Did i say something wrong? –  Michael Chernick May 27 '12 at 19:25
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@Michael This last comment of yours makes little sense. You first asked about the maths (and I answered to that) but now you say that what really interests you is the downvotes, please make up your mind! Concerning the downvotes which you do not understand, some quotes might serve as clues: causes confusion, the proper way, confusing the issue, belaboring the point, the correct complete answer... –  Did May 27 '12 at 20:39
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You can't look at point probabilities when dealing with absolutely continuous random variables. Both $X$ and $Y$ are absolutely continuous (i.e. they have a density wrt. the lebesgue measure) and hence $P(X=x)=P(Y=x)=0$ for all $x\in\mathbb{R}$. Instead try looking at $P(Y\leq y)$ and see if it matches the corresponding CDF of a $\mathcal{E}(1)$ distribution. Edited.

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Did not address the question. Okay to point out the mistake but then try to solve the problem if you are going to make this an answer. –  Michael Chernick May 27 '12 at 18:41
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First of all, how can you say that the OP obviously meant $P(Y=y)$ for the density $f(y)$, because obviously it went over my head? Second, I did give him a hint on how to solve his problem. And since when do you need to give a full solution in order to make an answer? I've seen plenty of good answers only including hints, and as the OP seemed to have no problem doing the calculations him-/herself, and chose only to give that hint. –  Stefan Hansen May 27 '12 at 18:49
    
My answer completely answers the question. It does not clculate the answer for the OP. As you said the solution to the problem is not what is asked for. Didier did more than answer the question he gave the solution. But the answer to the question is to explain what the OP did wrong. Treating the density as a probability was one mistake forgetting the Jacobian was the other. You didn't explain everything he did wrong. –  Michael Chernick May 27 '12 at 19:32
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Alright, I just thought that downvotes was for answers that are wrong/false and not for answers that does not meet your criterias. You could have pointed this out without downvoting. But what the heck... –  Stefan Hansen May 27 '12 at 19:58
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Upvoted. $ $ $ $ –  Did May 27 '12 at 20:40
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