Find the distribution of a transformation 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?
 A: 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.
A: 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.
