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?

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.
