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?