Deriving probability density function

Question

I'd like to derive the probability density of $\eta:=e^{-\log(\xi)}$ for $\xi$ being uniformly distributed on $[0,1]$.

I tried to figure it out by calculating according to

https://www.cl.cam.ac.uk/teaching/2003/Probability/prob11.pdf

which has led to $f_\eta(y)=\frac{1}{y^2}$ but I'm not sure if this is right and if so, which distribution it represents.

Answer 1

As Kavi Rama Murthy commented, $\eta = \frac1{\xi}$

$F_\eta(x) = \mathbb P(\eta \le x) = \mathbb P(\xi \ge \frac1x) = 1-\frac1x$, so $f_\eta(x)=\frac1{x^2}$ as you found

Note that your density is for $x \ge 1$, since $0 \le \frac1x\le 1$

You have a Pareto distribution with shape parameter $\alpha=1$ and minimum $1$

Answer 2

$\eta=\xi^{-1}$ so that for $x\geq1$ we find: $$P(\eta\leq x)=P(\xi\geq x^{-1})=1-x^{-1}$$ Differentiating this CDF we find a corresponding PDF prescribed by:$$x\mapsto x^{-2}\text{ if }x\geq1\text{ and }x\mapsto0\text{ otherwise}$$

Answer 3

As per the fact that $f_{\xi}=1$ to derive $f_{\eta}$ it is enough to calcualate the derivative

$$f_{\eta}=\Bigg|\frac{d \xi}{d \eta}\Bigg|=\frac{1}{\eta^2}$$

Obviously being $\xi \in (0;1]$ it is also $\eta \in [1;+\infty)$