Finding a density function

Question

As part of an exercise I was given the following question and would love if anyone could help me.

Assume we raffle $X$~$U(0,1)$ a random variable uniform and continuous in the interval $(0,1)$ and then we compute $Y$ by $$Y = \frac{\alpha}{X^\frac{1}{\beta}}$$

a. Find the density function of Y (PDF)

b. Find $P(Y\geq3\alpha | Y\geq2\alpha )$

Answer 1

Note that $Y=\frac{\alpha}{X^{1/\beta}} \;$ implies $y \geq \alpha$.

To calculate the cdf of $Y$,

$Pr(Y \leq y)= Pr(\frac{\alpha}{X^{1/\beta}} \leq y)=Pr(X \geq e^{-\frac{1}{\beta} log(\frac{y}{\alpha})})$

Then, since it is given that $X \sim U(0,1)$

$F_Y(y)=Pr(X \geq e^{-\frac{1}{\beta} log(\frac{y}{\alpha})}) = 1 - e^{-\frac{1}{\beta} log(\frac{y}{\alpha})}$, so that the pdf of $Y$ is given by

$f_Y(y)= \frac{1}{\alpha \beta} (\frac{y}{\alpha})^{-\frac{1}{\beta}-1} \;$ for $y \geq \alpha$

Answer 2

1. Plug in your definition for $Y$ in the term $P(Y\leq y)$
2. Rewrite the expression to derive at the form $P(X \leq \dots)$
3. Use the definition of the uniform distribution, you may need to distinguish the cases where the argument is larger\less than 1
4. Take the derivative w.r.t. $y$