Let $X$ be a continuous random variable, uniformly distributed on the interval $(0, 1)$.
Let $Y = \sqrt{X}$ and $Z = 1/X$.
Show that:
1. For $0 < a < 1$, $\mathrm{P}(Y<a) = a^{2}$ and so prove that $Y$ has density function $f_{Y}(y) = 2y, \quad (0 < y < 1)$.
2. $Z$ has range space $(1,\infty)$. Also find its density function.
$\mathrm{P}(X < a) = a$ for $0<a<1$.
Now $X = Y^{2}$ so $\mathrm{P}(Y^{2}<a) = a$
But what do I do next?
Follow the hint. We change notation slightly. For $0\lt y\lt 1$, we have $$F_Y(y)=\Pr(\sqrt{X}\le y)=\Pr(X\le y^2)=y^2.$$ Thus for $0\lt y\lt 1$, we have that $f_Y(y)=\frac{d}{dy}F_Y(y)=2y$.
Note that if $y\lt 0$, then $F_Y(y)=0$, and if $y\gt 1$ then $F_Y(y)=1$. Thus if $y\lt 0$ or $y\gt 1$ we have $f_Y(y)=0$.
For $Z$, use the same basic strategy. We have $Z\le z$ if and only if $\frac{1}{X}\le z$. For $z\ge 1$, this is the probability that $X\ge \frac{1}{z}$. But $\Pr(X\ge \frac{1}{z})=1-\frac{1}{z}$.
Thus for $z\ge 1$, we have $F_Z(z)=1-\frac{1}{z}$. For the density function, differentiate. We get $f_Z(z)=\frac{1}{z^2}$ (for $z\ge 1$).
