monotone functions and pdf's Could you please show me step by step? Also how does the probability integral transformation come into play? "If the random variable $X$ has pdf
$$
f(x)=
\begin{cases}
\tfrac{1}{2}(x-1)\quad \text{if }1< x< 3,\\
 0 \qquad\qquad\;\, \text{otherwise},
\end{cases}
$$
then find a monotone function $u$ such that random variable $Y = u(X)$ has a uniform $(0,1)$ distribution." The answer key says "From the probability integral transformation, Theorem 2.1.10, we know that if $u(x) = F_X(x)$, then $F_X(X)$ is uniformly distributed in $(0,1)$. Therefore, for the given pdf, calculate
$$
u(x) = F_X(x) =
\begin{cases}
0 \qquad\qquad \;\,\text{if }  x\leq 1,\\
\tfrac{1}{4}(x − 1)^2 \quad \text{if  }1 < x < 3, \\
1 \qquad\qquad\;\, \text{if } x\geq 3.
\end{cases}
$$
But what does this mean?
 A: $F_X(x)$ is the cumulative distribution function of $X$, given by
$$F_X(x)=\int_{-\infty}^xf(t)~dt\;.$$
Clearly this integral is $0$ when $x\le 1$. For $1\le x\le 3$ it’s
$$\begin{align*}
\int_{-\infty}^xf(t)~dt&=\int_{-\infty}^10~dt+\int_1^x\frac12(t-1)~dt\\\\
&=0+\frac12\left[\frac12(t-1)^2\right]_1^x\\\\
&=\frac14(x-1)^2\;,
\end{align*}$$
and for $x\ge 3$ it’s
$$\begin{align*}
\int_{-\infty}^xf(t)~dt7&=\int_{-\infty}^10~dt+\int_1^3\frac12(t-1)~dt+\int_3^x0~dt\\\\
&=\int_1^3\frac12(t-1)~dt\\\\
&=1\;,
\end{align*}$$
so altogether it’s
$$F_X(x) =
\begin{cases}
0,&\text{if }  x\leq 1\\\\
\tfrac{1}{4}(x − 1)^2,&\text{if  }1 \le x \le 3\\\\
1,&\text{if } x\geq 3\;.
\end{cases}$$
Now your Theorem 2.1.10 tells you that if you set $u(x)=F_X(x)$, then $Y=u(X)$ will be uniformly distributed in $(0,1)$.
A: You obtain $u(x)$ by integrating $f(x)$.  Note that $\frac{d}{dx}\frac{(x-1)^2}{4}= \frac{x-1}{2}$.
So the probability integral transformation  $Y=u(X)$ is uniform on $[0,1]$ where $X$ has the density $f$.
