[Math] monotone functions and pdf’s

Question

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?

Answer 1

$F_X(x)$ 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 $(0,1)$.