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 $(0,1)$.

Answer 2

You obtain u(x) by integrating f(x). Note that d/dx(x-1)$^2$/4= (x-1)/2. So the probability integral transformation Y=u(X) is uniform on [0,1] where X has the density f.