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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?

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

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    @user43126: You’re very welcome.2012-09-30
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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 $f$.

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    @DilipSarwate thank you for making that point.2012-09-30