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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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  • $\begingroup$ @user43126: You’re very welcome. $\endgroup$ – Brian M. Scott Sep 30 '12 at 1:36
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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.

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    $\begingroup$ Please, at least remind the student that when you say integral you really do mean the integral as in area under the curve instead of just the antiderivative. Your showing that the density in question is the derivative of $\frac{1}{4}(x-1)^2$ almost invites a beginner to use antiderivatives instead of integrals, a very common mistake made by those just learning the subject. $\endgroup$ – Dilip Sarwate Sep 30 '12 at 0:45
  • $\begingroup$ @DilipSarwate thank you for making that point. $\endgroup$ – Michael Chernick Sep 30 '12 at 1:43

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