Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 3 down vote accepted

$F_X(x)$ is the cumulative distribution function of $X$, given by


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 $(0,1)$.

share|cite|improve this answer
Thank you so very much! – user43126 Sep 30 '12 at 1:31
@user43126: You’re very welcome. – Brian M. Scott Sep 30 '12 at 1:36

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.

share|cite|improve this answer
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. – Dilip Sarwate Sep 30 '12 at 0:45
Thank you so very much! – user43126 Sep 30 '12 at 1:30
@DilipSarwate thank you for making that point. – Michael Chernick Sep 30 '12 at 1:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.