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

$X\sim U(1,3)$. Verify that X has cdf $F_X(x) = 2(x − 1)$ for $x \epsilon(1, 3)$ and thus that $F^{−1}_X (y) = 2y +1$ for $y \epsilon (0, 1)$.

My attempt for $F_X(x)=\int_{-\infty}^{\infty}\frac{1}{b-a}dx=\int_{1}^{x}\frac{1}{3-1}dx=\frac{1}{2}\int_{1}^{x}dx=\frac{1}{2}[x-1]$

But the result is $F_X(x) = 2(x − 1)$.

I couldn't solve $F^{−1}_X (y)$

share|cite|improve this question
up vote 1 down vote accepted

Surely the question that says $F_{X}(x) = 2(x-1)$ is wrong, since if you substitute 3 for $x$ you are getting $4$ as the answer.

share|cite|improve this answer
Thank you. My first part has been solved. – ABC Sep 22 '13 at 6:43
Second part is straight forward from your first answer. Invert the expression to get the value of $x$. To be more clear, the 2nd part asks for the value of $x$ when $y$ is the given CDF value. So $F_{X}(x)=\frac{x-1}{2}$ and the value of $x$ which gives $y$ as the CDF value is simply $2y+1.$ – Sudarsan Sep 22 '13 at 6:45

I believe that is a typo because your answer is indeed correct. For the second part you can set $x = (y-1)\frac{1}{2}$ and just solve for y. This will give you your inverse.

share|cite|improve this answer
Yes, there is a theorem that says that if U is uniformly distributed over the interval (0,1), then $X = F_X^{-1}(U)$ has c.d.f $F_X(x)$. So the inverse is distributed over (0,1). To see this $P(X\leq x) = P(F_X^{-1}(U)\leq x) = P(U\leq F_X(x)) = F_X(x)$. – RDizzl3 Sep 22 '13 at 6:57
Thank you. Your above comment is due to my another question in the comment[pardon me, which i have erased before posting your comment.]. Now The second part is clearer. – ABC Sep 22 '13 at 7:07

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.