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

A really simple one - I don't know why I got stuck with this \=

$$ 1-P(x>\dfrac{-ln\alpha+\Theta}{\Theta}) $$

When $f(x)=2xI_{0<x<1}$

share|cite|improve this question
What are $\alpha,\Theta$? What is the question? – Berci Apr 26 '13 at 10:24
up vote 1 down vote accepted

Let $t:=\displaystyle\frac{-\ln\alpha+\Theta}\Theta$, and suppose $t\in[0,1]$ (else the answer is trivially $0$ or $1$). $$1-P(X>t)=P(X\le t)=\int_0^t 2x\,dx=\left[x^2\right]_{x=0}^{x=t}\,=t^2\,.$$

share|cite|improve this answer

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.