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

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

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

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What are $\alpha,\Theta$? What is the question? – Berci Apr 26 '13 at 10:24

1 Answer 1

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 $1$). $$1-P(X>t)=P(X\le t)=\int_0^t 2x\,dx=\left[x^2\right]_{x=0}^{x=t}\,=t^2\,.$$

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