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I have this Weibull density function,

$$ f(x) = 0.25 \left| 1-x \right|^{-0.5} \exp(-\left| 1-x \right|^{0.5})$$

Because of the absolute value, this is split into 2 cases.

Its cumulative function

$$F(a)=\int f(x) \; dx = \begin{cases} \frac{1}{2} e^{-\left( 1-a \right)^{0.5}} -\frac{1}{2} e^{-1} & 0

I take a=$\infty$, this cdf doesn't integrate to 1??

$$\left[1-\frac{1}{2} e^{-(a-1)^{0.5}} - \frac{1}{2} e^{-1}\right]_1^\infty = 1 - \frac{1}{2} e^{-1}$$ What is wrong?

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For a cumulative distribution function, you must integrate from $-\infty$. That is, $$F(a)=\int_{-\infty}^a f(x)\,dx.$$ If you do that, you get the right answer $$F(a)=\begin{cases} \exp(-\sqrt{1-a})/2 & \text{if } a\leq 1 \\[8pt] 1-\exp(-\sqrt{a-1})/2 &\text{ otherwise.} \end{cases}$$

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    Hey Bryon, the interval for Weilbull is $x\ge 0$, correct? This see this on Wiki http://en.wikipedia.org/wiki/Weibull_distribution2012-02-13
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    @user1061210 Well, the function in your question is not a Weibull density, so I don't think the Wikipedia page is relevant to your question.2012-02-13
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    @Bryon, you know what, I'll take your word for it, even though it is clearly read Weibull Distribution on the page. It reason being is, it integrates to one! yes!2012-02-13
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    Here it is $$F(a)=\int{f(x)dx}=\left\{ \begin{array}{*{35}{l}} \frac{1}{2}{{e}^{-{{\left( 1-a \right)}^{0.5}}}} & a\le 1 \\ \frac{1}{2}+\frac{1}{2}{{e}^{-{{(a-1)}^{0.5}}}} & 12012-02-13
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    actually, I got it! Thank everyone for trying to help.2012-02-13