# Why doesn't this distribution function integrate to $1$?

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

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

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?

-

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 $$F(a)=\begin{cases} \exp(-\sqrt{1-a})/2 & \text{if } a\leq 1 \\[8pt] 1-\exp(-\sqrt{a-1})/2 &\text{ otherwise.} \end{cases}$$
Hey Bryon, the interval for Weilbull is $x\ge 0$, correct? This see this on Wiki en.wikipedia.org/wiki/Weibull_distribution – user1061210 Feb 13 '12 at 2:17
Here it is $$F(a)=\int{f(x)dx}=\left\{ \begin{array}{*{35}{l}} \frac{1}{2}{{e}^{-{{\left( 1-a \right)}^{0.5}}}} & a\le 1 \\ \frac{1}{2}+\frac{1}{2}{{e}^{-{{(a-1)}^{0.5}}}} & 1<a \\ \end{array} \right.$$ How do I find the inverse of it? I am taking it piece by piece. But there is a negative sign in the exponent. – user1061210 Feb 13 '12 at 2:55