1
$\begingroup$

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?

1 Answers 1

3

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 $F(a)=\begin{cases} \exp(-\sqrt{1-a})/2 & \text{if } a\leq 1 \\[8pt] 1-\exp(-\sqrt{a-1})/2 &\text{ otherwise.} \end{cases}$

  • 0
    actually, I got it! Thank everyone for trying to help.2012-02-13