Let X be a random variable with probability density function
$$f(x)=\left\{\begin{array}{ll}{\frac{c}{\sqrt{1-x^{2}}},} & {x \in(-1,1)} \\ {0,} & {x \notin(-1,1)}\end{array}\right.$$
now I want to find the cumulative distribution function of X. I know that $c=1/\pi$ to normalize this properly, and the CDF is the integral of the PDF, but I'm not sure exactly what I'm looking for in this case (i.e., what the bounds are, what the final form of the CDF should be written as, etc.) Thanks.
$\endgroup$2 Answers
$\begingroup$The definition of a CDF is as follows:$$F_X(x) = P(X \leq x) = \int_{-\infty}^{x} f_X(t)dt $$
for continuous random variables ($t$ is a dummy variable).
Since x is bounded by (-1,1) there is no 'probability density' outside of these bounds.
The lower limit of integration is -1 and the upper limit is $x$.
When $x \geq 1$ then the value of the CDF reduces to 1.
When $x \leq -1$ then the value of the CDF reduces to 0.
$\endgroup$ $\begingroup$The CDF, say $g(x),$ can be written as
$g(x)=\begin{cases}0,&x\leq -1\\ \frac{1}{\pi}\arcsin(x)+\frac{1}{2},&x\in(-1,1)\\1,&x\ge 1\end{cases}.$
$\endgroup$
