How to prove this is a probability density function?

Question

Being θ an aleatory variable with a uniform distribution function on the interval (-pi/2 , pi/2). Knowing that x = cos θ. How can I prove that this is a probability density function?

Can anyone here explain me how I get to know the analytic expression?

Answer 1

First calculate the CDF of your uniform rv.

Then, observe that

$$P(X> x)=F_{\Theta}(arccos (x))-F_{\Theta}(-arccos(x))=\frac{2}{\pi}arccos(x)$$

Thus, just using the definition of CDF you get

$$F_X(x)=1-\frac{2}{\pi}arccos(x)$$

Deriving you get the requested density

$$f_X(x)=\frac{2}{\pi\sqrt{1-x^2}}\mathbb{1}_{(0;1)}(x)$$

EDIT: if you do not like the use of the indicator function, the density can be expressed in the following form

$$ f_X(x) = \begin{cases} \frac{2}{\pi\sqrt{1-x^2}}, & \text{if $x \in (0;1)$ } \\ 0, & \text{elsewhere} \end{cases}$$

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To prove that $f_X(x)$ is a nice density you can verify that all the density's properties are attained

1. $f_X(x)\geq 0$, $\forall x$

2. $\int_{\infty}^{\infty}f_X(x)dx=1$

EDIT2: Here are the passages for the integral

$$\int_0^1\frac{2}{\pi\sqrt{1-x^2}}dx=\frac{2}{\pi}\Big[-arccos(x)\Big]_0^1=\frac{2}{\pi}\cdot\frac{\pi}{2}=1$$

The same properties can be verified directly on the CDF

1. $F(-\infty)=0$

2. $F(\infty)=1$

3. $\frac{d}{dx}F_X(x)\geq 0$, $\forall x$