MHB Given probability density function find its cumulative distribution function

[sofanglom]

Hi :) Here's my problem along with what I've done.

Here is the problem:

View attachment 8716

That is the p.d.f. of a random variable X.

I have to find the cdf. I don't know which I should do so I tried it two ways. First:

$\int_{-1}^{1} \ \frac{2}{\pi(1+x^{2})} dx = {{\frac{2}{\pi} arctan(x)]}^{1}}_{-1}=1$

Second:

$\int_{-1}^{x} \ \frac{2}{\pi(1+t^{2})} dt = {{\frac{2}{\pi} arctan(x)]}^{x}}_{-1}=\frac{2(arctan(x)+\frac{\pi}{4}}{\pi}$

Which one is the required CDF for X?

 

[Evgeny.Makarov]

Hi, and welcome to the forum!

Which one is the required CDF for X?

The second one, except for the missing closing parenthesis. That is, the CDF is $\dfrac{2}{\pi}\arctan x+\dfrac12$.