MHB Given probability density function find its cumulative distribution function

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The discussion revolves around finding the cumulative distribution function (CDF) from a given probability density function (PDF) of a random variable X. The user attempted two integration methods to derive the CDF, both involving the arctangent function. The correct CDF is identified as (2/π)arctan(x) + 1/2, with a note on a minor error in the closing parenthesis in the user's second attempt. The forum members confirm that the second method is the appropriate approach for determining the CDF. The conversation highlights the importance of careful notation in mathematical expressions.
sofanglom
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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?
 

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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 $\dfrac{2}{\pi}\arctan x+\dfrac12$.
 

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