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Consider the continuous distribution with density function $$ p(x) = \frac{1}{2}\cos(x) \;, \quad -\frac{\pi}{2} < x < \frac{\pi}{2}. $$

I want to derive the cumulative distribution function for this density function. In other examples I've seen, this is usually done by taking the anti-derivate of the density function which (in the examples that I've seen) leads to a valid CDF.

However, it seems that following the same approach for this example yields an invalid CDF, $$ \begin{cases} \: -0.5 , \quad \quad x \leq \frac{-\pi}{2} \\ \frac{1}{2}\sin(x), \: \: -\frac{\pi}{2} < x < \frac{\pi}{2} \\ \: \: \: 0.5, \quad \quad x \geq \frac{\pi}{2} \end{cases} $$ and that it somehow needs rescaling. What am I doing wrong here, and how does one go about finding the CDF for this density function?

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  • $\begingroup$ The integral for the CDF of a density $f$ is, generically, $F(x) =\int_{-\infty}^x f(t)\,\mathrm dt.$ Notice the lower limit. You started with a lower limit of $0.$ $\endgroup$
    – whuber ♦
    Commented Jun 5, 2023 at 13:22

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You forgot about the constant of integration: https://en.wikipedia.org/wiki/Constant_of_integration

Just add $C = 0.5$ and your anti derivative becomes a valid CDF with corresponding density $p(x)$.

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