# Analytical form of a Histogram of $1-\cos(X)$, $X\sim U(0, 2\pi)$

If I generate a uniform distribution of $$X$$ ranging from $$0$$ to $$2\pi$$ (so $$X\sim U(0, 2\pi)$$), then the probability distribution of $$1-\cos(X)$$ appears to be this function:

Is this an analytical function that I can work with directly? And if so, how do I prove this?

• Based on inspection and some clues in the text, I can guess that it's an arcsine distribution en.wikipedia.org/wiki/Arcsine_distribution but you'll need to provide more details to know for sure. How is $X$ distributed? If an "even distribution" means "uniform" then there are certain transformations of uniform distributions that are definitely arcsine distributions.
– Sycorax
Commented Apr 24, 2023 at 17:59
• I just simulated from $N(0, 1)$ and then applied your $1-\cos(x)$ transformation, leading to a somewhat different plot.
– Dave
Commented Apr 24, 2023 at 18:01
• @dave could you try it with a uniform distribution ranging from (0, 2pi)? Commented Apr 24, 2023 at 18:53
• You can find extensive explanations by searching our site.
– whuber
Commented Apr 24, 2023 at 20:11

We can evaluate the distribution function $$F(y)$$ of $$Y := 1 - \cos(X)$$ directly as follows.

To begin with, note that as $$X$$ ranges over $$[0, 2\pi]$$, the range of $$Y$$ is from $$0$$ (achieved at $$X = 0$$ or $$2\pi$$) to $$2$$ (achieved at $$\pi$$). Therefore, for $$y < 0$$, $$F(y) = 0$$ and for $$y \geq 2$$, $$F(y) = 1$$.

For $$y \in [0, 2)$$ (it is helpful to draw the graph of $$x \mapsto \cos(x)$$ on $$[0, 2\pi]$$ to determine the region $$\{x \in [0, 2\pi]: \cos(x) \geq 1 - y\}$$. Also keep in mind that the domain of $$x \mapsto \arccos(x)$$ is $$[-1, 1]$$ with range $$[0, \pi]$$, so mirroring is needed for angle that is greater than $$\pi$$): \begin{align} & F(y) = P[1 - \cos(X) \leq y] = P[\cos(X) \geq 1 - y] \\ =& P[X \in [0, \arccos(1 - y)] \cup [2\pi - \arccos(1 - y), 2\pi]] \\ =& \frac{\arccos(1 - y)}{\pi}. \end{align}

To summarize, the distribution of $$Y$$ is given by \begin{align} F(y) = \begin{cases} 0 & y < 0, \\ \frac{\arccos(1 - y)}{\pi} & 0 \leq y < 2, \\ 1 & y \geq 2. \end{cases} \end{align}

Taking derivative of $$F$$ yields the pdf of $$Y$$: \begin{align} f(y) = \begin{cases} \frac{1}{\pi\sqrt{1 - (1 - y)^2}} & 0 < y < 2, \\ 0 & \text{ otherwise.} \end{cases} \end{align}

A graph of $$f$$ looks as follows, which matches the histogram you simulated. As @Sycorax pointed out in the comment, this is Arcsine distribution with support $$(0, 2)$$.

You can get the result with the delta method: $$p(z)=\int_R p(x) \delta(z-(1-cos(x)))dx$$. https://en.m.wikibooks.org/wiki/Probability/Transformation_of_Probability_Densities