CDF of the X-Projection From a Uniform Point on a Circle
A point is chosen uniformly at random on the unit circle (the set of points at distance
- Find the CDF $F_X(x)$ for $x \in [-1, 1]$.
- Find the PDF $f_X(x)$ and identify the named distribution it corresponds to.
- Does the density have a maximum at $x = 0$? Explain intuitively why or why not.
Hints
- Parametrize the unit circle as $(\cos\theta, \sin\theta)$ with $\theta$ uniform on $[0, 2\pi)$. The $x$-coordinate is $X = \cos\theta$.
- For the CDF, find the set of $\theta$ values where $\cos\theta \leq x$. This is a single interval whose length you can compute.
- The resulting distribution has a name -- think about where the density should concentrate. Near $x = \pm 1$, small changes in $\theta$ produce tiny changes in $\cos\theta$, so the density piles up there.
Worked Solution
How to Think About It: A uniform point on the unit circle is parametrized by an angle $\theta \sim \text{Unif}[0, 2\pi)$, with coordinates $(\cos\theta, \sin\theta)$. The $x$-coordinate is $X = \cos\theta$. Before computing anything, think about where the density should pile up. Near $x = \pm 1$, the circle is nearly vertical -- a small change in $\theta$ barely moves $x$. So the point "lingers" near the edges, and the density should blow up at $\pm 1$. Near $x = 0$, the circle is nearly horizontal, so the point moves through quickly and the density is lower. This is the hallmark of the arcsine distribution.
Quick Estimate: By symmetry, $F_X(0) = 1/2$. At $x = 0.9$, we expect $F_X$ to be close to 1 but not too close, since the density is high near the boundary. Rough check: $\arccos(0.9) \approx 0.45$ rad, so $F_X(0.9) \approx 1 - 0.45/\pi \approx 0.86$. That feels right -- 86% of the probability mass is below $x = 0.9$.
Approach: Use the CDF method: express $P(X \leq x)$ in terms of the event on $\theta$ and use the uniform density of $\theta$.
Formal Solution:
Parametrize the circle as $\theta \sim \text{Unif}[0, 2\pi)$, so $X = \cos\theta$.
$F_X(x) = P(\cos\theta \leq x)$
The set $\{\theta \in [0, 2\pi) : \cos\theta \leq x\}$ is the interval $[\arccos(x),\, 2\pi - \arccos(x)]$, which has length
$F_X(x) = \frac{2\pi - 2\arccos(x)}{2\pi} = 1 - \frac{\arccos(x)}{\pi}$
Using the identity $\arccos(x) = \pi/2 - \arcsin(x)$:
$F_X(x) = \frac{1}{2} + \frac{\arcsin(x)}{\pi}, \quad x \in [-1, 1]$
Differentiating:
$f_X(x) = \frac{1}{\pi\sqrt{1 - x^2}}, \quad x \in (-1, 1)$
This is the arcsine distribution on $[-1, 1]$.
For part 3: No, the density does not have a maximum at $x = 0$. It actually has a minimum there ($f_X(0) = 1/\pi \approx 0.318$) and diverges to infinity as $x \to \pm 1$. Intuitively, near the "poles" ($x = \pm 1$), the circle runs nearly vertically, so the projection onto the $x$-axis changes very slowly -- the point spends a lot of "time" near the edges.
Answer: $F_X(x) = \dfrac{1}{2} + \dfrac{\arcsin(x)}{\pi}$ for $x \in [-1, 1]$, with PDF $f_X(x) = \dfrac{1}{\pi\sqrt{1 - x^2}}$. This is the arcsine distribution. The density is U-shaped, diverging at $\pm 1$.
Intuition
The arcsine distribution appears whenever you project circular or oscillatory motion onto a line. The key intuition is geometric: near $x = \pm 1$, the unit circle is nearly tangent to vertical lines, so the projection changes very slowly with $\theta$ and probability mass accumulates. Near $x = 0$, the circle crosses the $y$-axis steeply, so the projection sweeps through quickly and the density is thin. This is the opposite of what most people's first guess would be -- they expect a bell shape centered at 0, but it is actually U-shaped.
This distribution shows up in several places in quantitative finance. The fraction of time a symmetric random walk spends positive follows the arcsine law. If you simulate Brownian motion paths and look at what fraction of the interval $[0, T]$ the path is above zero, that fraction is arcsine-distributed -- it tends to be near 0 or near 1, not near 1/2. Traders who evaluate strategy performance over time should be aware that long winning or losing streaks are more likely than intuition suggests, precisely because of this distribution.