Probability of a vector being in a predefined interval in $\mathbb{R}^{2}$

Question

Suppose we have a two-dimensional vector of constant length that is equally likely to point in any direction specified by $\theta$, measured from the $x$-axis. Compute the probability that the $x$-component of $v_{x}$ lies between $v_{x}$ and $v_{x} + \mathop{dx}$.

Answer 1

The vector is given by $(r\cos\theta, r\sin\theta)$, where $r$ is fixed and $\theta$ is chosen uniformly at random. The $x$-component is in the interval $[x_0,x_1]$ (where $-r\le x_0 \le x_1 \le r$) if $$r\cos\theta\in[x_0, x_1] \iff \cos\theta\in\left[\frac{x_0}{r},\frac{x_1}{r}\right].$$ There will be two equal-sized intervals of $\theta$ satisfying this, one above the $x$-axis and one below: $\theta\in[\cos^{-1}(x_0/r), \cos^{-1}(x_1/r)]$ and its reflection. So the probability that the $x$-component is between $x_0$ and $x_1$ is $$ \frac{1}{\pi}\left(\cos^{-1}\left(\frac{x_1}{r}\right)-\cos^{-1}\left(\frac{x_0}{r}\right)\right). $$

Answer 2

It is not a double integral. However the probability has to be split into two parts. Specifically $(v=|\overrightarrow{v}|)$, let $p=P(v_x\le u)=P(vcos\theta \le u)=P(vcos\theta \le u,0\le \theta\le \pi)+P(vcos\theta \le u,\pi\le \theta\le 2\pi)$. Let $w=\frac{u}{v}$ Then $p=(\theta\ge arccos(w),0\le \theta\le \pi)+P(\theta\le arccos(w),\pi\le \theta \le 2\pi)=\frac{|\pi-arccos(w)|}{\pi}$

Note that the arccos is double valued over the interval $0,2\pi$, but $\pi-arccos(w)$ is the same for both values.

Answer 3

Probabilities are defined with respect to distributions, that is, you need to provide a way to go from a collection of possibilities (called an event) to a number between $0$ and $1$. In this context, a distribution over $\mathbb{R}^2$, you would typically provide a function of two variables whose double integral was $1$, and then the probability of an arbitrary event is the integral of that function over that region.

But there's no magical choice for the function, for the distribution, in this case. One could make an argument for a bivariate normal, but there's no definitive reason to choose that.