Probability of $n$ Random Points in a Circle Falling Within a Sector with Central Angle $ $

Joint Distribution

Starting from the last equivalent formulation, let the unit line segment be divided by n-1 random points into n segments with lengths x_1, x_2, \dots, x_n from left to right (note: this just defines an order, they are not sorted by size). Let their joint probability density function be p_n(x_1, x_2, \dots, x_n). The answer in the Zhihu link is based directly on the fact that “p_n(x_1, x_2, \dots, x_n) is a uniform distribution.” However, I feel that the fact that “p_n(x_1, x_2, \dots, x_n) is a uniform distribution” is not immediately obvious, so I will derive it here.

First, note two facts:

1. x_1, x_2, \dots, x_n are subject to the constraint x_1 + x_2 + \dots + x_n = 1, so there are only n-1 degrees of freedom. We take the first n-1 variables as independent variables.

2. p_n(x_1, x_2, \dots, x_n) is a probability density, not a probability; p_n(x_1, x_2, \dots, x_n) dx_1 dx_2 \dots dx_{n-1} is the probability.

To understand why p_n(x_1, x_2, \dots, x_n) is a uniform distribution, we start with n=2. In this case, a single point is chosen uniformly in (0, 1) to divide the segment into two parts. Since the point is sampled from a uniform distribution (density is 1) and the point corresponds one-to-one with (x_1, x_2), we have p_2(x_1, x_2) = 1.

Next, consider n=3. The corresponding probability is p_3(x_1, x_2, x_3) dx_1 dx_2. There are two possibilities:

1. Sample one point first, dividing the segment into two parts (x_1, x_2 + x_3). The probability for this is p_2(x_1, x_2 + x_3) dx_1. Then sample another point to divide the x_2 + x_3 segment into (x_2, x_3). The probability for this part is dx_2. Multiplying them gives p_2(x_1, x_2 + x_3) dx_1 dx_2.

2. Sample one point first, dividing the segment into two parts (x_1 + x_2, x_3). The probability for this is p_2(x_1 + x_2, x_3) dx_2. Then sample another point to divide the x_1 + x_2 segment into (x_1, x_2). The probability for this part is dx_1. Multiplying them gives p_2(x_1 + x_2, x_3) dx_1 dx_2.

Adding them together: \begin{equation} p_3(x_1, x_2, x_3) dx_1 dx_2 = p_2(x_1, x_2 + x_3) dx_1 dx_2 + p_2(x_1 + x_2, x_3) dx_1 dx_2 = 2 dx_1 dx_2 \end{equation} Thus p_3(x_1, x_2, x_3) = 2, which is also a uniform distribution. By induction, we can obtain: \begin{equation} p_n(x_1, x_2, \dots, x_n) = (n - 1) p_{n-1}(x_1, x_2, \dots, x_{n-1}) = \dots = (n - 1)! \end{equation}

Marginal Distribution

With the joint distribution, we can find the marginal distribution for any k variables. This prepares us for using the “Inclusion-Exclusion Principle” in the next section. Since the joint distribution is uniform, the variables are symmetric. Therefore, without loss of generality, we can find the marginal distribution of the first k variables.

By definition: \begin{equation} p_n(x_1, x_2, \dots, x_k) = \int \dots \int p_n(x_1, x_2, \dots, x_n) dx_{k+1} \dots dx_{n-1} \end{equation} Note the integration limits. The lower limit is naturally 0. Due to the constraint x_1 + x_2 + \dots + x_n = 1, the upper limit is different for each variable. Given x_1, x_2, \dots, x_i, the value of x_{i+1} can only range from 0 to 1 - (x_1 + x_2 + \dots + x_i). Thus, the precise form is: \begin{equation} \int_0^{1 - (x_1 + \dots + x_k)} \dots \left[ \int_0^{1 - (x_1 + \dots + x_{n-2})} p_n(x_1, \dots, x_n) dx_{n-1} \right] \dots dx_{k+1} \end{equation} Since p_n(x_1, \dots, x_n) = (n-1)! is a constant, the above expression can be integrated sequentially. The final result is: \begin{equation} p_n(x_1, x_2, \dots, x_k) = \frac{(n-1)!}{(n - k - 1)!} [1 - (x_1 + x_2 + \dots + x_k)]^{n-k-1} \end{equation} Now we can calculate the probability that x_1, x_2, \dots, x_k are each greater than given thresholds c_1, c_2, \dots, c_k (where c_1 + c_2 + \dots + c_k \leq 1): \begin{equation} \begin{aligned} & P_n(x_1 > c_1, x_2 > c_2, \dots, x_k > c_k) = \int_{c_1}^{1 - (c_2 + \dots + c_k)} \dots \\ & \left[ \int_{c_{k-1}}^{1 - (x_1 + \dots + x_{k-2} + c_k)} \left[ \int_{c_k}^{1 - (x_1 + \dots + x_{k-1})} p_n(x_1, \dots, x_k) dx_k \right] dx_{k-1} \right] \dots dx_1 \end{aligned} \end{equation} The integration limits are derived similarly under the constraint \sum x_i = 1. Like p_n(x_1, \dots, x_n), this can be integrated sequentially, and the final result is surprisingly simple: \begin{equation} P_n(x_1 > c_1, x_2 > c_2, \dots, x_k > c_k) = [1 - (c_1 + c_2 + \dots + c_k)]^{n-1} \end{equation}

Inclusion-Exclusion Principle

Now everything is ready for the “Inclusion-Exclusion Principle” to deliver the final blow. Recall that our goal is to find the probability of the longest segment length, i.e., for a threshold x, calculate P_n(\max(x_1, x_2, \dots, x_n) > x). The event \max(x_1, x_2, \dots, x_n) > x is the union of several events: \begin{equation} \max(x_1, x_2, \dots, x_n) > x \quad \Leftrightarrow \quad x_1 > x \text{ \color{red}{or} } x_2 > x \text{ \color{red}{or} } \dots \text{ \color{red}{or} } x_n > x \end{equation} Crucially, when x < \frac{1}{2}, these events are not mutually exclusive. Therefore, we cannot simply sum the probabilities of each part; we must use the Inclusion-Exclusion Principle: \begin{equation} \begin{aligned} P(\text{at least one } x_i > x) = & \sum P(\text{any one } x_i > x) \\ & \color{red}{-} \sum P(\text{any two } x_i, x_j > x) \\ & \color{red}{+} \sum P(\text{any three } x_i, x_j, x_l > x) \\ & \color{red}{-} \dots \end{aligned} \end{equation} According to the results of the previous section, the probability that any k chosen points are each greater than x is (1-kx)^{n-1}, and the number of ways to choose k points is \binom{n}{k}. Substituting these into the formula gives: \begin{equation} \begin{aligned} P_n(\max(x_1, x_2, \dots, x_n) > x) &= \sum_{k=1, 1 - kx > 0}^n (-1)^{k-1} \binom{n}{k} (1-kx)^{n-1} \\ &= \binom{n}{1} (1 - x)^{n-1} - \binom{n}{2} (1 - 2x)^{n-1} + \dots \end{aligned} \end{equation}