关于百分位数的概率闭式求解及概念规范化的技术问询

Hey there! Let's break down your question step by step since you're just starting out with probability—totally get where you're coming from with the Secretary Problem tangents, Algorithms to Live By is such a fun deep dive into real-world math!

First, let's validate your percentile definition #

Your operational definition is spot-on for sample percentiles, which is exactly how we define them in introductory stats for finite datasets:

Given a data set of size n, the percentile of a data point $X$ is value $ \pi(X) = \frac{|{Y | S(Y)\leq S(X)}|}{n}$ where $S:D\rightarrow \mathbb{R}$ gives the score of a data point.

This matches the standard empirical cumulative distribution function (ECDF) definition—where the ECDF at a value $t$ is the fraction of sample points ≤ $t$. Your $\pi(X)$ is just the ECDF evaluated at $S(X)$, which is a totally rigorous way to define percentiles for discrete samples.

Core Problem: Closed-form for $P_y(X) = P(\pi(X) > y)$ #

Let's formalize this properly, starting with your intuition (and fixing a tiny wording mix-up you had earlier).

First, rephrase the condition using your definition:
$$\pi(X) > y \iff \frac{|{Y | S(Y)\leq S(X)}|}{n} > y \iff |{Y | S(Y)\leq S(X)}| > n y$$

Case 1: All scores are distinct (no ties) #

Since we're picking $X$ uniformly at random from the n data points, every possible rank (where rank = number of points ≤ $X$) is equally likely. Let's denote $m = \lfloor n y \rfloor$ (the largest integer less than or equal to $n y$).

The number of data points where their rank is greater than $n y$ is $n - m$. Since each point has an equal chance of being picked, the probability is:
$$P(\pi(X) > y) = \frac{n - m}{n}$$

Let's test this with your example:

• n=100, y=0.95: $m = 95$, so $\frac{100-95}{100} = 0.05$ (5% chance, not 0.5%—your intuition had a tiny decimal slip-up here!). This makes sense: only the top 5 points have a percentile score above 0.95.
• n=100, y=0.2: $m=20$, so $\frac{100-20}{100}=0.8$ (80% chance), which aligns with your gut feeling that more data sits above lower percentiles.
• y=1: $m=100$, so probability 0 (impossible, since no point can have a percentile above 1).
• y=0: $m=0$, so probability 1 (every point is above the 0th percentile).

Case 2: Duplicate scores (ties) #

If multiple points share the same score, they'll have identical $\pi(X)$ values. For example, if 3 points have the highest score in a 100-point dataset, each has $\pi(X)=1$. In this case, the closed-form just becomes:
$$P(\pi(X) > y) = \frac{\text{Total number of data points with } \pi(X) > y}{n}$$

In probability, we often assume samples come from a continuous distribution (where ties have a 0 probability of occurring), so the distinct-scores formula is the go-to for most theoretical work.

How your intuition aligns with standard probability #

Your guess that $P(\pi(X)>y) \approx 1 - y$ is actually the large-sample limit of our closed-form! As n grows to infinity, $\frac{n - \lfloor n y \rfloor}{n}$ gets closer and closer to $1 - y$.

In formal probability (for infinite populations or continuous random variables), the y-th percentile (quantile) is defined as the value $q_y$ where $P(S ≤ q_y) = y$. The probability that a random draw is above $q_y$ is exactly $1 - y$—which is exactly the approximation you intuited!

Quick Recap #

• Your percentile definition is identical to the standard sample percentile (using the ECDF).
• For finite distinct-score samples: $P(\pi(X) > y) = \frac{n - \lfloor n y \rfloor}{n}$
• For large n, this approximates to $1 - y$, matching your initial intuition.
• Ties require counting qualifying points, but are negligible in continuous distributions.

备注：内容来源于stack exchange，提问作者Mani