NBA Assists

Author

Dr. Alexander Fisher

See libraries used in these notes 
library(tidyverse)
library(latex2exp)

Data: Boston Celtics

In basketball, an “assist” is attributed to a player that passes the ball to a teammate in a way that directly leads to a basket. The data below was scraped from espn.com September 2024. Each row of the data set is an individual Boston Celtics player during a particular game of the 2023-2024 season. The AST records the number of assists made by the player in the particular game. MIN records the number of minutes each player played in a particular game. vs records the opposing team.

# A tibble: 10 × 4
   player                 AST   MIN vs                   
   <chr>                <dbl> <dbl> <chr>                
 1 Payton Pritchard PG      4  23.5 Golden State Warriors
 2 Derrick White PG         4  33.5 Dallas Mavericks     
 3 Jaylen Brown SG          4  34   Brooklyn Nets        
 4 Luke Kornet C            2  21.3 Chicago Bulls        
 5 Xavier Tillman F *       2  15   Dallas Mavericks     
 6 Luke Kornet C            1  19   Dallas Mavericks     
 7 Kristaps Porzingis C     1  23.7 Brooklyn Nets        
 8 Kristaps Porzingis C     1  30   Chicago Bulls        
 9 Oshae Brissett SF        0   4   Dallas Mavericks     
10 Luke Kornet C            0  11   Golden State Warriors
set.seed(360)
yx = read_csv("../data/BostonCeltics_Assists_23-24_season.csv")

yx %>%
  slice_sample(n = 10) %>%
  arrange(desc(AST))

Question: how many assists on average do we expect a Celtics player to make per minute played?

Bayesian framework

  1. Define a data generative model and write down the likelihood (often assuming exchangeability and invoking de Finetti’s theorem)
  2. Choose a prior distribution for all things unknown
  3. Compute or approximate the posterior
  4. Make inference

Data-generative model

To write down a data generative model, let’s assume players accumulate assists \(y\) at some per-minute rate, \(\theta\). We will further assume that the expected number of assists by a player in a given game is then \(\theta x\) where \(x\) is the number of minutes played.

Given \(y\) takes integer values \(\{0, 1, 2, \ldots \}\), a Poisson distribution might make sense. If \(Y | \lambda\) is Poisson\((\lambda)\), then

\[ p(y |\lambda) = \frac{(\lambda)^y e^{-\lambda}}{y!} \]

Here, \(\lambda = \theta x\).

Assuming this conditionally independent model for generating \(y\)s, we write the likelihood,

\[ p(y_1,\ldots y_n | \theta) = \prod_{i = 1}^n \frac{(\theta x_i)^{y_i} e^{-\theta x_i}}{y_i!} \]

What assumptions are we making about assists in the data generative model above?

Prior beliefs

  • Notice the support: \(\theta > 0\)
  • A gamma prior may be suitable

\[ \begin{aligned} \theta &\sim gamma(a, b)\\ p(\theta | a, b) &= \frac{b^{a}}{\Gamma(a)} \theta^{a - 1} e^{-b \theta} \end{aligned} \] ### Compute the posterior

Compute the posterior.

\[ \begin{aligned} \theta | y_1,\ldots y_n &\sim gamma(\alpha, \beta)\\ \alpha &= a + \sum_{i=1}^n y_i\\ \beta &= b + \sum_{i=1}^n x_i \end{aligned} \]

Hint: we have conjugacy. You can see this if you view the likelihood and the prior each as a function of \(\theta\), the kernel of each has the same functional form.

Follow-up:

  1. what is \(E[\theta | y_1,\ldots, y_n]\)? How does it compare to \(E[\theta]\)?
  2. what is \(Var[\theta | y_1,\ldots y_n]\)?

a = 9
b = 3
sumY = sum(yx$AST)
sumX = sum(yx$MIN)

data.frame(x = c(0, 0.5)) %>%
  ggplot(aes(x = x)) + 
  stat_function(fun = dgamma, args = list(shape = sumY + a,
                                          rate = sumX + b)) + 
  labs(x = TeX("$\\theta$"), y = TeX("p($\\theta | y_1, \\ldots, y_n$)")) +
  theme_bw()

Given our prior of \(a = 9, b = 3\), \(E[\theta | y_1,\ldots, y_n] =\) 0.1194

Exponential families

Definition: a sufficient statistic is a function of the data \((y_1,\ldots y_n\)) that is sufficient to make inference about unknown parameters (\(\theta\)).

If density \(p(y|\theta)\) can be written \(h(y) c(\phi) e^{\phi t(y)}\) for some transform \(\phi = f(\theta)\) we can say \(p(y|\theta)\) belongs in the exponential family.

Note

\(t(y)\) is referred to as the sufficient statistic.

If \(p(y|\theta)\) belongs in the exponential family, then the conjugate prior is \[ p(\phi | n_0, t_0) = \kappa(n_0, t_0)c(\phi)^{n_0} e^{n_0 t_0 \phi}, \] where \(\kappa(n_0, t_0)\) is a normalizing constant.

Notes about the prior

  1. the conjugate prior is given over \(\phi\) and we’d have to transform back if we care about \(p(\theta)\).

  2. \(n_0\) is interpreted as the prior sample size and \(t_0\) is the prior guess.

The resulting posterior is

\[ p(\phi | y_1,\ldots y_n) \propto p \left(\phi | n_0 + n, \frac{n_0 t_0 + n \sum t(y_i)/n}{n_0 + n}\right) \] In words, one can show that the kernel of the posterior of \(\phi\) is proportional to the kernel of the prior of \(\phi\) with specific parameters. Thereby, we have conjugacy.

Example offline: Poisson density