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Wikipedia has an article about the Brier score whose notation confuses me.

The article starts out easy enough by defining the Brier score to be:

$$ BS = \dfrac{1}{N}\overset{N}{\underset{i = 1}{\sum}}\left( f_t - o_t\right)^2 $$

$N$ is the sample size.

$o_t$ is true value $t$, either $0$ or $1$.

$f_t$ is probability forecast $t$ giving the predicted probability of an outcome of category $1$.

This is the usual way I would think of Brier score as mean squared error.

Where the article loses me is when it gets into decomposing the Brier score into reliability/resolution/uncertainty and calibration/refinement. The notation is as follows, remarking about an abuse of the equals sign.

$$ BS = REL - RES + UNC = \left[\dfrac{1}{N}\overset{K}{\underset{K = 1}{\sum}}n_k\left(f_k - \bar o_k\right)^2\right] +\left[ -\dfrac{1}{N}\overset{K}{\underset{K = 1}{\sum}}n_k\left(\bar o_k - \bar o\right)^2 \right] + \left[ \bar o\left(1 - \bar o\right) \right] $$$$ BS = CAL + REF = \left[\dfrac{1}{N}\overset{K}{\underset{K = 1}{\sum}}n_k\left(f_k - \bar o_k\right)^2\right] + \left[ \dfrac{1}{N}\overset{K}{\underset{K = 1}{\sum}}n_k\left(\bar o_k\left(1 - \bar o_k\right)\right) \right] $$

The notation also puts these in bold, which I think it to handle the case of having more than just two outcome categories (addressed earlier in the article).

Could someone unpack this notation to give the explicit equations for reliability, resolution, uncertainty, refinement, and calibration, valid for both binary and multi-class outcomes?

This goes through the intuition but not the explicit computations that I seek here.

EDIT

I have gone through the calculations proposed in an answer by picky_porpoise, and they do not match up with what I should be getting.

library(SpecsVerification)
set.seed(2023)
N <- 100
# p <- rep(seq(0, 1, 1/50), 3)
# y <- rbinom(length(p), 1, p)

p <- rbeta(N, 1, 1)
y <- rbinom(N, 1, p)

res <- function(y, p){
  
  # Determine the unique values in the p vector
  #
  u <- unique(p)
  
  # Loop over the unique values
  #
  summands <- rep(NA, length(u)) # Hold the loop results
  for (i in 1:length(u)){
    
    # Calculate the o_k
    #
    o_k <- sum(y[p == u[i]])
    
    # Calculate the n_k
    #
    n_k <- length(which(p == u[i]))
    
    # Calculate the o-bar
    #
    o_bar <- mean(y)
    
    # Put the pieces together
    #
    summands[i] <- n_k * ((o_k/n_k) - o_bar)^2
    
  }
  
  # Add up the summands
  #
  my_sum <- sum(summands)
  
  # Divide the sum by the sample size
  #
  return(my_sum/length(y))
}


rel <- function(y, p){
  
  # Determine the unique values in the p vector
  #
  u <- unique(p)
  
  # Loop over the unique values
  #
  summands <- rep(NA, length(u)) # Hold the loop results
  for (i in 1:length(u)){
    
    # Calculate the o_k
    #
    o_k <- sum(y[p == u[i]])
    
    # Calculate the n_k
    #
    n_k <- length(which(p == u[i]))
    
    # Calculate the p_k
    #
    p_k <- u[i]
    
    # Put the pieces together
    #
    summands[i] <- n_k * ((o_k/n_k) - p_k)^2
    
  }
  
  # Add up the summands
  #
  my_sum <- sum(summands)
  
  # Divide the sum by the sample size
  #
  return(my_sum/length(y))
  
  
}

unc <- function(y, p){
  return(mean(y) * (1 - mean(y)))
}


rel(y, p) # 0.1881283
res(y, p) # 0.2499
unc(y, p) # 0.2499
SpecsVerification::BrierDecomp(p, y) # 0.011184570, 0.07068499, 0.2499000

rel(y, p) -
  res(y, p) +
  unc(y, p) -
  mean((y - p)^2) # Zero difference between the Brier score decomposition
                  # calculation and the Brier score!
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  • $\begingroup$ Did you read the cited article that constructed this decomposition? doi.org/… $\endgroup$ Nov 15, 2023 at 5:36

1 Answer 1

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The appendix of the paper Simplifying and generalising Murphy's Brier score decomposition by Stefan Siegert provides a simpler proof than the original work by Murphy. Let me just give the intuition here. I will use Siegert's notation in order to make the transition from my summary to his proof as simple as possible.

The empirical Brier score for binary outcomes $y_1, \ldots, y_N$ and forecast probabilities $p_1, \ldots, p_N$ is given by $$ B(p) = \frac{1}{N} \sum_{t=1}^{N} (y_t - p_t)^2 . $$ The idea of the decomposition is to look at the performance of the forecasts for each possible value they took. To do this, the following notation is needed:

  • The set of $K$ distinct values which the forecast probabilities assume, denoted via $\{P_1, \ldots, P_K\}$.
  • The number $n_k$ for any $k=1, \ldots, K$ which states how often the forecast took the value $P_k$
  • The number $o_k$ for any $k=1, \ldots, K$ which states how often success was observed, i.e. $y_t=1$ when the forecast took the value $P_k$.
  • The base rate $\bar o = \frac{1}{N} \sum_{t=1}^N y_t$

With these we get $$ B(p) = \mathrm{REL} - \mathrm{RES} + \mathrm{UNC} $$ where \begin{align*} \mathrm{REL} &= \frac{1}{N} \sum_{k=1}^K n_k \left( \frac{o_k}{n_k} - P_k \right)^2 \\ \mathrm{RES} &= \frac{1}{N} \sum_{k=1}^K n_k \left( \frac{o_k}{n_k} - \bar o \right)^2 \\ \mathrm{UNC} &= \bar o (1 - \bar o) \end{align*} Now the key insight to understand why this decomposition holds is to define new 're-calibrated' probability forecasts $q_1, \ldots, q_N$ via $$ q_t = \frac{o_{k(t)}}{n_{k(t)}} $$ where $k(t)$ is the index $k$ such that $p_t = P_k$. The $q_t$ are calibrated in the sense that they always hit the exact event frequencies for the $K$ different forecast cases we consider. With this notation, the decomposition follows from calculating that \begin{align*} \mathrm{REL} &= B(p) - B(q) \\ \mathrm{RES} &= B(\bar o) - B(q) \\ \mathrm{UNC} &= B(\bar o) \end{align*} such that the new terms $B(q)$ and $B(\bar o)$ cancel out and only $B(p)$ remains. Apart from enabling an easy proof of the decomposition, these equations also provide an intuitive interpretation of the components:

  • The reliability (REL) component is the difference in performance (measured via Brier score) between the forecast and a re-calibrated version of it, where calibration means that the values have been fit to the true event frequencies. It thus measures the performance loss due to getting the event frequencies wrong.
  • The resolution (RES) component is the difference in performance between the base rate forecast and the calibrated version of the forecast. It thus quantifies how good the forecast distinguishes events from non-events, but does so by ignoring the reliability question (this is already in the REL component)
  • The uncertainty (UNC) is the performance of a simple constant base rate forecast. The forecast improves on this by increasing resolution, while keeping the reliability component low, i.e. being reliable

Although intuitive, this type of score decomposition has the disadvantage that the 're-calibration' is trivial. In particular, if all forecast probabilities are distinct (such that $K=N$), which is realistic in many applications, then it becomes useless. An improved decomposition, which should probably be preferred, is derived in Stable reliability diagrams for probabilistic classifiers by Dimitriadis et al.

Edit

  • The decomposition naturally extends to the multi-class setting. The $p_t$ are then probability vectors over the classes and $K$ is the number of distinct probability vectors. The number of successes and the base rate have to be computed per class.

  • The function BrierDecomp from the package SpecsVerification does not yield the decomposition described here. The reason for this is that they calculate re-calibrated forecasts via binning.

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  • 1
    $\begingroup$ In most situations, $k$ will be equal to the number of predictions made, correct? (I guess I expect all of the predictions to be at least a little bit different.) $\endgroup$
    – Dave
    Nov 16, 2023 at 20:14
  • $\begingroup$ @Dave Probably yes, if there is no restriction on how precise you can make your probabilities, there will be no ties, so $K=N$. $\endgroup$ Nov 16, 2023 at 20:19
  • $\begingroup$ I've tried to implement your equations (edited into my question) but have only gotten the uncertainty to work. Are there errors in your equations? Have a made a coding error? $\endgroup$
    – Dave
    Nov 16, 2023 at 21:10
  • $\begingroup$ @Dave I think when you compute the $o_k$ you have to use sum instead of mean . Otherwise the sum of successes is divided by $n_k$ two times. $\endgroup$ Nov 17, 2023 at 11:33
  • 1
    $\begingroup$ So Wikipedia’s $f_k$ is what you call $P_k$, and $\bar o_k$ is what you call $o_k/n_k$, right? $\endgroup$
    – Dave
    Nov 17, 2023 at 17:34

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