How to calculated DRPS (Discrete Rank Probability Score)

Question

I am working on replicating the scoring rule found in a paper Forecasting the intermittent demand for slow-moving inventories: A modelling approach

The paper describes the scoring rule as follows:

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This is my attempt

y <- rpois(n = 100, lambda = 10) # forecasted distribution
x <- 10 # actual value

drps_score <- function(x = value, y = q){
  # x = actual value (single observation); y = quantile forecasted value (vector)
  Fy = ecdf(y) # cdf function
  indicator <- ifelse(y - x > 0, 1, 0) # Heaviside
  score <- sum((indicator - Fy(y))^2)
  return(score)
}

> drps_score(x = x, y = y)
[1] 53.028

This seems to work well until I provide a vector of 0s as follows:

y <- rep(x = 0, 100)
> drps_score(x = x, y = y)
[1] 0

I know that one of their methods used in this paper was a 0s forecast and their results did not show 0 for DRPS. This makes me think that the calculation is off.

Answer 1

I think there are a few issues at play here.

First off, I don't think you are computing the correct sum inside the scoring function. The score asks you to sum across all possible values of y (i.e. across all positive integers) not across all forecasted samples of y.

Second, I don't think the above definition gives the desired result, with \hat F (y) defined to be 0 when y=x then you don't get a zero score for a forecast with a point mass at the true value. (Yes, I'm saying that source is "wrong", or at least has a definition that doesn't give the desired result.) Here is a re-formulated function that I think fixes both issues:

x <- 10 # actual value

drps_score <- function(x = value, y = q, nsum=100){
    # x = actual value (single observation); y = quantile forecasted value (vector)
    Fy = ecdf(y) # cdf function
    ysum <- 0:nsum
    indicator <- ifelse(ysum - x >= 0, 1, 0) # Heaviside
    score <- sum((indicator - Fy(ysum))^2)
    return(score)
}



> drps_score(x = x, y = rpois(n = 1000, lambda = 8))
[1] 1.248676
> drps_score(x = x, y = rpois(n = 1000, lambda = 9))
[1] 0.878183
> drps_score(x = x, y = rpois(n = 1000, lambda = 10))
[1] 0.692667
> drps_score(x = x, y = rep(10, 100))
[1] 0
> drps_score(x = x, y = rpois(n = 1000, lambda = 11))
[1] 0.883333

The above shows that the distribution that is centered on the true value (lambda=10) has the lowest score for distributions that aren't a point mass.