r - R 中的 TrueSkill 实现

Question

我正在尝试编写基本的 Vanilla TrueSkill (3.1) R 中的算法，但得到了一些奇怪的结果。

我的代码如下:

# A simple test between two players repeatedly laying one another
betaSq = 0.1
obs = 10000

p1_skills = 0.333
p2_skills = 0

p1_draws = rnorm(obs, p1_skills, sqrt(betaSq))
p2_draws = rnorm(obs, p2_skills, sqrt(betaSq))

p1_pred_mu = rep(NA, obs+1)
p1_pred_sigmaSq = rep(NA, obs+1)
p2_pred_mu = rep(NA, obs+1)
p2_pred_sigmaSq = rep(NA, obs+1)

# Initial values
p1_pred_mu[1] = 0
p1_pred_sigmaSq[1] = 1
p2_pred_mu[1] = 0
p2_pred_sigmaSq[1] = 1

results = p1_draws > p2_draws
probs = rep(NA, obs)

# Run TrueSkill
for (i in seq(2,obs+1)) {
  probs[i-1] = predictProb(p1_pred_mu[i-1], p1_pred_sigmaSq[i-1], p2_pred_mu[i-1], p2_pred_sigmaSq[i-1], betaSq)
  out = updateSkill(p1_pred_mu[i-1], p1_pred_sigmaSq[i-1], p2_pred_mu[i-1], p2_pred_sigmaSq[i-1], betaSq, results[i-1])

  # Now update based on the out  
  p1_pred_mu[i] = out$mu1
  p1_pred_sigmaSq[i] = out$sigmaSq1
  p2_pred_mu[i] = out$mu2
  p2_pred_sigmaSq[i] = out$sigmaSq2
}

# Output results
dev.new()
mu = p1_pred_mu
lower = qnorm(0.05, p1_pred_mu, p1_pred_sigmaSq)
upper = qnorm(0.95, p1_pred_mu, p1_pred_sigmaSq)
plot(mu, ylim = c(min(lower), max(upper)), main = "p1")
lines(lower)
lines(upper)

dev.new()
mu = p2_pred_mu
lower = qnorm(0.05, p2_pred_mu, p2_pred_sigmaSq)
upper = qnorm(0.95, p2_pred_mu, p2_pred_sigmaSq)
plot(mu, ylim = c(min(lower), max(upper)), main = "p2")
lines(lower)
lines(upper)

a = filter(probs, rep(1, 20))/20
dev.new()
plot(a)

print(sprintf("Mean p1: %g", mean(p1_pred_mu)))
print(sprintf("Mean p2: %g", mean(p2_pred_mu)))
print(sprintf("Mean results: %g", mean(results)))
print(sprintf("Mean predicted results: %g", mean(probs)))

调用的函数有:

# Functions
updateSkill <- function(mu1, sigmaSq1, mu2, sigmaSq2, betaSq, result) {
  # http://papers.nips.cc/paper/3331-trueskill-through-time-revisiting-the-history-of-chess.pdf
  c = 2*betaSq + sigmaSq1 + sigmaSq2

  if (result == 1) {
    # Player 1 wins
    v = dnorm((mu1-mu2)/c)/pnorm((mu1-mu2)/c)
    w = v*(v+(mu1-mu2)/c)

    mu1 = mu1 + (sigmaSq1/c)*v
    mu2 = mu2 - (sigmaSq2/c)*v

    sigmaSq1 = sigmaSq1 * sqrt(1 - (sigmaSq1/c^2)*w)
    sigmaSq2 = sigmaSq2 * sqrt(1 - (sigmaSq2/c^2)*w)
  } else if (result == 0) {
    # Player 2 wins
    v = dnorm((mu2-mu1)/c)/pnorm((mu2-mu1)/c)
    w = v*(v+(mu2-mu1)/c)

    mu1 = mu1 - (sigmaSq1/c)*v
    mu2 = mu2 + (sigmaSq2/c)*v

    sigmaSq1 = sigmaSq1 * sqrt(1 - (sigmaSq1/c^2)*w)
    sigmaSq2 = sigmaSq2 * sqrt(1 - (sigmaSq2/c^2)*w)    
  }

  return(list(mu1=mu1, mu2=mu2, sigmaSq1=sigmaSq1, sigmaSq2=sigmaSq2))
}

predictProb <- function(mu1, sigmaSq1, mu2, sigmaSq2, betaSq) {
  # Try to predict the probability of player 1 beating player 2 using Trueskill model
  mean1 = mu1
  mean2 = mu2
  var1 = sigmaSq1 + betaSq
  var2 = sigmaSq2 + betaSq

  # Now the dist of player1 - player2 is N(mean1 - mean2, sqrt(var1 + var2))
  prob1Wins = pnorm(0, mean2 - mean1, sqrt(var1 + var2))

  return(prob1Wins)
}

我讨厌发布大量代码块，但我真的无法弄清楚哪里出了问题。

该程序运行并且预测的技能(分配给 N(mu, sigma))收敛。然而，他们给出的预测概率并没有收敛到结果的真实概率!

示例输出是:

[1] "Mean p1: 0.0762161"
[1] "Mean p2: -0.0762161"
[1] "Mean results: 0.7733"
[1] "Mean predicted results: 0.631424"

知道出了什么问题吗？

Answer 1

这不起作用的原因是因为在 updateSkills 函数的第 3 行它应该读取

c = sqrt(2*betaSq + sigmaSq1 + sigmaSq2)

不是

c = 2*betaSq + sigmaSq1 + sigmaSq2