I'm working on a hobby project that involves tournament-style player rankings, and I'm using the TrueSkill system developed by Microsoft for online gaming. It's a great system, but much of the math is far above my head. Even without understanding it, I've got most of what I need figured out, except for two equations which are never explicitly defined. The document that best outlines them is http://research.microsoft.com/pubs/67956/NIPS2006_0688.pdf.

(1) On the first page, equation 1 references psi, which denotes "the cumulative density of a zero-mean unit-variance Gaussian."

(2) In the preceding paragraph, there is mention of a function, N, representing player performance.

Having never taken a day of statistics in my life, and certainly not the graduate-level math that it seems is necessary for understanding this paper, it'd be great if somebody who does understand could just give me a numerical formula for these two functions that I could plug into my code (programming project). If you're curious, or it helps to answer the question, another great article about the system which is a little more approachable is http://research.microsoft.com/en-us/projects/trueskill/details.aspx.

Thanks for whatever help anyone

  • $\begingroup$ en.wikipedia.org/wiki/… for $\Phi$. $\cal N$ does not give the player's performance, but rather the "probability distribution" of the performance. I'm fairly certain that for $\cal N$, use $f(p_i; s_i,\beta^2)$, where $F$ is as in the wiki link right above $\Phi$. $\endgroup$ Nov 21, 2011 at 10:12
  • $\begingroup$ Whoever wrote the words "cumulative density" was at best not paying attention to what they wrote and not an expert on what they wrote. There is no such thing as "cumulative density". It's an oxymoron, and whoever wrote it is probably a moron. The word "cumulative" contradicts the word "density". $\endgroup$ Nov 21, 2011 at 17:38

1 Answer 1



The phi function $\Phi(x)$ you saw in the paper is defined as the integral

$$\Phi(x)=\frac1{\sqrt{2\pi}}\int_{-\infty}^x e^{-\frac{t^2}{2}} \mathrm dt$$

where the function

$$\phi(x)=\frac1{\sqrt{2\pi}} e^{-\frac{x^2}{2}}$$

is what they were referring to as the "zero-mean unit-variance Gaussian". One might alternatively see $\Phi(x)$ being expressed in terms of the error function $\mathrm{erf}(x)$:


(in short, what they called "cumulative density" (more properly, "cumulative distribution", as Michael points out.) here means taking the integral of the "probability density"; that is, you are integrating your Gaussian function.)

Any good statistics books should have a discussion of the Gaussian/normal distribution.

For methods on evaluating the error function on a computer, see this or this.


$p_i \sim \mathcal N(p_i;s_i,\beta^2)$ was in fact already explained in the sentence that brought it up: that is, the performance $p_i$ follows a Gaussian/normal distribution around the skills $s_i$, with $\beta^2$ being the variance.

  • $\begingroup$ I rarely downvote anything, but I'm making an exception here. The source of the phrase "cumulative density" is students who take courses and try to parrot things without understanding. The word "cumulative" contradicts the word "density" and I'm surprised to see an otherwise intelligent person condoning that phrase. "PDF" = "probability density function". "CDF" = "cumulative distribution function". Do not ever say or write "cumulative density function". $\endgroup$ Nov 21, 2011 at 17:47
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    $\begingroup$ I hope the fix was to your satisfaction, @Michael. :) Thanks for the nudge. $\endgroup$ Nov 21, 2011 at 17:49
  • $\begingroup$ I've rescinded my downvote. I've also written to the three guys at microsoft who wrote that thing on the web that was linked to. $\endgroup$ Nov 21, 2011 at 17:56
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    $\begingroup$ My appologies for writing this - I've never taken statistics at all, so I'm really going in blind, although by your description, I see how wrong that is. Thanks for writing the Microsoft guys - sounds like an important fix. $\endgroup$
    – joshlf
    Dec 1, 2011 at 20:45
  • $\begingroup$ @MichaelHardy If you hadn't seen, and are interested, I believe you were referenced here :) $\endgroup$
    – pjs36
    May 16, 2015 at 23:04

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