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Distribution of Bernoulli estimate RRS feed

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    In the Infer.net model I'm using I have a boolean variable that I wish to work out the distribution of Infer.net's estimation of the Bernoulli distribution for that variable (which I think is called the conjugate prior, in this case the beta distribution?). Currently I infer from the model the Bernoulli variable like so:

    Bernoulli P1WinEstimate = predictmodel.engine.Infer<Bernoulli>(predictmodel.P1Win);

    However I'm unsure what I should do to get the Beta distribution for this Bernoulli variable from here. What is the appropriate way to do this in Infer.net?

    Wednesday, September 24, 2014 3:48 PM

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    Given skills for the two players, the probability that player 0 will win is MMath.NormalCdf((Skill[0]-Skill[1])/Math.Sqrt(2*Beta)).  Call this p.  Infer.NET can't compute the distribution of p, but you can easily compute moments of p.  The expected value of p is the probability that P1Win is true, as you have already computed.  The expected value of p^2 is the probability that player 0 will win 2 games.  You can make a second model that computes this.  Finally, the variance of p is p(win 2 games) - p(win 1 game)^2.
    • Marked as answer by MarkG87 Sunday, September 28, 2014 1:37 PM
    Friday, September 26, 2014 12:48 PM
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    The Beta distribution would be for the parent variable of P1Win.  How have you defined P1Win in the model?
    Wednesday, September 24, 2014 5:38 PM
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    It's the TrueSkill model - full model code below:

    public PredictModel() //Build model
        {
            BetaPrior = Variable.New<Gamma>();
            Beta = Variable.Random<double, Gamma>(BetaPrior);
            for (int i = 0; i < 2; i++)
            {
                SkillPrior[i] = Variable.New<Gaussian>();
                Skill[i] = Variable.Random<double, Gaussian>(SkillPrior[i]);
                Performance[i] = Variable.GaussianFromMeanAndPrecision(Skill[i], Beta);
            }
            PerformanceDiff = Performance[0] - Performance[1];
            P1Win = (PerformanceDiff) > 0;
            P2Win = (-PerformanceDiff) > 0;
        }

    Thursday, September 25, 2014 11:39 PM
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    There is no Beta distribution in this model.  Perhaps you could clarify the question?
    Friday, September 26, 2014 11:45 AM
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    Essentially I'm trying to infer how confident the model is in its prediction of the Bernoulli value - so something like the distribution of that prediction or confidence intervals would be very useful. I tried variance but that simply gives p(1-p) which doesn't capture what I'm trying to understand.
    Friday, September 26, 2014 12:03 PM
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    Given skills for the two players, the probability that player 0 will win is MMath.NormalCdf((Skill[0]-Skill[1])/Math.Sqrt(2*Beta)).  Call this p.  Infer.NET can't compute the distribution of p, but you can easily compute moments of p.  The expected value of p is the probability that P1Win is true, as you have already computed.  The expected value of p^2 is the probability that player 0 will win 2 games.  You can make a second model that computes this.  Finally, the variance of p is p(win 2 games) - p(win 1 game)^2.
    • Marked as answer by MarkG87 Sunday, September 28, 2014 1:37 PM
    Friday, September 26, 2014 12:48 PM
    Owner