Beta-Bernoulli model

[mateusjoffily]

Hi all,

Why is the prediction of y in the Beta-Bernoulli model below different from a/(a+b), when y is missing?

@model function beta_bernoulli(y, a, b) 
    θ ~ Beta(a, b)
    y ~ Bernoulli(θ) 
end

result = infer(
    model = beta_bernoulli(a = 2, b = 1),
    data  = (y =  missing, )
)

println(result.posteriors[:θ])
println(result.predictions[:y])

Beta{Float64}(α=2.0, β=1.0)
Bernoulli{Float64}(p=0.7310585786300049)

[bvdmitri]

Hey, yes this is a consequence of default variational factorization for the data points, RxInfer treats y as actual data point, thus it is not being included in the resulting variational distributions and you get "worse" estimate. If you want to change that you should use y = UnfactorizedData(missing), e.g.

julia> result = infer(
           model = beta_bernoulli(a = 2, b = 1),
           data  = (y =  UnfactorizedData(missing), )
       )
Inference results:
  Posteriors       | available for (θ)
  Predictions      | available for (y)


julia> println(result.posteriors[:θ])
Beta{Float64}(α=2.0, β=1.0)

julia> println(result.predictions[:y])
Bernoulli{Float64}(p=0.6666666666666666)

Long story short UnfactorizedData usually gives you better estimate, but works in less scenarios. We could document it better. It only mentioned once in the documentation.

[mateusjoffily]

Great, thank you very much for your help!

Best,
Mateus