9
$\begingroup$

Question summary

I had recently asked this question where problems encoding a Bayesian Network were linked to the use of MultinomialDistribution. While that problem can be avoided using EmpiricalDistribution, there remains an issue with using ProbabilityDistribution for larger networks as it seems: While Probability can be used for inference with 4 nodes, it will not evaluate for the "full" example network of 5 nodes -- which still is far removed from real application demands. Why is this so? What can be done about it?

Bayesian Network Example

Again I would like to use the (simple) example that is given on page 53 in Probabilistic Graphical Models (2009), by Daphne Koller and Neir Friedman:

BayesianNetwork

The network has five nodes (random variables):

  • Difficulty of a class taken by a student (0 = easy, 1 = hard)
  • Intelligence of the student (0 = low, 1 = high)
  • Grade achieved by the student (1 = A, 2 = B, 3 = C)
  • SAT score of the student (0 = low, 1 = high)
  • Letter of recommendation by the teacher (0 = False, 1 = True)

We would like to use this network to do probabilistic inference (causal or evidential) like: "What is the probability of the student achieving an A, given that he is intelligent?"

Encoding the Bayesian Network in Mathematica

Essentially the Bayesian Network is a sparse way to define the joint probability distribution function for the random variables using the chain rule of probability theory:

$ \begin{align} P(I,D,G,S,L) = P(I) \times P(D) \times P(G|I,D) \times P(S|I) \times P(L|G) \end{align} $

I am encoding this in Mathematica as follows:

(* nodes without parents *)
distI = BernoulliDistribution[ 0.3 ]; (* prior probability of high intelligence *)
distD = BernoulliDistribution[ 0.4 ]; (* prior probability of hard class *)

(* nodes with parents = conditional probability distributions *)
(* conditional distribution of the grade *)
cpdG = Function[ { i, d },
    With[
      {
        p = Piecewise[
                {
                  { {  0.3,  0.4,  0.3  }, i == 0 && d == 0 },
                  { {  0.05, 0.25, 0.7  }, i == 0 && d == 1 },
                  { {  0.9,  0.08, 0.02 }, i == 1 && d == 0 },
                  { {  0.5,  0.3,  0.2  }, i == 1 && d == 1 }
                }
            ]
      },
      EmpiricalDistribution[ p -> Range[3] ]
    ]
];

(* conditional distribution for the SAT score *)
cpdS = Function[ i,
   With[
    {
     θ = Piecewise[
           {
             { 0.05, i == 0 },
             { 0.8,  i == 1 }
           }
         ] (* probability of a high SAT score *)
     },
     BernoulliDistribution[θ]
   ]
];

(* conditional probability function for the Letter *)
cpdL = Function[ g,
   With[
     {
       θ = Piecewise[
             {
               { 0.9,  g == 1 },
               { 0.6,  g == 2 },
               { 0.01, g == 3 } 
             }
           ]
     },
     BernoulliDistribution[θ]
   ]
];

(* BayesNetwork = Joint Probability Distribution Function *)
(* B4 = P(I,D,G,L) *) 
distB4 = ProbabilityDistribution[
   PDF[ distI, i] PDF[ distD, d] PDF[ cpdG[i,d], g] PDF[ cpdL[g], l],
   {i, 0, 1, 1},
   {d, 0, 1, 1},
   {g, 1, 3, 1},
   {l, 0, 1, 1}
];

(* B5 = P(I,D,G,S,L) *)
distB5 = ProbabilityDistribution[
   PDF[ distI, i] PDF[ distD, d] PDF[ cpdG[i,d], g] PDF[ cpdS[i], s] PDF[ cpdL[g], l],
   {i, 0, 1, 1},
   {d, 0, 1, 1},
   {g, 1, 3, 1},
   {s, 0, 1, 1},
   {l, 0, 1, 1}
];

Doing Inference

Now we would like to ask the question as stated above:

Probability[ g == 1 \[Conditioned] i == 1, {i,d,g,l} \[Distributed] distB4 ]

0.74

Probability[ g == 1 \[Conditioned] i == 1, {i,d,g,s,l} \[Distributed] distB5 ]

Probability[ ] is returned unevaluted.

Why is this the case? What can be done about it - after all 5 nodes should not be too far a stretch?

$\endgroup$
2
  • $\begingroup$ One can also verify that the network P(I,D,G,S) will work without any problems, so it seems to be the number of nodes that causes trouble. $\endgroup$
    – gwr
    Commented Oct 17, 2016 at 12:06
  • $\begingroup$ The problem also arises using Which instead of Piecewise or using functions with pattern tests, e.g. cpdL[ g_?NumericQ] := ... . $\endgroup$
    – gwr
    Commented Nov 16, 2016 at 14:56

1 Answer 1

11
$\begingroup$

Beware of Piecewise

As indicated in the answer given by WRI here, the interplay of Piecewise and ProbabilityDistribution is tricky and -- so my temporary verdict -- is best avoided.

Indeed, using indicator functions, e.g. Boole, as a replacement for Piecewise solves the issue:

(* nodes without parents remain unchanged *)
(* CPDs are redefined using Boole instead of Piecewise *)

(* conditional distribution of the grade *)
cpdG = Function[ {i,d},
   With[
      {
         p = Plus[
                {0.3 , 0.4 , 0.3 } Boole[ i == 0 && d == 0 ],
                {0.05, 0.25, 0.7 } Boole[ i == 0 && d == 1 ],
                {0.9 , 0.08, 0.02} Boole[ i == 1 && d == 0 ],
                {0.5 , 0.3 , 0.2 } Boole[ i == 1 && d == 1 ]
             ]
      },
      EmpiricalDistribution[ p -> Range[3] ]
   ]
];

(*conditional distribution for the SAT score*)
cpdS = Function[ i,
   With[
      {
         θ = Plus[
                0.05 Boole[ i == 0 ],
                0.8  Boole[ i == 1 ]
             ] (*probability of a high SAT score*)
      },
      BernoulliDistribution[θ]
   ]
];

(*conditional probability function for the Letter*)
cpdL = Function[ g,
   With[
      {
         θ = Plus[
                0.9  Boole[ g == 1 ],
                0.6  Boole[ g == 2 ],
                0.01 Boole[ g == 3 ]
             ]
      },
      BernoulliDistribution[θ]
   ]
];

(* B5 = P(I,D,G,S,L) complete BN *)
distB5 = ProbabilityDistribution[
   PDF[ distI, i] PDF[ distD, d] PDF[ cpdG[i,d], g] PDF[ cpdS[i], s] PDF[ cpdL[g], l],
   {i, 0, 1, 1}, {d, 0, 1, 1}, {g, 1, 3, 1}, {s, 0, 1, 1}, {l, 0, 1, 1}
];

Now doing inference for the complete joint probability distribution as specified by the Bayesian Network works out fine:

Probability[ g == 1\[Conditioned] i == 1, {i,d,g,s,l} \[Distributed] distB5 ]

0.74

$\endgroup$
2
  • 6
    $\begingroup$ Maybe (as a reminder) it is worth pointing out that within Probability the order of the variables must match the one given by the distribution. E.g. Probability[ i == 1 \[Conditioned] d ==1 && g ==2, ... ] will evaluate, but Probability[ i == 1 \[Conditioned] g==2 && d==1, ... ] will not. $\endgroup$
    – gwr
    Commented Oct 26, 2016 at 16:57
  • 1
    $\begingroup$ You might also consider using Simplify`PWToUnitStep[] to convert any Piecewise[] instance, if you find that representation troublesome. $\endgroup$ Commented Jun 5, 2020 at 13:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.