Machine Learning for Computer Vision Winter term November 2017 Topic: Probabilistic Graphical Models
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1 TU München Fakultät für Informatik PD Dr. Rudolph Triebel John Chiotellis, Max Denninger Machine Learning for Computer Vision Winter term November 207 Topic: Probabilistic Graphical Models Exercise : Reading a graphical model We have the following graphical model: Abbildung : Graphical model. a Write the joint probability distribution corresponding to the graphical model depicted in Fig.. p(a, B, C, D, E p(ap(cp(b A, Cp(D Cp(E B b What are the conditional indepence assumptions of this model? A C, D A, B C, E A, C, D B. c Which of the following assertions are true, and why? Algorithm to check, whether X is d-seperated from Y by Z (X,Y,Z sets of nodes:
2 boolean is_dsep(x,y,z{ foreach x X, y Y foreach path p connecting x and y if (!is_blocked(p,z return false; ; ; return true; } boolean is_blocked(p,z{ foreach n p if (type(n hh if (n / Z m / Z n... m return true; //case (b else //type(n ht or type(n tt if (n Z return true; //case (a return false; } B is d-separated from D by C: true (case (a, A is d-separated from C by E: false (case (b fails as B E, A is d-separated from C by D: true (case (b, E is d-separated from D by B: true (case (a, E is d-separated from D by A: false. 2
3 Exercise 2: Markov Chain We have the following Markov Chain: a Write the joint probability distribution associated to this Markov Chain. p(a, B, C, D Z ψ A,B(A, Bψ B,C (B, Cψ C,D (C, D b Each variable can take value 0 or, and we want to express that it is 9 times more probable that neighboring variables have equal values than they have different value. Give the potential functions of this Markov Chain. All three potential functions are the same: V V Notice that the values need not be normalized in any way. c Compute the probability distributions p(a and p(c. µ α and µ β can be calculated recursively: µ α (x n x n ψ n,n (x n, x n µ α (x n µ β (x n x n+ ψ n,n+ (x n, x n+ µ β (x n+ With our potentials this yields: ( µ α (A (initialization, optional, µ α (B A µ ( ( α(aψ A,B (A, A µ α(aψ A,B (A, µ α (C B µ ( ( α(bψ B,C (B, B µ α(bψ B,C (B, µ α (D C µ ( ( α(cψ C,D (C, C µ α(cψ C,D (C, ( µ β (D (initialization, optional, 3
4 µ β (C D µ ( ( β(dψ C,D (0, D D µ β(dψ C,D (, D µ β (B C µ ( ( β(cψ B,C (0, C C µ β(cψ B,C (, C µ β (A B µ ( ( β(bψ A,B (0, B B µ β(bψ A,B (, B Then we compute the normalization factor Z at any point, for example B: Z B µ α (B.µ β (B 2000 Finally we can compute the marginal distributions requested: p(a Z.µ α(a.µ β (A ( ( p(c Z.µ α(c.µ β (C ( ( The assumptions were only that neighboring nodes should be equal. The marginal on A and C both say that we have no idea on their value. That was to be expected. d Now, we observe that D is, recompute the distributions over A and C: p(a [D ] and p(c [D ]. We ve learned that we could compute marginal distributions by decomposing the inference into messages to be passed between nodes. How can we adapt this mecanism to observations? If the chain contained only C and D, we would have: p(c [D ] Z ( ψc,d (0, ψ C,D (, This can be written in the same message passing form: p(c [D ] (. Z D µ β (Dψ C,D(0, D D µ β (Dψ C,D(, D ( 0 with µ β (D. Actually, you just have to replace the µ (X with a Dirac in order to account for an observation of the value of X (and recompute the normalization factor: ( µ α (A, 4
5 µ α (B A µ ( ( α(aψ A,B (A, A µ α(aψ A,B (A, µ α (C B µ ( ( α(bψ B,C (B, B µ α(bψ B,C (B, µ α (D C µ ( ( α(cψ C,D (C, C µ α(cψ C,D (C, ( 0 µ β (D (observation, µ β (C D µ β (Dψ ( ( C,D(0, D D µ β (Dψ C,D(, D µ β (B C µ β (Cψ ( ( B,C(0, C C µ β (Cψ B,C(, C µ β (A B µ β (Bψ ( ( A,B(0, B B µ β (Bψ A,B(, B As above, we can also compute Z 000 and then: p(a [D ] Z.µ α(a.µ β(a 000 ( p(c [D ] Z.µ α(c.µ β(c ( ( ( Now, we know that node D equals and we see it has become more probable for A and C to be equal to (the more for C which is nearer D than A. At least the result makes sense. e Compute p(c [A 0], [D ]. With the same way, we can recompute µ α (which is symmetric to µ β : ( µ α(a, 0 ( ( µ α(b A µ α(aψ A,B (A, A µ α(aψ A,B (A, ( ( µ α(c B µ α(bψ B,C (B, B µ α(bψ B,C (B, ( ( µ α(d C µ α(cψ C,D (C, C µ α(cψ C,D (C, Now Z 244 and: p(c [D ] Z.µ α(c.µ β(c ( (
6 ( It would be the reverse for B: chain, both µ α and µ β. It is not exactly 2. Actually, with a longer 3 would (exponentially converge to uniforms as we consider node further ( from their origin. Therefore ( for a long chain, the probability ( will come 00% 50% 0% from to rest at the uniform before setting to. In 0% 50% 00% order to straighten the values, we could lower the probability of being different from neighboring nodes. Note that the known nodes are at the boundary of our chain. If it was not the case, the d-separation property would have allowed us to split the chain in two indepent subchains having both a copy of the observed variable as the new boundary. 6
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Copyright Richard J. Povinelli rev 1.0, 10/1//2001 Page 1 Probabilistic Reasoning Systems Dr. Richard J. Povinelli Objectives You should be able to apply belief networks to model a problem with uncertainty.