Topic: Probabilistic Reasoning
Related Reading: Ch. 14
Due:

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1. (20 points)

   We have a bag of three biased coins, a, b, and c, with probabilities of coming up heads of 20%, 60%, and 80%, respectively. One coin is drawn randomly from the bag (with equal likelihood of drawing each of the three coins), and then the coin is flipped three times to generate the outcomes X₁, X₂, X₃.

   1. Draw a Bayesian network corresponding the this setup and define the necessary CPTs.

      General topology is

                      Coin
                    /   |  \
                   X1   X2  X3
      		

      The variable Coin has (unconditional) probability <1/3, 1/3, 1/3> of respectively being <a, b, c>.

      Each of the X_k variables has the identical CPT

      Coin	P(Xk=head | Coin)
      a	0.2
      b	0.6
      c	0.8

   2. Calculate the conditional probability P(Coin | X₁=head, X₂=head, X₃=tail). That is, although there was a prior probability that each coin was picked with probability 1/3, what is the posterior probability that the coin is a, b, or c.

      By Bayes rule, P(Coin | X₁=head, X₂=head, X₃=tail) = P(X₁=head, X₂=head, X₃=tail | Coin)*P(Coin)/P(X₁=head, X₂=head, X₃=tail) and we can calculate that by normalizing αP(X₁=head, X₂=head, X₃=tail | Coin)*P(Coin)

      Given that there are three discrete values for Coin, and that P(Coin) is uniform, we get that the desired conditional probability is α<P(X₁=head, X₂=head, X₃=tail | Coin=a), P(X₁=head, X₂=head, X₃=tail | Coin=b), P(X₁=head, X₂=head, X₃=tail | Coin=c)> = α<0.2*0.2*0.8, 0.6*0.6*0.4, 0.8*0.8*0.2> = α<0.032, 0.144, 0.128> = <0.10526, 0.47368, 0.42105> = <2/19, 9/19, 8/19>

2. (30 points)

   Let H_x be a random variable denoting the handedness of a person x, with possible values l for x being left-handed and r for x being right-handed. A popular hypothesis is that left- or right-handedness is inherited by a simple mechanism; that is, perhaps there is a gene G_x, also with values l or r, and actual handedness turns out the same (with some probability s) as the gene that the person possesses. Furthermore, the gene itself is equally likely to be inherited from either of a person's parents, with a small non-zero probability m of a random mutation flipping the handedness.

   Consider the following three Bayesian Networks.

   (1)	(2)	(3)

   1. Which of the three networks claims that P(G_father, G_mother, G_child) = P(G_father) P(G_mother) P(G_child) ?

      The three variables are clearly modeled as independent in Network (3). In the other two networks, P(G_child) is modeled with a direct dependency on the other two.

   2. Which of the three networks make independence claims that are consistent with the hypothesis about the inheritance of handedness?

      Networks (1) and (2) both model the inheritance portion of the hypothesis, as they suggest that a child's gene is dependent solely on the parent's genes.

   3. Which of the three networks is the best description of the hypothesis?

      Network (1) best models the genetic hypothesis as well as the hypothesis that an individual's handedness is predicted based solely on the underlying gene.

      Note that Network (2) suggests that the handedness of the child is not only dependent on that child's gene, but also on the manifested handedness of the parents (e.g, that the right-handed parent may always put toys in the child's right hand)

   4. Write down the CPT for the G_child node in network (1), in terms of s and m.

      Gmother	Gfather	P(Gchild=l)
      l	l	1-m
      l	r	0.5
      r	l	0.5
      r	r	m

      We note that the above values can be calculated by separately considering the two equally likely cases of the gene being passed by the mother versus the father, and then the subsequent possibility of mutation in either case. Note that for the row with G_mother = l and G_father = r, the child might end up left handed either by having that gene inherited from the mother and then not mutated (thus with probability 0.5 * (1-m) ), or perhaps by inheriting r from the father but then having it mutated (thus with probability 0.5 * m). Therefore, the overall probability in that case is 0.5*(1-m) + 0.5*m = 0.5.

   5. Suppose that P(G_mother = l) = P(G_father = l) = q. In network (1), derive an expression for P(G_child = l) in terms of m and q only, by conditioning on its parent nodes.

      Using the CPT from the previous question, the desired probability is q*q*(1-m) + q*(1-q)*0.5 + q*(1-q)*0.5 + (1-q)*(1-q)*m = q*q*(1-m) + q*(1-q) + (1-q)*(1-q)*m. Simplifying, we get q + m - 2mq

      This expression can also be deduced by recognizing that no matter which parent a gene is inherited from, there will be probability q that it is L. The probability of the child ending up with L is either by inheriting L and not mutating it or by inheriting R and mutating it. So the overall probability is thus q*(1-m) + (1-q)*m = q + m - 2qm.

   6. Under conditions of genetic equilibrium, we expect the distribution of genes to be the same across generations. Use this to calculate the value of q, and given what you know about handedness in humans, explain why the hypothesis described at the beginning of the question must be wrong.

      To get equilibrium, we should have P(G_mother = l) = P(G_father = l) = P(G_child = l) = q. Based on the previous part, this requires that q = q + m - 2mq and thus q = 0.5. However, we do not observer lefthandedness with probability 0.5, and so it is unlikely that a single underlying gene has an equilibrium probability of 0.5.