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solhw43825

Course: ECONOMICS 2301, Spring 2012
School: UT Dallas
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Word Count: 877

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FOR SOLUTIONS HOMEWORK 5, STAT 4382 1. Exerc. 3.1.1. Well, the probability in question is Pr(X0 = 0, X1 = 1, X2 = 2). This joint probability can be written via conditional ones as follows, Pr(X0 = 0, X1 = 1, X2 = 2) = Pr(X0 = 0)Pr(X1 = 1|X0 = 0)Pr(X2 = 2|X1 = 1, X0 = 0). Then, using Markov property of the chain we can simplify the right side, Pr(X0 = 0, X1 = 1, X2 = 2) = Pr(X0 = 0)Pr(X1 = 1|X0 = 0)Pr(X2 = 2|X1 =...

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FOR SOLUTIONS HOMEWORK 5, STAT 4382 1. Exerc. 3.1.1. Well, the probability in question is Pr(X0 = 0, X1 = 1, X2 = 2). This joint probability can be written via conditional ones as follows, Pr(X0 = 0, X1 = 1, X2 = 2) = Pr(X0 = 0)Pr(X1 = 1|X0 = 0)Pr(X2 = 2|X1 = 1, X0 = 0). Then, using Markov property of the chain we can simplify the right side, Pr(X0 = 0, X1 = 1, X2 = 2) = Pr(X0 = 0)Pr(X1 = 1|X0 = 0)Pr(X2 = 2|X1 = 1). Now we can plug in numbers, Pr(X0 = 0, X1 = 1, X2 = 2) = (.3)(.2)(0) = 0. 2. Exerc. 3.1.2. Well, here we have a similar situation. Write Pr(X2 = 1, X3 = 1|X1 = 0) = Pr(X3 = 1|X2 = 1, X1 = 0)Pr(X2 = 1|X1 = 0) [and continue using Markov property] = Pr(X3 = 1|X2 = 1)Pr(X2 = 1|X1 = 0). Now plug in the given values from the transition probability matrix, Pr(X2 = 1, X3 = 1|X1 = 0) = (.6)(.2) = .12. For the second probability in question the formula is the same, Pr(X1 = 1, X2 = 1|X0 = 0) = Pr(X2 = 1|X1 = 1)Pr(X1 = 1|X0 = 0) = .12. Remark: Note that the probabilities are the same. It should be clear why: there is just a unit shift in index (time) and, because the Markov chain is stationary -- we have this equality. 3. Exerc. 3.1.3. Note that the probability in question is not as it is written in the text but Pr(X0 = 1, X1 = 0, X2 = 2|X0 = 1) because it is given that X0 = 1. In the strict sense, the problem is not written correctly in the text. But in any case, now we can continue, using the Markov property of the chain, Pr(X0 = 1, X1 = 0, X2 = 2|X0 = 1) == Pr(X2 = 1|X1 = 1)Pr(X1 = 1|X0 = 0)Pr(X0 = 0|X0 = 0) [and then we plug in numbers and continue] = (.3)(.1) = .03. 1 (1) 4. Exerc. 3.1.4. Note that here again the two probabilities are just shifts in time, and thus due to stationarity must be the same. Write, Pr(X1 = 1, X2 = 1|X0 = 0) = Pr(X2 = 1|X1 = 1)Pr(X1 = 1|X0 = 0) = (.2)(.1) = .02. The second probability is the same. 5. Exerc. 3.1.5. Write, Pr(X0 = 1, X1 = 1, X2 = 0) = Pr(X0 = 1)Pr(X1 = 1|X0 = 1)Pr(X2 = 0|X1 = 1, X0 = 1). Using the Markov property of the chain we can simplify the right side, Pr(X0 = 1, X1 = 1, X2 = 0) = Pr(X0 = 1)Pr(X1 = 1|X0 = 1)Pr(X2 = 0|X1 = 1) = (.5)(.1)(.5) = .025. (2) The second probability is more involved because we do not have Pr(X1 = i) and must calculate it. In any case, similarly to (2) we write Pr(X1 = 1, X2 = 1, X3 = 0) = Pr(X1 = 1)Pr(X2 = 1|X1 = 1)Pr(X3 = 0|X2 = 1) = Pr(X1 = 1)(.1)(.5). Now we need to calculate Pr(X1 = 1). Using Total Probability Theorem we get P (X1 = 1) = P (X1 = 1, X0 = 0) + P (X1 = 1, X0 = 1) + P (X1 = 1, X0 = 2) = P (X0 = 0)P (X1 = 1|X0 = 0)+P (X0 = 1)P (X1 = 1|X0 = 1)+P (X0 = 2)P (X1 = 1|X0 = 2) = (.5)(.2) + (.5)(.1) + (0)(.2) = .1 + .05 = .15. Now we can finish. Plug in the result in (3) and get Pr(X1 = 1, X2 = 1, X3 = 0) = (.15)(.1)(.5) = .0075. 6. Problem 3.1.1. Note that we are asked only about the transition probability matrix. Let Xn be the number of diseased the at end of nth iteration. Then it is clear that P0,0 = 1, P5,5 = 1 and Pk,k-1 = 0 for any k = 1, 2, 3, 4, 5. As a result, we need to find only two transition probabilities: Pk,k+1 and Pk,k for k = 1, 2, 3, 4. Note that Pk,k+1 = 1 - Pk,k so it is enough to find only one of them. For instance, Pk,k+1 = P ( choose, among k diseased and (N-k) healthy, 1 diseased and 1 healthy) P (disease is transferred) The first probability is hypergeometric, the second is equal to . This yields Pk,k+1 = k N C1 C1 -k , N C2 (3) 2 m where Ck = m!/[k!(m - k)!]. We defined all transition probabilities. 7. Problem 3.1.4. Let us solve it in turn "row-by-row". We have P0,0 = P (X1 X0 ) = P (X1 = 0) = .1. Further, for k = 1, 2, or 3 we can write P0,k = P (X1 = k) = P ( = k). The last probability is from the table, and thus we have numbers for the first row of the transition matrix. Now let us find values for the second row. Plainly P1,0 = 0 because Xn cannot decrease according to its definition. Further, P1,1 = P (Xn = 1|Xn-1 = 1) = P ( 1) = .1 + .3 = .4. Further, P1,2 = P (Xn = 2|Xn-1 = 1) = P ( = 2) = .2. Further, P1,3 = P (Xn = 3|Xn-1 = 1) = P ( = 3) = .4. Now let us consider elements for the third row. We have P3,0 = P3,2 = 0 because the Markov chain does not decrease. At the same time P2,2 = P (Xn = 2|Xn-1 = 2) = P ( 2) = .6. Because the total in each row must be 1, we get P2,3 = .4 (or you can calculate it directly). Finally, for the fourth row we get P3,k = 0 for k = 0, 1, 2 and P3,3 = 1. Combining the obtained results we get the transition matrix .1 0 0 0 .3 .4 0 0 .2 .2 .6 0 .4 .4 .4 1 P= 8. Exerc. 3.2.1. The two-step transition matrix allows us to calculate probabilities for Xn+2 to be in a particular state if the state of Xn is given. The formula is very simple, P(2) = P P = P2 and using matrix multiplication we get P (2) = .47 .13 .4 .42 .14 .44 .26 .17 .57 3 9. Exerc. 3.2.3. Well, we need to calculate two multistep-transition matrices. For this particular example it is OK to use software to multiply matrices. P(3) = P3 = .478 .264 .258 .360 .256 .384 .570 .180 .250 .4636 .254 .2824 .444 .2256 .3304 .5240 .2222 .2540 P(4) = P4 = Remark: Always check that the sum in rows is 1!!! Using these transition matrices we get that P (X3 = 1|X0 = 0) = .264 and P (X4 = 1|X0 = 0) = .254. 10. Exerc. 3.2.5 Well, here again we need to multiply matrices, but it is simpler -- we need just multiply two matrices. Write, P (2) =P = 2 .27 .27 .46 .24 .24 .52 .21 .21 .58 Using this result we conclude that P (X3 = 1|X1 = 0) = P0,1 = .27. The same two-step transition matrix can be used for answering the second question, P (X2 = 1|X0 = 0) = P0,1 = .27. Note that the two probabilities in question must be the same due to stationarity of a Markov chain, that is, a shift in time does not change the transition probabilities. (2) (2) 4
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