ST3236: Stochastic Process Tutorial

1. ST3236: Stochastic ProcessTutorial TA: Mar Choong Hock g0301492@nus.edu.sg

2. Set Theory Revision • Interception is associative: AB = BA • Interception is distributive over union: • A(B U C) = AB U AC • AA = A • AW=A • A  = 

3. Question 2 If A and B are independent, prove A and Bcare also independent. Solution: P(ABc) = P(A(W-B)) = P(A-AB) = P(A) - P(AB) = P(A) - P(A)P(B) = P(A)(1 - P(B)) = P(A)P(Bc) W

4. Question 3 Two fair dice are thrown. Let A denotes the event that the sum of the dice is 7. Let B denotes the event that the first die equals 4 and let C be the event that the second die equals 3. Let (w1,w2) denotes the sample point. A = {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) } B = {(4,1), (4,2), (4,3), (4,4), (4,5), (4,6) } C = {(1,3), (2,3), (3,3), (4,3), (5,3), (6,3) }

5. Question 3a Show that A and B are independent P(A) = P( {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1) }) P(B)

6. Question 3a Note that AB = BC = AC. Thus P(AB) = P(BC) = P(B)P(C) Finally, P(AB) = P(A)P(B).

7. Question 3b Show that A and C are independent P(A) P(C) P(AC) = P(BC) = P(B)P(C) Finally, P(AC) = P(A)P(C).

8. Question 3c P(ABC)= P({4,3}) P(A)P(BC) Therefore, A not independent of BC.

9. Question 4a If E is independent of F and E is independent of G, then E is independent of (F U G) (Not True)

10. Question 4b If E is independent of F and E is independent of G and FG = , then E is independent of (F U G) Now, P(EFG) = 0 (Proved)

11. Exercise 2 - Question 1 Toss a fair coin independently four times. Let For i = 1, 2, 3, 4

12. Exercise 2 - Question 1a Write out the sample space and the joint distribution for (X1, X2, X3, X4)

13. Exercise 2 - Question 1b Check that X2 and X3 are independent. Note: The probability is the same for all sample points

14. Exercise 2 - Question 1c Let Yi be the total number of heads for tosses 1 to i. Find the joint distribution of (Y1, Y2). Show thatY1 and Y2 are NOT independent.

15. Exercise 2 - Question 1c

16. Exercise 2 - Question 1d A special case for item (c) is that Y1 and Y3 are NOT independent. However, given Y2, Y1 and Y3 are conditionally independent. If i > j or j > k or i > k, then

17. If i > j, then P(Y1 = i |Y2 = j) = P(Y1 = i,Y2 = j) / P(Y2 = j) = 0 If j > k, then P(Y3 = k |Y2 = j) = P(Y3 = k, Y2 = j) / P(Y2 = j) = 0 Note that i > k implies i > j or j > k. Thus, if i > j or j > k or i > k, then P(Y1 = i, Y3 = k |Y2 = j) =P(Y1 = i |Y2 = j)P(Y3 = k |Y2 = j) = 0