ST3236: Stochastic Process Tutorial TA: Mar Choong Hock

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

Question 2 If A and B are independent, prove A and B c are also independent. Solution: P(AB c ) = P(A  B)) = P(A-AB) = P(A) - P(AB) = P(A) - P(A)P(B) = P(A)(1 - P(B)) = P(A)P(B c ) B AB A 

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      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) }

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)

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

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).

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

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

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) GF

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

Exercise 2 - Question 1a Write out the sample space and the joint distribution for (X 1, X 2, X 3, X 4 )

Exercise 2 - Question 1b Check that X 2 and X 3 are independent. Note: The probability is the same for all sample points

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

Exercise 2 - Question 1c

Exercise 2 - Question 1d A special case for item (c) is that Y 1 and Y 3 are NOT independent. However, given Y 2, Y 1 and Y 3 are conditionally independent. If i > j or j > k or i > k, then

If i > j, then P(Y 1 = i |Y 2 = j) = P(Y 1 = i,Y 2 = j) / P(Y 2 = j) = 0 If j > k, then P(Y 3 = k |Y 2 = j) = P(Y 3 = k, Y 2 = j) / P(Y 2 = j) = 0 Note that i > k implies i > j or j > k. Thus, if i > j or j > k or i > k, then P(Y 1 = i, Y 3 = k |Y 2 = j) =P(Y 1 = i |Y 2 = j)P(Y 3 = k |Y 2 = j) = 0