ST3236: Stochastic ProcessTutorial

TA: Mar Choong Hock

g0301492@nus.edu.sg

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 Bc are alsoindependent.

Solution:

P(ABc) = P(AB)) = P(A-AB) = P(A) - P(AB)= P(A) - P(A)P(B) = P(A)(1 - P(B))= P(A)P(Bc)

B

AB

A

Question 3Two 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)

6

1

36

6

6

1

Question 3a

Note that AB = BC = AC. Thus

P(AB) = P(BC) = P(B)P(C)

Finally,

P(AB) = P(A)P(B).

36

1

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).

6

1

6

1

36

1

Question 3c

P(ABC)= P({4,3})

P(A)P(BC)

Therefore, A not independent of BC.

36

1

36

1

6

1

Question 4a

EFGPEGPEFPEGEFPGFEP )()(

EFGPEPGPEPFPEP )()(

EFGPGPFPEP )()(

)( GFPEP

If E is independent of F and E is independent of G, then E is independent of (F U G)

(Not True)

Question 4b

EFGPEGPEFPEGEFPGFEP )()(

)()( GPFPEP

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) )( GFPEP

GF

Exercise 2 - Question 1Toss a fair coin independently four times. Let

For i = 1, 2, 3, 4

otherwise0

head a is ith throw if1iX

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

1,0,,,:,,, lkjilkji

16

1

2

1

2

1

2

1

2

1,,, 43214321 XPXPXPXPXXXXP

Exercise 2 - Question 1bCheck that X2 and X3 are independent.

Note: The probability is the same for all sample points

2

1

2

1

4

1

16

4, 3232 XPXPkXjXP

Exercise 2 - Question 1cLet 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.

4321

43214

3213

212

11

YYYY

XXXXY

XXXY

XXY

XY

Exercise 2 - Question 1c

otherwise

kjjkjkXjXPkYjYP0

2,1,4

1,, 2121

8

1

4

1

2

1

0,01

01

00,1

211

21

21

XXPXP

YPYP

YYP

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

0, 231 jYkYiYP

jYP

jkXijXiXPjYkYiYP

2

321231

,,,

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, thenP(Y1 = i, Y3 = k |Y2 = j)=P(Y1 = i |Y2 = j)P(Y3 = k |Y2 = j) = 0

jYP

jkXijXiXPjYkYiYP

2

321231

,,,

kjiif

jXXP

jXXPjkXPjYiYP

jkXPjYiYP

jYP

jkXPijXiXP

21

21321

321

2

321 ,

jYkYPjYiYP

jYP

kYjYPjYiYP

jXXP

kXXXjXXPjYiYP

jXXP

jkXjXXPjYiYP

2321

2

3221

21

3212121

21

32121

,

,

,