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
,
,
,
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