Probability theory
Lesson 16
Variance Of A Random Variable
16.1- What is the Variance of a Random Variable?
16.1 - Problem 1:
Let X equal the number of tosses.
Let Fk equal the event that five occurs on the k th toss.
Step 1: P{X = 2} = P{F11F2} = (1/6)(1/6) = 1/36
Step 2: P{X = 3} = P{(F11F2N1F3)c(F1N1F2 1F3) } = P{(F1 1 F2N1F3)+ P(F1N1F21F3)} =
(1/6)(5/6)(1/6) + (5/6)(1/6)(1/6) = 10/216
Step 3: Since P{X = 2} + P{X = 3} + P{X = 4} = 1,
we have P{X = 4} = 1 - 1/36 - 10/216 = 200/216.
The following is the table for computing the mean:
X
P{X = x}
xP{X = x}
2
3
4
1/36
10/216
200/216
2/36
30/216
800/216
Total 1
E(X) = 832/216
The following is the table for computing the variance F2:
X=x
P{X = x} = P{X = x2}
x2
x2P{X = x2}
2
3
4
1/36
10/216
200/216
Total 1
4
9
16
24/216
90/216
3200/216
2
E(X ) = 3314/216
Since F2 = E(X2) - E(X)2 , from the 2 tables above we have
246
Probability Theory
Lesson 16
Variance of a Random Variable
F2 = 3314/216 - (832/216)2 . 0.51
16.1 - Problem 2:
Let X equal the amount he wins/losses.
Wk: the event he wins the k th game (k = 1,2,3).
P{X = $610} = P(W11W21W3) = (0.60)3 = 0.216
P{X = - $100} = 1 - 0.216 = 0.784
X=x
$610
-$100
P{X = x}
xP{X = x}
0.216
0.784
$131.76
- $78.40
Total 1
E(O) = $53.36 average winnings
The table for computing the variance is
X=x
P{X = x}
x2
$610
-$100
0.216
0.784
372100
10000
x2P{X = x2}
80373.60
7840.00
Total = 88213.60
Total = 1
F2 = 88213.60 - 53.362 . $85366.31
Supplementary Problems
1.
A: The event that option A doubles in value.
B: The event that option B doubles in value.
C: The event that option C doubles in value.
Let Z equal her total possible returns.
P{Z = - $1,500} = P(AN1BN1CN) = (0.30)(0.40)(0.45) = 0.054
P{Z = - $500} = P[(A1BN1CN)c(AN1B1CN)c(AN1BN1C)] =
247
Probability Theory
Lesson 16
Variance of a Random Variable
P(A1BN1CN)+ P(AN1B1CN)+ P(AN1BN1C)] =
(0.70)(0.40)(0.45) + (0.30)(0.60)(0.45) + (0.30)(0.40)(0.55) = 0.126 + 0.081 + 0.066 = 0.273
P{Z = $500} = P[(A1B1CN)c(A1BN1C)c(AN1B1C)] =
P(A1B1CN)+ P(A1BN1C)+ P(AN1B1C)] = (0.70)(0.60)(0.45) + (0.70)(0.40)(0.55) +
(0.30)(0.60)(0.55) = 0.189 + 0.154 + 0.099 = 0.442
P{Z = $1,500} = P(A1B1C) = (0.70)(0.60)(0.55) = 0.231
Z=z
- $1500
- $500
$500
$1500
P{Z = z}
zP{Z = z}
0.054
0.273
0.442
0.231
- $81.00
- $136.50
$221.00
$346.50
Total 1
E(Z) = $350 average gain
The table for computing the variance is
X=x
P{X = x}
x2
-$1500
-$500
$500
$1500
0.054
0.273
0.442
0.231
2250000
250000
250000
2250000
x2P{X = x}
121500
68250
110500
519750
Total = 820000
Total 1
F2 . $820,000 - $3502 = $697,500
2.
From 14.2 - Example 2, Lesson 14, we have the following distribution:
248
Probability Theory
x
P{X = x}
xP{X = x}
2
3
4
5
6
1/36
2/36
3/36
4/36
5/36
2/36
6/36
12/36
20/36
30/36
7
8
9
10
11
12
6/36
5/36
4/36
3/36
2/36
1/36
42/36
40/36
36/36
30/36
22/36
12/36
Total 1
X = x P{X = x}
Lesson 16
E(X) = 7
x2
x2P{X = x}
2
3
4
5
1/36
2/36
3/36
4/36
4
9
16
25
4/36
18/36
48/36
100/36
6
7
8
9
10
11
12
5/36
6/36
5/36
4/36
3/36
2/36
1/36
36
49
64
81
100
121
144
180/36
294/36
320/36
324/36
300/36
242/36
144/36
Total
F2 .
- 72 =
249
Variance of a Random Variable
Probability Theory
Lesson 16
Variance of a Random Variable
3.
'a.
The following table computes the variance of X:
X=x
P{X = x}
x2
x2P{X = x}
$900
$1500
$2000
0.60
0.30
0.10
810000
2250000
4000000
486000
675000
400000
Total = 1561000
F2X = $1561000 - 11902 = $144,900
Total = 1
The following table computes the variance of Y:
Y=y
P{Y = y}
y2
y2P{Y = y}
$500
$1100
$1500
0.70
0.25
0.05
250000
1210000
2250000
175000
302500
112500
Total = 590000
F2X = $590000 -7002 = $100,000
Total 1
'b.
F2X + F2y = $144,900 + $100,000 = $244,900
From Supplementary problem 4, Lesson 15:
Z=X+Y=z
$1400
$2000
$2400
$2500
$2600
$3000
$3100
$3500
P{Z = z}
zP{Z = z}
0.42
0.36
0.03
0.07
0.075
0.015
0.025
0.005
$588
$720
$72
$175
$195
$45
$77.5
$17.5
Total 1
E(Z) = $1,890
250
Probability Theory
Lesson 16
Variance of a Random Variable
Z=z
P{Z = z}
z2
z2P{Z = z}
$1400
$2000
$2400
$2500
$2600
$3000
$3100
$3500
0.42
0.36
0.03
0.07
0.075
0.015
0.025
0.005
1960000
4000000
5760000
6250000
6760000
9000000
9610000
12250000
823200
1440000
172800
437500
507000
135000
240250
61250
Total = 3817000
F2Z = $3817000 -18902 = $244,900
Total 1
Therefore, F2Z = F2X + F2Y
4.
'a.
Let X equal the number of possible diamonds drawn.
From Supplementary problem 5, Lesson 15,
X
P{X = x}
xP{X = x}
0
1
2
1482/2652
1014/2652
156/2652
0
1014/2652
312/2652
Total = 1
E(X) =
= 0.5
251
Probability Theory
Lesson 16
Variance of a Random Variable
X=x
P{X = x}
X2
x2P{X = x}
0
1
2
1482/2652
1014/2652
156/2652
0
1
4
0
1014/2652
624/2652
Total =
F 2X =
'b.
Let Y equal the possible number of clubs drawn.
Since clubs and diamonds are exactly interchangeable,
F 2Y =
'c.
C1: The first card drawn is a club.
C2: The second card drawn is a club.
D1: The first card drawn is a diamond.
D2: The second card drawn is a diamond.
Let Z = X + Y, the total possible number of diamonds and clubs.
P{Z = 0 + 0 = 0} = P[(C1N1D1N)1(C2N1D2N)] =
P(C1N1D1N)P[(C1N1D1N)(C2N1D2N)*(C1N1D1N)] =
P{Z = 1} = P[(C1)1(C2N1D2N)] + P[(C2)1(C1N1D1N)] +
P[(D1)1(C2N1D2N)] + P[(D2)1(C1N1D1N)] =
P{Z = 2} = P(C11C2) + P(D11D2) + P(D11C2) + P(C11D2) =
Since we are only drawing 2 cards,
252
Probability Theory
Lesson 16
Variance of a Random Variable
P{Z = 3} = 0
P{Z = 4} = 0
Z
P{Z = z}
zP{Z = z}
0
650/2652
0
1
1352/2652
1352/2652
2
650/2652
1300/2652
3
0
0
4
0
0
Total = 1
E(Z) = 1
X=x
P{X = x}
X2
x2P{X = x}
0
650/2652
0
0
1
1352/2652
1
1352/2652
2
650/2652
4
2600/2652
3
0
9
0
4
0
16
0
Total 1
F2X = 3952/2652 - 12 = 325/663
F2X +F2Y =
…
5.
Strategy 1.
W1: The event that he wins first game.
W2: The event that he wins second game.
W3: The event that he wins third game.
Strategy 1.
Let X equal the wins/losses possible.
253
= F 2Z
Probability Theory
Lesson 16
Variance of a Random Variable
P{X = - $55 - $55 - $55 = - $165} = P(W1N1W2N1W3N) = (0.40)3 = 0.064
P{X = $50 -$55 -$55 = -$60} =
P(W11W2N1W3N) +P(W1N1W21W3N) + P(W1N1W2N1W3) = 3(0.6)(0.4)2 = 0.288
P{X = $50 + $50 - $55 = $45} =
P(W11W21W3N) +P(W1N1W21W3) + P(W11W2N1W3) = 3(0.4)(0.6)2 = 0.432
P{X = $50 + $50 + $50 = $150} = P(W11W21W3) = (0.6)3 = 0.216
E(X) = (-165)(0.064) + (-60)(0.288) + (45)(0.432) + (150)(0.216)
= -10.56 -17.28 + 19.44 + 32.4 = $24
X=x
P{X = x}
X2
x2P{X = x}
- $165
0.064
27225
1742.4
- 60
0.288
3600
1036.8
45
0.432
2025
874.8
150
0.216
22500
4860
Total = 8514
Total = 1
F2X = 8514 - 242 = $7938
Strategy 2:
Let X equal the wins/losses possible.
P{X = $130 + $130 + $130 = $ 390} = P(W11W21W3) = (0.6)3 = 0.216
P{X = $130 - $100 = $30} = P(W11W21W3N) + P(W11W2N1W3) + P(W1N1W21W3) = 3(0.6)2(0.4)
= 0.432
P{X = -$150} = 1 - 0.216 - 0.432 = 0.352
E(X) = ($390)(0.216) + ($30)(0.432) + (-$150)(0.352) = $84.24 + 12.96 -52.8 = $44.40
254
Probability Theory
Lesson 16
Variance of a Random Variable
X=x
P{X = x}
X2
x2P{X = x}
$390
30
-150
0.216
0.432
0.352
152100
900
22500
32853.6
388.8
7920.0
41162.4
Total
Total 1
F2X = $41162 - $44.42 = $39191.04
6.
'a.
Step 1:: = 16.15
F2 = 0.08
Step 2: If X > 16.25 or X < 16.05 then * X - 16.15* > 0.10 $ a = 0.10.
Letting a = 0.10,
we have P(* X - :* $ a) = P(* X - 16.15* > 0.10) #
= 0.64.
'b.
Since we are only using X > 16.25, then by symmetry we are only computing half the interval. There
the probability is 0.64/2 - 0.32.
7.
'a.
F is called the standard deviation.
Let a = 2F.
P(* X - :* $ a) = P(* X - :* $ 2F) #
= 0.25
'b.
Let a = 3F.
P(* X - :* $ a) = P(* X - :* $ 3F) #
. 0.11
255
Probability Theory
Lesson 16
Variance of a Random Variable
8.
A game is called a "fair game" if E(X) = 0 where X equals the win/losses resulting from the game.
'a.
X=x
$100
-$110
P{X = x}
X2
x2P{X = x}
0.523
0.477
10000
12100
5230.0
5771.7
F2X = $1101.70 - $02 = $1101.70
Total = 1
'b.
Number of teams in the parley
Winning
2
3
4
5
$ 260
$ 600
$1,000
$2,000
For each of these parleys, p equals the probability of winning each game that the game is fair.
Number of teams in the parley
Winning
P
2
3
4
5
$ 260
$ 600
$1,000
$2,000
0.527
0.523
0.549
0.544
'Two team parley
X=x
P{X = x}
X2
x2P{X = x}
$260
-$100
0.278
0.722
67600
10000
18792.80
7220.00
Total = 26012.80
Total 1
F2X = $26012.80 - $02 = $26,012.80
'Three team parley
256
Probability Theory
Lesson 16
X=x
P{X = x}
X2
x2P{X = x}
$600
-$100
0.143
0.857
360000
10000
51480
8570
60050
Total
Variance of a Random Variable
F2X = $60,050- $02 = $60,050
Total 1
'Four team parley
X=x
P{X = x}
X2
x2P{X = x}
$1000
-$100
0.09
0.91
1000000
10000
$90000
$ 9100
Total = 99100
F2X = $99,100- $02 = $99,100
Total 1
'Five team parley
X=x
P{X = x}
X2
x2P{X = x}
$2000
-$100
0.048
0.952
4000000
10000
$192,000
$ 9,520
Total = $201,520
Total 1
9.
FX+c2 = E[(X + c)2] - [E(X + c)]2
F2X = $201,520- $02 = $201,520
= E(X2) + E(2cX) + E(c2) - [E(X) + E(c)]2 =
E(X2) + 2cE(X) + E(c2) - {[E(X)]2 + 2cE(X) + [E(c)]2 } =
E(X2) + 2cE(X) + c2 - [E(X)]2 - 2cE(X) - c2 =
E(X2) - [E(X)]2 = FX2
10.
257
Probability Theory
Lesson 16
Variance of a Random Variable
11.
E(X) = 1p + 0(1 - p) = p
F2 = E(X2) - [E(X)]2
E(X2) = 12p + 0(1 - p) = p
F2 = p - p2 = p(1 - p)
12.
'a.
F2 = P{X = x1}[x1 - E(X)]2 + P{X = x2}[x2 - E(X)]2 + ... + P{X = xn}(xn - E(X))2 =
P{X = x1}[x12 - 2x1E(X) + E(X)2] + P{X = x2}[x22 - 2x2E(X) + E(X)2] + ... +
P{X = xn}[xn2 - 2xnE(X) + E(X)2] =
x12P{X = x1} + x22P{X = x2}+...+ xn2P{X = xn} - [2x1E(X)P{X = x1}+ ... + 2xnE(X)P{X = xn}] +
E(X)2P{X = x1} + E(X)2P{X = x2} + ... + E(X)2P{X = xn} =
E(X2) - 2E(X)E(X) + E(X)2 = E(X2) - E(X)2
'b.
From a, we have F2 = E(X2) - E(X)2.
Therefore, E(X2) = F2 + E(X)2 = F2 + :2.
13.
'a.
Step 1: First we show S2, Xk are independent (k > 2).
P{S2 = x1 + x2; X3 = x3} = P[{X1 = x1 }1{X2 = x2}1{Xn+1 = x3}] =
P{X1 = x1 }P{X2 = x2}P{X3 = x3} = P[{X1 = x1 }1{X2 = x2}]P{X3 = x3}] =
P{X1 = x1 ;X2 = x2}P{X3 = x3} = P{S2 = x1 + x2}P{X3 = x3}
Step 2: Using induction, we assume Sn - 1, Xn ,Xn+1 are independent.
P{Sn - 1 + Xn= sn - 1 + xn ; Xn + 1 = xn + 1} = P{Sn = sn ; Xn + 1 = xn + 1 } =
P{Sn - 1 = sn - 1; Xn = xn; Xn + 1 = xn + 1} = P[{Sn - 1 = sn - 1}1{Xn = xn}1{Xn + 1 = xn + 1}] =
P{Sn - 1 = sn - 1}P{Xn = xn}P{Xn + 1 = xn + 1 } =
258
Probability Theory
Lesson 16
Variance of a Random Variable
P[{Sn - 1 = sn - 1}1{Xn = xn }] P{ Xn + 1 = xn + 1 } = P{Sn = sn - 1 + xn }P{ Xn + 1 = xn + 1 }
'b.
F2(S2) = F2(X1 + X2) = E[(X1 + X2)2] - [E(X1 + X2)]2 =
E(X12 + 2X1X2 + X22) - [E(X1) +E(X2)]2 =
E(X12) + 2E(X1X2) + E(X22) - [E(X1)2 + 2E(X1)E(X2) + E(X2)2] =
E(X12) + 2E(X1)E(X2) + E(X22) - [E(X1)2 + 2E(X1)E(X2) + E(X2)2] =
E(X12) + E(X22) - E(X1)2 - E(X2)2 = [E(X12) - E(X1)2] + [E(X22) - E(X2)2] =
F2(X1) + F2(X2) = F2(S2)
By induction, we assume F2(Sn) = F2(X1) + ... + F2(Xn)
By Problem a. we know Sn and Xn + 1 are independent. Therefore,
F2(Sn + 1) = F2(Sn + Xn + 1) + F2(Sn) + F2 (Xn + 1) = F2(X1) + ... + F2(Xn + 1)
'c.
F2(cX) = E[(cX)2] - E[(cX)]2 = c2E[X2] - c2 [E(X)]2
'd.
=
'e.
Step 1:
Step 2:
Step 3:
Step 4: E(X12) = F2 + :2
=
259
Probability Theory
Lesson 16
Variance of a Random Variable
Step 5:
Since each X1 - 0 has the same distribution, we sum each term if E(nS2) and get
=
nE(S2) =
= nF2 + n:2 -2[F2 + n:2] + F2 + n:2 = (n - 1)F2
Therefore, E(S2) = [(n - 1)/n] F2.
14.
From lesson 15, P{X = k} = [k3 - (k - 1)3]216-1, k = 1,..,6 and E(X) . 4.96.
F2 = E(X2) - E(X)2
P(X2 = k2) = P{X = k} = [k3 - (k - 1)3]216-1, k = 1,..,6.
E(X2) = 12[13 - (1 - 1)3]216-1 +
22[23 - (2 - 1)3]216-1 + 32[33 - (3 - 1)3]216-1 +
42[43 - (4 - 1)3]216-1 + 52[53 - (5 - 1)3]216-1 + 62[63 - (6 - 1)3]216-1 =
[1 + 4(7) + 9(19) + 16(37) + 25(61) + 36(91)]216-1 = [1 + 28 + 171 +592 + 1525 + 3276 ]216-1.
25.89
F2 = E(X2) - E(X)2 = 25.89 - 4.962 . 1.29
15.
Hk : head occurs on the kth toss.
Kk: tails occurs on the kth toss.
{X = 1} = H1
{X = 2}= T11H2
260
Probability Theory
Lesson 16
Variance of a Random Variable
{X = 3} = T11T21H3
{X = 4} = T11T21T31H4
{X = 5} = (T11T21T31T41H5)c(T11T21T31T41T5)
P{X = 1} = P(H1) = 1/2
P{X = 2}= P(T11H2) = P(T1)P( H2) = (1/2) (1/2) = 1/4
P{X = 3} = P(T11T21H3) = P(T1)P(T2)P(H3) = (1/2)(1/2)(1/2) = 1/8
P{X = 4} = P(T11T21T31H4) = P(T1 )P(T2)P(T3)P(H4) = (1/2)(1/2)(1/2)(1/2) = 1/16
P{X = 5} = P{(T11T21T31T41H5)c(T11T21T31T41T5)} =
P(T1)P(T2)P(T3)P(T4)P(H5) + P(T1)P(T2)P(T3)P(T4)P(T5) = 1/32 + 1/32 = 1/16
Step 1: E(X) = 1(1/2) + 2(1/4) + 3(1/8) + 4(1/16) + 5(1/16) = 1/2 + 2/4 + 3/8 + 4/16 + 5/16 = 31/16
Step 2: P{X2 = 12 = 1} = P(H1) = 1/2
P{X2 = 22 = 4}= P(T11 H2) = (1/2) (1/2) = 1/4
P{X2 = 32 = 9} = P(T11 T21 H3) = (1/2)(1/2)(1/2) = 1/8
P{X2 = 42 = 16} = P(T1 )P(T2)P(T3)P(H4) = (1/2)(1/2)(1/2)(1/2) = 1/16
P{X2 = 52 = 25} = P(T1)P(T2)P(T3)P(T4)P(H5) + P(T1)P(T2)P(T3)P(T4) P(T5) = 1/16
E(X2) = 1(1/2) + 4(1/4) + 9(1/8) + 16(1/16) + 25(1/16) = 1/2 + 4/4 + 9/8 + 16/16 + 25/16 = 83/16
F2 = E(X2) - [E(X)]2 = 83/16 - (31/16)2 = 1328/256 - 961/256 = 367/256
16.
Rk : The kth marble selected is red.
Bk: The kth marble selected is black.
S = (X = 2)c(X = 3)c(X = 4)
Step 1:
( X = 2) = (R11B2)c(B11R2)
(X = 3) = (R11R21B3)c(B11B21R3)
(X = 4) = (R11R21R31B4)c(B11B21B31R4)
261
Probability Theory
Lesson 16
Variance of a Random Variable
Step 2:
P( X = 2) =
P(R11B2) + P(B11R2) = P(R1)P(B2|R1) + P(B1)P(R2|B1) = (3/6)(3/5) + (3/6)(3/5) = 18/30
P(X = 3) =
P(R11R21B3) + P(B11B21R3) = P(R1)P(R2|R1)P(B3|R11R2) + P(B1)P(B2|B1)P(R3|B11B2) =
(3/6)(2/5)(3/4) + (3/6)(2/5)(3/4) = 36/120
P(X = 4) =
P(R11R21R31B4) + P(B11B21B31R4)
Since P(S) = P(X = 2) + P(X = 3) + P(X = 4) = 1,
18/30 + 36/120 + P(X = 4) = 1
P(X = 4) = 1 - 18/30 - 36/120 = 1 - 108/120 = 12/120 = 1/10
Step 3: E(X) = 2(18/30) + 3(36/120) + 4(1/10) = 2(6/10) + 3(3/10) + 4(1/10) = 25/10 = 5/2 = 2.5
Step 4: ( X2 = 4) = (R11B2)c(B11R2)
(X2 = 9) = (R11R21B3)c(B11B21R3)
(X2 = 16) = (R11R21R31B4)c(B11B21B31R4)
P( X2 = 4) = 18/30
P(X2 = 9) = 36/120
P(X2 = 16) = 1/10
E(X2) = 4(18/30) + 9(36/120) + 16(1/10) = 72/30 + 324/120 + 16/10 = 67/10
F2 = E(X2) - E(X)2 = 67/10 - (5/2)2 = 67/10 - 25/4 = 9/20
17.
Ak: urn A is selected. on the k th toss (k = 1,2).
Bk: urn B is selected on the k th toss. (k = 1,2).
Rk: red marble is selected on the k th toss (k = 1,2).
262
Probability Theory
Lesson 16
Variance of a Random Variable
Wk: white marble is selected on the k th toss. (k = 1,2).
Step 1: S = (X = 1)c(X = 2)c(X = 3), since the maximum number of selections is 3.
(X = 1) = (A11R1)c(B11R1)
(X = 2) = [(A11W1)c(B11W1)]1[(A21R2)c(B21R2)] =
{(A11W1)1[(A21R2)c(B21R2)]}c{(B11W1)1[(A21R2)c(B21R2)]}=
[(A11W1)1(A21R2)]c[(A11W1)1(B21R2)]c[(B11W1)1(A21R2)]c[(B11W1)1(B21R2)]
Step 2: P(X = 1) = P(A11R1) + P(B11R1) = P(A1)P(R1|A1) + P(B1)P(R1|B1) =
(1/2)(1/2) + (1/2)(1/2)= 1/2.
P(X = 2) =
P[(A11W1)1(A21R2)] + P[(A11W1)1(B21R2)] + P[(B11W1)1(A21R2)] + P[(B11W1)1(B21R2)] =
P(A11W1)P(A21R2|A11W1) + P(A11W1)P(B21R2|A11W1) +
P(B11W1)P(A21R2|B11W1) + P(B11W1)P(B21R2|B11W1)
From lesson 13, problem 9, we have
P(A21R2|A11W1) = P(A2|A11W1) P[R2|(A11W1)1A2] = (1/2)(1) = 1/2
P(B21R2|A11W1) = P(A2|A11W1) P[R2|(A11W1)1B2] = (1/2)(1/2) = 1/4
P(A21R2|B11W1) = P(A2|B11W1) P[R2|(B11W1)1A2] = (1/2)(1/2) = 1/4
P(B21R2|B11W1) = P(B2|B11W1) P[R2|(B11W1)1B2] = (1/2)(1) = 1/2
P(X = 2) =
P(A11W1)P(A21R2|A11W1) + P(A11W1)P(B21R2|A11W1) +
P(B11W1)P(A21R2|B11W1) + P(B11W1)P(B21R2|B11W1) =
P(A1)P(W1|A1)P(A21R2|A11W1) + P(A1)P(W1|A1)P(B21R2|A11W1) +
P(B1)P(W1|B1)P(A21R2|B11W1) + P(B1)P(W1|B1)P(B21R2|B11W1) =
263
Probability Theory
Lesson 16
Variance of a Random Variable
(1/2)(1/2)(1/2) + (1/2)(1/2)(1/4) + (1/2)(1/2)(1/4) + (1/2)(1/2)(1/2) =
1/8 + 1/16 + 1/16 + 1/8 = 6/16 = 3/8
Step 3: S = (X = 1)c(X = 2)c(X = 3), since only a maximum of 3 selections are possible.
P(S) = P(X = 1) + P(X = 2) + P(X = 3) = 1
P(X = 3) = 1 - P(X = 1) - P(X = 2) = 1 - 1/2 - 3/8 = 1/8
Step 4: E(X) = 1(1/2) + 2(3/8) + 3(1/8) = 13/8
Step 5: E(X2) = 12(1/2) + 22(3/8) + 32(1/8) = (1/2) + 4(3/8) + 9(1/8) = 25/8
F2 = E(X2) - E(X)2 = 25/8 - (13/8)2 = 31/64
18.
Cov(X,Y) = E[(X - :X)(Y - :Y)] = E[XY - :YX - :XY + :X:Y] =
E(XY) - E(:YX) - E(:XY) + E(:X:Y) = E(XY) - :Y E(X) - :X E(Y) + E(:X:Y) =
E(XY) - :Y :X - :X :Y + :X:Y = E(XY) - :X :Y
19.
Since X,Y are independent, we have shown (Lesson 15, supplementary problems)
E(XY) = E(X)E(Y) = :X :Y
By problem 18, we have
Cov(X,Y) = E(XY) - :X :Y = E(X)E(Y) - :X :Y = :X :Y - :X :Y = 0.
20.
Define Z = X1 + X2 and W = X1 - X2.
'a.
Step 1: S = {(h,h), (t,t) (h,t), (t,h)}
E(X1) = 1(1/2) + 0(1/2) = 1/2
E(X2) = 1(1/2) + 0(1/2) = 1/2
E(X12) = 12(1/2) + 02(1/2) = 1/2
E(X22) = 12(1/2) + 02(1/2) = 1/2
:Z = E(Z) = E(X1 + X2) = E(X1) + E(X2) = 1/2 + 1/2 = 1
:W = E(W) = E(X1 - X2) = E(X1) - E(X2) = 1/2 - 1/2 = 0
264
Probability Theory
Lesson 16
Variance of a Random Variable
Step 2: ZW = (X1 + X2)(X1 - X2) = X12 - X22
E(ZW) = E(X12 - X22) = E(X12) - E(X22) = 1/2 - 1/2 = 0
Cov(Z,W) = E(ZW) - :Z:W = 0 - 1(0) = 0.
'b.
{Z = X1 + X2 = 2} = {(h,h)}
{W = X1 - X2 = 1} = {(h,t)}
Since {W = X1 - X2 = 1}1{Z = X1 + X2 = 2}= N, P{{Z = X1 + X2 = 2}|{W = X1 - X2 = 1}}= N
Z and W are dependent.
21.
The results of problems 19, and 20 show the following:
If 2 random variables are independent, then the covariance will equal 0. However, the converse is
not true. In problem 20, random variables Z,W have covariance 0 but they are not independent.
22.
Step 1:
X = 0,1,2; possible number of diamonds drawn.
Y = 0,1,2; possible number of clubs drawn.
XY = 0,1; possible since there are only 2 cards selected.
D1: diamond drawn on the first drawing.
D2: diamond drawn on the second drawing.
C1: club drawn on the first drawing.
C2: club drawn on the second drawing.
(XY = 1) = (D11C2)c(C11D2)
P(XY = 1) = P[(D11C2)c(C11D2)] = P(D11C2) + P(C11D2) = (13/52)(13/51) + (13/52)(13/51) =
338/2652
P(XY = 0) = 1 - P(XY = 1) = 1 - 338/2652 = 2314/2652
Step 2: E(XY) = 0(2314/2652) + 1(338/2652) = 338/2652
(X = 0) = (D1N1D2N)
(X = 1) = (D11D2N)c(D1N1D2)
265
Probability Theory
Lesson 16
Variance of a Random Variable
(X = 2) = (D11D2)
P(X = 0) = P(D1N1D2N) = (39/52)(38/51) = 1482/2652
P(X = 1) = P[(D11D2N)c(D1N1D2)] = P[(D11D2N) + P(D1N1D2)] = (13/52)(39/51) + (39/52)(13/51)=
(1014)/(2652)
P(X = 2) = P(D11D2) = (13/52)(12/51) = 156/2652
:X = E(X) = 0P(X = 0) + 1P(X = 1) + 2P(X = 2) = 1014/2652 + 312/2652 = 1326/2652 = 1/2
Using the same argument,
:y = E(Y) = 0P(Y = 0) + 1P(Y = 1) + 2P(Y = 2) = 1014/2652 + 312/2652 = 1326/2652 = 1/2
Cov(X,Y) = E(XY) - :X :y = 338/2652 - (1/2)2 = 169/1326 - 1/4 = (676 - 1326)/5304 = - 325/2652
23.
F2X + Y + Z = E[(X + Y + Z)2] - [E(X + Y + Z)]2
E[(X + Y + Z)2] = E[X2 + XY + XZ + YX + Y2 + YZ + ZX + ZY + Z2] =
E(X2) + E(Y2) + E(Z2) + 2E(XY) + 2E(XZ) + 2E(YZ)
[E(X + Y + Z)]2 = [E(X) + E(Y) + E(Z)]2 =
E(X)2 + E(Y)2 + E(Z)2 + 2E(X)E(Y) + 2E(X)E(Z) + 2E(Y)E(Z)
F2X + Y + Z = E[(X + Y + Z)2] - [E(X + Y + Z)]2 =
E(X2) + E(Y2) + E(Z2) + 2E(XY) + 2E(XZ)+2E(YZ) +
- E(X)2 - E(Y)2 - E(Z)2 - 2E(X)E(Y) - 2E(X)E(Z) - 2E(Y)E(Z) =
[E(X2) - E(X)2] + [E(Y2) - E(Y)2] + [E(Z2) - E(Z)2 ] +
2{[E(XY)- 2E(X)E(Y)] + [E(XZ)- E(X)E(Z)] + [E(YZ) - E(Y)E(Z)]} =
FX2 +FY2 + FZ2 + 2[Cov(X,Y) + Cov(X,Z) + Cov(Y,Z)]
24.
From problem 23, we have
F2X + Y + Z = FX2 +FY2 + FZ2 + 2[Cov(X,Y) + Cov(X,Z) + Cov(Y,Z)]
Let Z = 0, we have
266
Probability Theory
Lesson 16
Variance of a Random Variable
F2X + Y = F2X + Y + 0 = FX2 +FY2 + F02 + 2[Cov(X,Y) + Cov(X,0) + Cov(Y,0 )] =
FX2 +FY2 + 2Cov(X,Y)
Therefore, Cov(X,Y) = [F2X + Y - (FX2 + FY2 )]/2
25.
Since X,Y are independent, we have
E(XY) = E(X)E(Y) (See Lesson 25, problem 9).
Therefore,
Cov(X,Y) = E(XY) - E(X)E(Y) = E(X)E(Y) - E(X)E(Y) = 0.
Using the same argument,
C(X,Z) = 0 and C(Y,Z) = 0
since each random variables are pair-wise independent.
From , prolem 23, F2X + Y + Z = FX2 +FY2 + FZ2 + 2[Cov(X + Y) + Cov(X + Z) + Cov(Y + Z)] =
F2X + Y + Z = FX2 +FY2 + FZ2 + 2[ 0 + 0 + 0] = FX2 +FY2 + FZ2
26.
Assume a, b are constants. Show
From problem 24, we have
Cov(X,Y) = [F2X + Y - (FX2 + FY2 )]/2
Cov(X +a,Y + b ) = [F2(X +a) + (Y + b) - (FX + a2 + FY +
From problem 9,
F2(X +a) + (Y + b) = F2X
+Y
F 2X + a = F 2X = F X 2
F 2Y + b = F Y 2.
Therefore,
267
2
b
)]/2
Probability Theory
Cov(X +a,Y + b ) = . [F2X
Lesson 16
+Y
Variance of a Random Variable
- FX 2 + FY 2 )]/2 = Cov(X,Y).
27.
Step 1: We first show this is true for special random variables X*, Y* where
P(X* = x) = p1
P(X* = 0) = 1 - p1
P(Y* = y) = p2
P(Y* = 0) = 1 - p2.
Step 2: Next we show independence of X* and Y*.
Since Cov(X*,Y*) = E(X*Y*) - E(X*)E(Y*) = 0, we have E(X*Y*) = E(X*)E(Y*).
E(X*Y*) = xyP[(X* = x)1(Y* = y)] + 0x P[(X* = x)1(Y* = 0)] + 0y P[(X* = 0)1(Y* = y)] =
xyP[(X* = x)1(Y* = y)]
E(X*) = xP(X* = x) + 0P(X* = 0) = xP(X* = x)
E(Y*) = yP(Y* = y) + 0P(Y* = 0) = yP(X* = x)
Since E(X*Y*) = E(X*)E(Y*), we have
xyP[(X* = x)1(Y* = y)] = xP(X* = x)yP(Y* = y) = xyP(X* = x)P(Y* = y).
Canceling xy from the above equation we have
P[(X* = x)1(Y* = y)] = P(X* = x)P(Y* = y).
Therefore, the events (X* = x) and (Y* = y) are independent events.
(X* = x)N = (X* = 0) and (Y* = y)N = (Y* = 0)
are the other remaining sets. From lesson 12,
problem 16, these 2 events are also independent.
Step 3: For the general case of the random variable X,Y: (X = x1), (X = x2), (Y = y1), (Y = y2),
Define X* = X - x1 and Y* = Y - y1 .
We assume Cov(X,Y) = 0 . Therefore, from problem 26, Cov(X*,Y*) = Cov(X,Y) = 0
268
Probability Theory
Lesson 16
Variance of a Random Variable
Therefore,
X - x1, and Y - y1 are independent.
Step 4: Now (X = x1) = (X* = X - x1 = 0) and (Y = y1) = (Y* = Y - y1 = 0).
Therefore,
P[(X = x1)1(Y = y1)] = P[(X* = 0)1(Y* = 0)] = P(X* = 0)P(Y* = 0) = P(X = x1)P(Y = y1)
and independents of X,Y is shown.
28.
From problem 23,
F2x1 + x2 + x3 = 3F2 + 2[3Cov(X1,X2) ] = 3F2 + 6Cov(X1,X2).
Using mathematical induction, we assume the above is true for n -1 and prove it for n :
F2x1 + x2 + ... + xn = nF2 + n(n-1)Cov(X1,X2).
Define S = X1 + X2 + ... + Xn
F2S = nF2 + n(n-1)Cov(X1,X2).
F2x1 + x2 + ... + xn+ 1 = F2(S + xn+ 1) = F2S + F2 xn+ 1 + 2Cov(S,Xn + 1) =
nF2 + n(n-1)Cov(X1,X2) + F2 xn+ 1 + 2Cov(S,Xn + 1)
29.
In our solution, we will use the following formulas on summation:
1 + 2 + ... + N = (N)(N + 1)/2
12 + 22 + ... + N2 = N(N + 1)(2N + 1)/6
Xk : The number selected on the kth ball drawn. (k = 1,2,...,r).
Step 1: Computing the mean of S.
S = X1 + X2 + ... + Xr, the sum of the numbers.
P(Xk = 1) = 1/10, where k = 1,2,..., r.
269
Probability Theory
Lesson 16
Variance of a Random Variable
E(Xk) = 1P(Xk = 1) + 2P(Xk = 2) + ... + 10P(Xk = 10) = 1/10 + 2/10 ... + 10/19 =
(1 + 2 + ... + 10)/10 = 10(10 + 1)/2(10) = (11)/2
E(S) = E(X1 + X2 + ... + Xr) = E(X1 )+ E(X2) + ... + E(( Xr) = r(11)/2
Step 2: Computing the variance of S.
F2S = E(S2) - [E(S)]2 = E(S2) - [r(10 + 1)/2] 2 = E(S2) - r2(121)/4
E(S2) = E[(X1 + X2 + ... + Xr)2] = E(X12 ) + E(X22 ) + ... + E(Xr2 ) +
E(X1 X2 ) + E(X1 X3) + ... + E(X1 Xr) +
E(X2 X1 ) + E(X2 X3) + ... + E(X2 Xr) +
::::::::::::::::::::::::::::::::::::::::::::::::::::::::
E(Xr X1 ) + E(Xr X3) + ... + E(Xr Xr - 1)
E(Xk2) = 12P(Xk2 = 12) + 22 P(Xk2 = 22) + ... + 102P(Xk2 = 102) =
12P(1/10) + 22 (1/10) + ... + 102 (1/10 ) = (12 + 22 + ... + 102 )(1/10) = [ 10(10 + 1)(2(10) + 1)/6]/10
= (10 + 1)(2(10) + 1)/6
E(Xj Xk) = (1)(2)P[Xj = 1;Xk = 2] + (1)(3)P[Xj = 1;Xk = 3] + ... + (1)(10)P[Xj = 1;Xk = 10] +
(2)(1)P[Xj = 2;Xk = 1] + (2)(3)P[Xj = 2;Xk = 3] + ... + (1)(10)P[Xj = 2;Xk = 10] + ... +
(10)(1)P[Xj = 10;Xk = 1] + (10)(2)P[Xj =10;Xk = 2] + ... + (10)(10 - 1)P[Xj = 10;Xk = 10 - 1] =
1{2/[10(10 - 1)] + 3/ [10(10 - 1)] + ... + 10 /[10(10 - 1)]} +
2{1/[10(10 - 1)] + 3/[10(10 - 1)] + ... + 10/[10(10 - 1)]} + ... +
10{1/[10(10 - 1)] + 2/[10(10 - 1)] + ... + (10 - 1)/[10(10 - 1)]} =
1{2 + 3 + ... + 10}/[10(10 - 1)]} + 2{1 + 3 + .. + 10}/[10(10 - 1)] + ... +
10{1+ 2 + ... + (10 - 1)}/[10 (10 - 1)] =
1{ (1 + 2 + 3 + ... + 10) - 1}/[10(10 - 1)] + 2{(1 +2 + 3 + .. + 10) - 2}/[10(10 - 1)] + ... +
10{(1+ 2 + ... + (10 - 1) + 10) - 10 }/[10 (10 - 1)] =
270
Probability Theory
Lesson 16
Variance of a Random Variable
1{10(10 + 1)/2 - 1}/[10(10 - 1)] + 2{10(10 + 1)/2 - 2}/[10(10 - 1)] + ... +
10{10(10 + 1)/2 - 10 }/[10 (10 - 1)] =
{10(10 + 1)/2 - 12}/[10(10 - 1)] + {210(10 + 1)/2 - 22}/[10(10 - 1)] + ... +
{(10)(10)(10 + 1)/2 - 102 }/[10 (10 - 1)] =
{(1 + 2 + ... + 10)10(10 + 1)/2 - [10(10 + 1)(2(10) + 1)/6]}/[10(10 - 1)] =
{10(10 + 1)/210(10 + 1)/2 - [10(10 + 1)(2(10) + 1)/6]}/[10(10 - 1)] =
{102(10 + 1)2/4 - [10(10 + 1)(2(10) + 1)/6]}/[10(10 - 1)] = E(XjXk)
E(S2) = E(X12 ) + E(X22 ) + ... + E(Xr2 ) +
E(X1 X2 ) + E(X1 X3) + ... + E(X1 Xr) +
E(X2 X1 ) + E(X2 X3) + ... + E(X2 Xr) +
::::::::::::::::::::::::::::::::::::::::::::::::::::::::
E(Xr X1 ) + E(Xr X3) + ... + E(Xr Xr - 1) =
E(S2) = r(10 + 1)(2(10) + 1)/6 + r2 {102(10 + 1)2/4 - [10(10 + 1)(2(10) + 1)/6]}/[10(10 - 1)]
F2S = E(S2) - [E(S)]2 = E(S2) - r2(10 + 1)2/4 =
r(10 + 1)(2(10) + 1)/6 + r2 {102(10 + 1)2/4 - [10(10 + 1)(2(10) + 1)/6]}/[10(10 - 1)] - r2(10 + 1)2/4
r(11)(20) + 1)/6 + r2 {100(121 )/4 - [10(11 )(21)/6]}/[10(9)] - r2(11)2/4 =
r231/6 + r2 {12100 /4 - [2310/6]}/90 - r2121/4 =
r77/2 + r2 {3025 - 385}/90 - r2121/4 = r77/2 + r2 88/3 - r2121/4 = (-11/12)r2 + (77/2)r
30.
'a .
E(Xk) = 1P(Xk = 1) + 2P( Xk = 2) + 3P(Xk = 3) + 4P(Xk = 4) + 5P(Xk = 5) + 6P(Xk = 6) =
1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6 = 3.5
= 100(3.5)/100 = 35/10 = 7/2
'b .
E(Xk2) = 12P (X = 12) + 22P (X = 22) + 32P (X = 32) + 42P (X = 42) + 52P (X = 52) + 62P (X = 62) =
271
Probability Theory
Lesson 16
Variance of a Random Variable
1P (X = 1) + 4P (X = 4) + 9P (X = 9) + 16P (X = 16) + 25P (X = 25) + 36P (X = 36) = 91/6
(k = 1,2,...,100)
[E(Xk)]2 = 3.52 = 12.25
F2(Xk) = E(Xk2) - [E(Xk)]2 = 91/6 - (7/2)2 = 91/6 - 49/4 = 35/12
From problem 13d, we have
= (35/12)(1/100) = 7/240
'c.
From problem 13,
.
Since the distribution of the random variables are all equal F2, from b. we have F2 = 7/240.
n = 100
= (99/100)(7/240) = 33/100(7/80) = 231/8000 . 7/240
Therefore,
31.
'a.
Xk = x
P(Xk = x)
0
1/2
1
1/2
'b.
= (X1 + X2)/2
=x
P(
= x)
0
1/4
1
1/2
2
1/4
'c.
X1(h,h) = 1
X2(h,h) = 1
272
Probability Theory
Lesson 16
(h,h) = (1 + 1)/2 = 1
S2 = 0
X1(h,t) = 1
X2(h,t) = 0
(h,t) = (1 + 0)/2 = 1/2
S2 = [(1 - 1/2)2 + (0 - 1/2)2]/2 = 1/4
X1(t,h) = 0
X2(t,h) = 1
(t,h) = (0 + 1)/2 = 1/2
S2 = [(0 - 1/2)2 + (1 - 1/2)2]/2 = 1/4
X1(t,t) = 0
X2(t,t) = 0
(t,t) = (0 + 0)/2 = 0
S2 = [(0 - 0)2 + (0 - 0)2]/2 = 0
S2 =s
P(S2 = s)
0
1/2
1/4
1/2
'd.
S2 =s
P(S2 = s) sP(S2 = s)
0
1/2
0
1/4
1/2
1/8
1
E(S2 ) = 1/8
E(Xk) = 1(1/2) + 0(1/2) = 1/2
E(X2k) = 1(1/2) + 0(1/2) = 1/2
273
Variance of a Random Variable
Probability Theory
Lesson 16
F2 = E(X2k) - [E(Xk)]2 = 1/2 - (1/2)2 = 1/4
= (1/2)(1/4) = 1/8
274
Variance of a Random Variable