1
Chapter5. Joint random variables
Problem PP143
1. When you roll one more time you receive 𝑠 + 𝑘 when you roll k (probability = 1⁄2𝑛)
with the die provided 𝑠 + 𝑘 ≤ 2𝑛, otherwise you receive nothing. Hence,
𝐸𝐶 (𝑠) =
2.
1
4𝑛
1
2𝑛
∗ [(𝑠 + 1) + (𝑠 + 2) + … + 2𝑛];
∗ (2𝑛 − 𝑠) ∗ (2𝑛 + 𝑠 + 1) = 𝑠 or
𝑠 2 + (4𝑛 + 1) ∗ 𝑠 − 4𝑛2 − 2𝑛 = 0
1
Solving the quadratic equation: 𝑠𝑓 = 2 ∗ (−4𝑛 − 1 + √32𝑛2 + 16𝑛 + 1) (the second
root is negative).
Notice that continuing to roll is better than stopping when the left hand side of the
quadratic equation is smaller than 0. Also the quadratic function has a convex shape,
hence 𝑠 ∗ = ⌊𝑠𝑓 ⌋.
Problem PP144
1. Straightforward;
2. Let 𝑆 = 𝑋1 + 𝑋2 and 𝑇 = 𝑋1. The joint pdf of S and T is
𝑓(𝑠, 𝑡) = 2 ∗ (𝑠 − 𝑡) , 0 ≤ 𝑡 ≤ 1 𝑎𝑛𝑑 𝑡 ≤ 𝑠 ≤ 𝑡 + 1
The marginal pdf of S:
𝑠
𝑓(𝑠) = ∫ 2 ∗ (𝑠 − 𝑡) ∗ 𝑑𝑡 = 𝑠 2 , 0 ≤ 𝑠 ≤ 1
0
1
= ∫𝑠−1 2 ∗ (𝑠 − 𝑡) ∗ 𝑑𝑡 = 𝑠 ∗ (2 − 𝑠) , 1 ≤ 𝑠 ≤ 2
1
2
4. 𝐸(𝑆) = ∫0 𝑠 3 ∗ 𝑑𝑠 + ∫1 𝑠 2 ∗ (2 − 𝑠) ∗ 𝑑𝑠 = 7⁄6 = 1.16667
or: 𝐸(𝑆) = 𝐸(𝑋1 ) + 𝐸(𝑋2 ) = 7⁄6
5. Straightforward
2
6. Let M denote the median. Using the result in Step 5:
− 𝑀3 ⁄3 + 𝑀2 − 1⁄3 = 1⁄2. Use the Solver to find 𝑀 = 1.168;
Problem PP145
1
1−𝑥1
1. ∫0 ∫0
𝑘 ∗ 𝑥1 ∗ 𝑥2 ∗ 𝑑𝑥2 ∗ 𝑑𝑥1 = 1
⟹
𝑘 = 24
1−𝑥
2. 𝑓(𝑥1 ) = ∫0 1 24 ∗ 𝑥1 ∗ 𝑥2 ∗ 𝑑𝑥2 = 12 ∗ 𝑥1 ∗ (1 − 𝑥1 )2 , 0 ≤ 𝑥1 ≤ 1: beta pdf with
parameters 𝛼 = 2 and 𝛽 = 3. Because of symmetry the pdf of 𝑋2 has the same pdf;
3. 𝐸(𝑋1 ) = 𝐸(𝑋2 ) =
𝛼
𝛼𝛽
𝛼+𝛽
= 2⁄5 , 𝑣𝑎𝑟(𝑋1 ) = 𝑣𝑎𝑟(𝑋2 ) = (𝛼+𝛽)2
∗(𝛼+𝛽+1)
= 1⁄25;
4. Let 𝑆 = 𝑋1 + 𝑋2 and 𝑇 = 𝑋2. Then 𝑋1 = 𝑆 − 𝑇 and 𝑋2 = 𝑇. The J(acobian) is 𝐽 = 1.
𝑓(𝑠, 𝑡) = 24 ∗ 𝑡 ∗ (𝑠 − 𝑡) , 0 ≤ 𝑡 ≤ 𝑠 ≤ 1
𝑠
𝑓(𝑠) = ∫ 24 ∗ 𝑡 ∗ (𝑠 − 𝑡) ∗ 𝑑𝑡 = 4 ∗ 𝑠 3 , 0 ≤ 𝑠 ≤ 1
0
1
5. 𝐸(𝑆) = ∫0 𝑠 ∗ 4 ∗ 𝑠 3 ∗ 𝑑𝑠 = 4⁄5 = 0.8
2⁄75
1
1−𝑥1 𝑥1
6. 𝐸(𝑋1⁄𝑋2 ) = ∫0 ∫0
𝑥2
24∗𝑥1 ∗𝑥2
7. 𝑓(𝑥2 |𝑥1 ) = 12∗𝑥
2
1 ∗(1−𝑥1 )
𝑣𝑎𝑟(𝑆) = 𝐸(𝑆 2 ) − [𝐸(𝑆)]2 = 2⁄3 − 16⁄25 =
∗ 24 ∗ 𝑥1 ∗ 𝑥2 ∗ 𝑑𝑥2 ∗ 𝑑𝑥1 = 2
𝑥
= 2 ∗ (1−𝑥2
2
1)
, 0 ≤ 𝑥2 ≤ 1 − 𝑥1
Problem PP146
𝑛
𝑓(𝑥, 𝜃) = ( ) ∗ 𝜃 𝑥 ∗ (1 − 𝜃)𝑛−𝑥 ∗ 1 , 0 ≤ 𝜃 ≤ 1 ,
𝑥
𝑥 = 0, 1, 2, … , 𝑛
1
𝑛
𝑝(𝑥) = ( ) ∗ ∫ 𝜃 𝑥 ∗ (1 − 𝜃)𝑛−𝑥 ∗ 𝑑𝜃 =
𝑥
0
1
𝑛 Γ(𝑥 + 1) ∗ Γ(𝑛 + 𝑥 + 1)
Γ(𝑛 + 2)
( )∗
∗∫
∗ 𝜃 𝑥 ∗ (1 − 𝜃)𝑛−𝑥 ∗ 𝑑𝜃 =
𝑥
Γ(𝑛 + 2)
Γ(𝑥
+
1)
∗
Γ(𝑛
+
𝑥
+
1)
0
3
= (𝑛𝑥) ∗
Γ(𝑥+1)∗Γ(𝑛+𝑥+1)
Γ(𝑛+2)
1
= 𝑛+1 , 𝑥 = 0, 1, 2, … , 𝑛
(integrand is a beta pdf)
Problem PP148
1. 𝑓(𝑙1 ) = 1 , 1 ≤ 𝑙1 ≤ 2 and likewise for 𝑙2 .
𝐹(𝑙1 ) = 𝑃(𝐿1 ≤ 𝑙1 ) = 𝑙1 − 1 ,
1 ≤ 𝑙1 ≤ 2
𝑃(𝐿 ≤ 𝑙) = 𝐹(𝑙) = 𝑃(𝐿1 ≤ 𝑙 ⋂ 𝐿2 ≤ 𝑙) = (𝑙 − 1)2 , 1 ≤ 𝑙 ≤ 2
𝑓(𝑙) = 2 ∗ (𝑙 − 1) , 1 ≤ 𝑙 ≤ 2
2
2. 𝐸(𝐿) = 2 ∗ ∫1 𝑙 ∗ (𝑙 − 1) ∗ 𝑑𝑙 = 5⁄3
2
2)
𝐸(𝐿
= 2 ∗ ∫ 𝑙 2 ∗ (𝑙 − 1) ∗ 𝑑𝑙 = 17⁄6
1
𝑣𝑎𝑟(𝐿) = 𝐸(𝐿2 ) − [𝐸(𝐿)]2 = 1⁄18
3. 𝐿𝑠 = 𝐿1 + 𝐿2 , 2 ≤ 𝐿𝑠 ≤ 4
Let 𝑙𝑠 = 𝑙1 + 𝑙2 and 𝑙𝑡 = 𝑙1 . Then 𝑙1 = 𝑙𝑡 and 𝑙2 = 𝑙𝑡 − 𝑙𝑠 , 𝐽(𝑎𝑐𝑜𝑏𝑖𝑎𝑛) = 1
𝑓(𝑙𝑡 , 𝑙𝑠 ) = 1 , 1 ≤ 𝑙𝑡 ≤ 2 , 𝑙𝑡 + 1 ≤ 𝑙𝑠 ≤ 𝑙𝑡 + 2
𝑙𝑠 −1
𝑓(𝑙𝑠 ) = ∫
=
𝑑𝑙𝑡 = 𝑙𝑠 − 2 , 2 ≤ 𝑙𝑠 ≤ 3
1
2
∫𝑙 −2 𝑑𝑙𝑡
𝑠
= 4 − 𝑙𝑠 , 3 ≤ 𝑙𝑠 ≤ 4
4. 𝐸(𝐿𝑠 ) = 𝐸(𝐿1 + 𝐿2 ) = 𝐸(𝐿1 ) + 𝐸(𝐿2 ) = 1
𝑣𝑎𝑟(𝐿𝑠 ) = 𝑣𝑎𝑟(𝐿1 ) + 𝑣𝑎𝑟(𝐿2 ) = 1⁄6
Problem PP149
1. 𝐸(2 ∗ 𝐿 + 2 ∗ 𝑊) = 2 ∗ 𝐸(𝐿) + 2 ∗ 𝐸(𝑊) = 2 ∗ 100 + 2 ∗ 70 = 340
𝑣𝑎𝑟(2 ∗ 𝐿 + 2 ∗ 𝑊) = 4 ∗ 𝑣𝑎𝑟(𝐿) + 4 ∗ 𝑣𝑎𝑟(𝑊) = 4 ∗ 100 + 4 ∗ 25 = 500
C is a linear combination of Normally distributed random variables, hence also normally distributed;
4
2. 𝑃(300 ≤ 𝐶 ≤ 350) = 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(350,340, 𝑆𝑄𝑅𝑇(500), 1) −
𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(300,340, 𝑆𝑄𝑅𝑇(500), 1) = 0.6358;
3. 𝐸(𝐴) = 𝐸(𝐿 ∗ 𝑊) = 𝐸(𝐿) ∗ 𝐸(𝑊) = 100 ∗ 70 = 7000 𝑚2
𝑣𝑎𝑟(𝐴) = 𝑣𝑎𝑟(𝐿 ∗ 𝑊) = 𝐸(𝐿2 ∗ 𝑊 2 ) − 𝐸 2 (𝐿 ∗ 𝑊) = 𝐸(𝐿2 ) ∗ 𝐸(𝑊 2 ) − 70002
= [𝑣𝑎𝑟(𝐿) + 𝐸 2 (𝐿)] ∗ [𝑣𝑎𝑟(𝑊) + 𝐸 2 (𝑊)] − 70002 = 742500 𝑚4
Problem PP150
1. Let 𝐹1 (𝐹2 ) be the end of operation 1 (operation 2). Then 𝐹1 − 𝐹2 is normally
distributed with 𝐸 (𝐹1 − 𝐹2 ) = −1 and 𝑣𝑎𝑟(𝐹1 − 𝐹2 ) = 4.
𝑃(𝐹1 ≤ 𝐹2 ) = 𝑃(𝐹1 − 𝐹2 ≤ 0) = 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(0, −1,2,1) = 0.6914;
2. 𝑃(−1 ≤ 𝐹1 − 𝐹2 ≤ 0) = 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(0, −1,2,1) − 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(−1, −1,2,1) =
0.1915;
3. Let X be the time that operation 2 starts after the start of operation 1 (in minutes).
𝐹1 − 𝐹2 ∼ 𝑁(5 − 𝑋, 2).
The probability that part 1 will be available before part 2 but by no more than 3
minutes:
𝑃(−3 ≤ 𝐹1 − 𝐹2 ≤ 0) = 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(0,5 − 𝑋, 2,1) − 𝑁𝑂𝑅𝑀. 𝐷𝐼𝑆𝑇(−3,5 − 𝑋, 2,1)
which should be maximized.
Problem PP151
1. Let 𝑋𝑖 be the weight of sweet i. Then 𝑋𝑖 ∼ 𝑁(5 , √0.5 ). The sum ∑𝑛𝑖=1 𝑋𝑖 of the
weight of n sweets is normal with expected value 5 ∗ 𝑛 and variance 0.5 ∗ 𝑛. The
number of sweets required is the smallest integer such that
𝑛
∑𝑛𝑖=1 𝑋𝑖 − 5 ∗ 𝑛 200 − 5 ∗ 𝑛
𝑃 (∑ 𝑋𝑖 > 200) = 𝑃 (
>
) ≥ 0.999
𝑖=1
√0.5 ∗ 𝑛
√0.5 ∗ 𝑛
n is the smallest integer larger than the solution of
200 − 5 ∗ 𝑛
√0.5 ∗ 𝑛
= −3.09
or 5 ∗ 𝑛 − 3.09 ∗ √0.5 ∗ 𝑛 − 200 = 0 ⟹ 𝑛 = 45
5
Problem PP152
1. Let A (B) denote the arrival time of A (B). Let DTA (DTB) denote the journey time of A
(B).
𝐸(𝐴) = 4 𝑝𝑚
𝐸(𝐷𝑇𝐵) = 5 + 1⁄2 + 2 + 1⁄2 + 2.5 = 10.5 ℎ𝑟𝑠 ⟹ 𝐸(𝐵) = 4.30 𝑝𝑚
2. DTB is normally distributed with expected value 630 minutes and variance 900 + 100 +
225 = 1225 𝑚𝑖𝑛𝑢𝑡𝑒𝑠 2 .
𝑃(𝐴 < 5 𝑝𝑚) = 𝑃(𝐷𝑇𝐴 < 660 𝑚𝑖𝑛) = 0.84
𝑃(𝐵 < 5 𝑝𝑚) = 𝑃(𝐷𝑇𝐵 < 660 𝑚𝑖𝑛) = 0.804
3. 𝑃(𝐵 𝑎𝑟𝑟𝑖𝑣𝑒𝑠 𝑏𝑒𝑓𝑜𝑟𝑒 𝐴) = 𝑃(𝐷𝑇𝐵 − 𝐷𝑇𝐴 < 0) = 𝑃 (𝑧 <
0−30
) = 0.333
√1225+3600
Problem PP153
1. Let 𝑋𝑘 be the length of step k with pmf
𝑃(𝑋𝑘 = 1) = 0.5 and 𝑃(𝑋𝑘 = −0.8) = 0.5
𝐸(𝑋𝑘 ) = 0.1 𝑚 , 𝑣𝑎𝑟(𝑋𝑘 ) = 0.81 𝑚2
The sum ∑900
𝑘=1 𝑋𝑘 is approximately normally distributed with expected value 90 and
variance 729. Hence, approximately,
900
𝑃 (∑
𝑋𝑘 ≥ 100) = 𝑃 (𝑍 ≥
𝑘=1
100 − 90
√729
) = 0.355
2. The pdf of 𝑋𝑘 is uniform:
𝑝(𝑋𝑘 ) = 1⁄1.8 ,
−0.8 ≤ 𝑥𝑘 ≤ 1
𝐸(𝑋𝑘 ) = 0.1 𝑚 , 𝑣𝑎𝑟(𝑋𝑘 ) = 0.27 𝑚2
900
𝑃 (∑
𝑋𝑘 ≥ 100) = 𝑃 (𝑍 ≥
𝑘=1
3. The pmf of 𝑋𝑘 is
𝑝(𝑥𝑘 ) = 1⁄4 , 𝑥𝑘 = 1
= 1⁄2 , 𝑥𝑘 = 0
= 1⁄4 , 𝑥𝑘 = −1
100 − 90
√243
) = 0.2606
6
𝐸(𝑋𝑘 ) = 0 𝑚 , 𝑣𝑎𝑟(𝑋𝑘 ) = 0.5 𝑚2
900
𝑃 (∑
𝑋𝑘 ≥ 10) = 𝑃 (𝑍 ≥
0 − 10
𝑘=1
√450
) = 0.3187
Problem PP156
1. We solve the problem for general parameters 𝜆1 and 𝜆2 .
Let 𝑦 = 𝑥1 ⁄𝑥2 and 𝑡 = 𝑥2 . Then 𝑥1 = 𝑦 ∗ 𝑡 and 𝑥2 = 𝑡. The Jacobian of the
transformation: 𝐽 = 𝑡.
𝑓(𝑦, 𝑡) = 𝜆1 ∗ 𝜆2 ∗ 𝑒 −𝜆1 ∗𝑦∗𝑡 ∗ 𝑒 −𝜆2 ∗𝑡 ∗ 𝑡
,
𝑦, 𝑡 > 0
∞
𝑓(𝑦) = 𝜆1 ∗ 𝜆2 ∗ ∫ 𝑡 ∗ 𝑒 −𝑡∗(𝜆2 +𝑦∗𝜆1 ) ∗ 𝑑𝑡
0
Notice that the integrand is, except for the constant (𝜆2 + 𝑦 ∗ 𝜆1 )2 , a gamma density
with parameters 𝛼 = 2 and 𝛾 = 𝜆2 + 𝑦 ∗ 𝜆1 . Hence,
𝑓(𝑦) =
𝜆1 ∗ 𝜆2
, 𝑦>0
(𝜆2 + 𝑦 ∗ 𝜆1 )2
2.
∞
3. 𝑃(𝑌 > 𝑘) = 𝜆1 ∗ 𝜆2 ∗ ∫𝑘
1
(𝜆2 +𝑦∗𝜆1
)2
∗ 𝑑𝑦 = 𝜆
𝜆2
2 +𝑘∗𝜆1
.
For 𝜆1 = 4, 𝜆2 = 2 and 𝑘 = 2:
𝑃(𝑌 > 2) = 0.2
Problem PP157
1. 𝑇 = 𝑚𝑎𝑥(𝑋, 𝑌) where X (Y) is the time to finish project 1 (project 2).
𝑥 1
1
𝐹𝑋 (𝑥) = ∫10 10 ∗ 𝑑𝑥 = 10 ∗ (𝑥 − 10) , 10 < 𝑥 < 20. Similarly for Y.
𝑃(𝑇 ≤ 𝑡) = 𝑃(𝑋 ≤ 𝑡 ∩ 𝑌 ≤ 𝑡) =
1
∗ (𝑡 − 10)2 , 10 < 𝑡 ≤ 20
150
1
= 1 ∗ 15 ∗ (𝑡 − 10) , 20 < 𝑡 ≤ 25
7
𝑓(𝑡) =
1
∗ (𝑡 − 10) , 10 < 𝑡 ≤ 20
75
1
= 15
, 20 < 𝑡 ≤ 25;
1
43
2. 𝑃(𝑇 > 18) = 1 − 150 ∗ (18 − 10)2 = 75
𝐸(𝑓𝑖𝑛𝑒) = 75000 ∗
43
= 43000
75
𝐸(𝑝𝑟𝑜𝑓𝑖𝑡) = 50000 − 43000 = € 7000
Problem PP159
1. Let 𝑤 = 𝑥 2 ∗ y and 𝑣 = 𝑦. Then 𝑥 = √𝑤 ⁄𝑣 and 𝑦 = 𝑣. The J(acobian) is
1
𝐽 = 2 ∗ (𝑣 ∗ 𝑤)−1⁄2.
𝑓(𝑣, 𝑤) = 6 ∗ [1 − (𝑤 ⁄𝑣)1⁄2 ] , 0 ≤ 𝑤 ≤ 𝑣 ≤ 1
1
𝑓(𝑤) = 6 ∗ ∫ [1 − (𝑤 ⁄𝑣)1⁄2 ] ∗ 𝑑𝑣 = 6 ∗ (1 + 𝑤 − 2 ∗ √𝑤) , 0 ≤ 𝑤 ≤ 1
𝑤
1
2. 𝐸(𝑊) = 6 ∗ ∫0 𝑤 ∗ (1 + 𝑤 − 2 ∗ √𝑤) ∗ 𝑑𝑤 = 1⁄5
Problem PP160
1. 𝑓(𝑦|𝑥) =
𝑓(𝑥, 𝑦) =
𝑓(𝑦) =
1
√2𝜋
1
2
∗ 𝑒 −2∗(𝑦−2∗𝑥)
, −∞ < 𝑦 < +∞
1 2
1
1
1
2
2
2
∗ 𝑒 −2∗[𝑥 +(𝑦−2∗𝑥) ] =
∗ 𝑒 −2∗(5∗𝑥 +𝑦 −4∗𝑥∗𝑦) ,
2𝜋
2𝜋
−∞ < 𝑥, 𝑦 < +∞
+∞
5
1 2
1
2 4
∗ 𝑒 −2∗𝑦 ∗ ∫ 𝑒 −2∗(𝑥 −5∗𝑥∗𝑦) ∗ 𝑑𝑥
2𝜋
−∞
+∞
5
1 2
2 2
1
2 4 2 4
=
∗ 𝑒 −2∗𝑦 ∗ 𝑒 5∗𝑦 ∗ ∫ 𝑒 − 2∗(𝑥 +25𝑦 −5∗𝑥∗𝑦) ∗ 𝑑𝑥
2𝜋
−∞
2
+∞
1
2
1 2
1
1
1
−
∗(𝑥− ∗𝑦)
− ∗𝑦
5
=
∗ 𝑒 10 ∗ √1⁄5 ∗ ∫
∗
∗ 𝑒 2∗(1⁄5)
∗ 𝑑𝑥
√2𝜋
−∞ √2𝜋 √1⁄5
8
=
1
1
√2𝜋∗√5
2
∗ 𝑒 −2∗5∗𝑦 : normal with expected value 0 and variance 5;
2. Let 𝑣 = 𝑥 + 𝑦 and 𝑤 = 𝑥. Then 𝑦 = 𝑣 − 𝑤 and 𝑥 = 𝑤. The Jacobian of the
transformation is 1.
𝑓(𝑣, 𝑤) =
𝑓(𝑣) =
=
=
1
1
2
2
∗ 𝑒 − 2∗[𝑤 +(𝑣−3𝑤) ] ,
2𝜋
−∞ < 𝑣, 𝑤 < +∞
+∞
10
1 2
1
2 6
∗ 𝑒 −2∗𝑣 ∗ ∫ 𝑒 − 2 ∗(𝑤 −10∗𝑣∗𝑤) ∗ 𝑑𝑤
2𝜋
−∞
1
∗𝑒
√2𝜋
1
2
− ∗𝑣 2
1
√2𝜋∗√10
∗𝑒
1
9
20
1
∗ √10 ∗
+∞ 1
∫−∞ √2𝜋
∗
1
√1⁄10
1
∗
3
−
∗(𝑤− ∗𝑣)
10
𝑒 2⁄10
2
∗ 𝑑𝑤
2
∗ 𝑒 −2∗10∗𝑣 : normal with expected value 0 and variance 10
Problem PP162
1. Call your arrival time 0, X the arrival time of the first arriving bus and Y the arrival
time of the first arriving cab. Then
𝑓(𝑥) = 1 , 0 ≤ 𝑥 ≤ 1 and
𝑓(𝑦) = 𝑒 − 𝑦 , 𝑦 > 0
𝑊 = 𝑚𝑖𝑛(𝑋, 𝑌)
𝑃(𝑊 > 𝑤) = 0 , 𝑤 > 1
= 𝑃(𝑋 > 𝑤 ∩ 𝑌 > 𝑤) = (1 − 𝑤) ∗ 𝑒 − 𝑤
𝐹(𝑤) = 1 − (1 − 𝑤) ∗ 𝑒 − 𝑤
𝑓(𝑤) = (2 − 𝑤) ∗ 𝑒 − 𝑤 , 0 ≤ 𝑤 ≤ 1
1
𝐸(𝑊) = ∫ 𝑤 ∗ (2 − 𝑤) ∗ 𝑒 − 𝑤 ∗ 𝑑𝑤 = 𝑒 −1 = 0.368
0
1
𝑥
2. 𝑃(𝑡𝑎𝑥𝑖 𝑏𝑒𝑓𝑜𝑟𝑒 𝑏𝑢𝑠) = 𝑃(𝑌 < 𝑋) = ∫0 𝑑𝑥 ∗ ∫0 𝑒 − 𝑦 ∗ 𝑑𝑦 = 𝑒 − 1 = 0.368
9
Problem PP163
1
Γ(𝛼+𝛽)
1. 𝑝𝑋 (𝑘) = ∫0 (𝑛𝑘) ∗ 𝑞 𝑘 ∗ (1 − 𝑞)𝑛−𝑘 ∗ Γ(𝛼)∗Γ(𝛽) ∗ 𝑞 𝛼−1 ∗ (1 − 𝑞)𝛽−1 ∗ 𝑑𝑞
𝑛!
1
Γ(𝛼+𝛽)
= 𝑘 !∗(𝑛−𝑘) ! ∗ Γ(𝛼)∗Γ(𝛽) ∗ ∫0 𝑞 𝛼+𝑘−1 ∗ (1 − 𝑞)𝑛+𝛽−𝑘−1 ∗ 𝑑𝑞
𝑛!
Γ(𝛼+𝛽)
= 𝑘 !∗(𝑛−𝑘) ! ∗ Γ(𝛼)∗Γ(𝛽) ∗
Γ(𝛼+𝑘)∗Γ(𝑛+𝛽−𝑘)
Γ(𝛼+𝛽+𝑛)
Problem PP166
1. 𝑓(𝑥, 𝑦) = 2⁄𝑥 , 0 < 𝑦 < 𝑥 2 , 0 < 𝑥 < 1
1
Let 𝑣 = 𝑥 2 − 𝑦 and 𝑤 = 𝑦. Then 𝑥 = (𝑣 + 𝑤)1⁄2 and 𝑦 = 𝑤. The J(acobian): 𝐽 = 2 ∗
(𝑣 + 𝑤)− 1⁄2
1
, 𝑣 > 0 ,𝑤 > 0 ,0 ≤ 𝑣 + 𝑤 ≤ 1
𝑣+𝑤
𝑓(𝑣, 𝑤) =
1−𝑣
𝑓(𝑣) = ∫
0
1
∗ 𝑑𝑤 = −𝑙𝑛(𝑣) , 0 < 𝑣 < 1
𝑣+𝑤
1
𝐸(𝑉) = ∫ 𝑣 ∗ ln(𝑣) ∗ 𝑑𝑣 = 1⁄4
0
1
2. Let 𝑣 = 𝑥 2 + 𝑦 and 𝑤 = 𝑦. Then 𝑥 = (𝑣 − 𝑤)1⁄2 and 𝑦 = 𝑤. The J(acobian): 𝐽 = 2 ∗
(𝑣 − 𝑤)− 1⁄2
1
, 𝑣 > 2 ∗ 𝑤 ,𝑣 ≤ 𝑤 + 1 ,0 ≤ 𝑤 ≤ 1 ,0 ≤ 𝑣 ≤ 2
𝑣−𝑤
𝑓(𝑣, 𝑤) =
𝑣⁄2
𝑓(𝑣) = ∫
0
𝑣⁄2
=∫
𝑣−1
2
1
∗ 𝑑𝑤 = 𝑙𝑛(2) , 0 ≤ 𝑣 ≤ 1
𝑣−𝑤
1
∗ 𝑑𝑤 = −𝑙𝑛(𝑣⁄2) = 𝑙𝑛(2) − 𝑙𝑛(𝑣) , 1 ≤ 𝑣 ≤ 2
𝑣−𝑤
2
𝐸(𝑉) = ∫ 𝑣 ∗ 𝑙𝑛(2) ∗ 𝑑𝑣 − ∫ 𝑣 ∗ 𝑙𝑛(𝑣) ∗ 𝑑𝑣 = 3⁄4
0
1
10
Problem PP167
∞
1. 𝑓(𝑥) = 𝜆2 ∗ ∫𝑥 𝑒 − 𝜆∗𝑦 ∗ 𝑑𝑦 = 𝜆 ∗ 𝑒 − 𝜆∗𝑥 , 𝑥 > 0 ;
𝑦
2. 𝑓(𝑦) = 𝜆2 ∗ ∫0 𝑒 − 𝜆∗𝑦 ∗ 𝑑𝑦 = 𝜆2 ∗ 𝑦 ∗ 𝑒 − 𝜆∗𝑦 , 𝑦 > 0 ;
3. 𝑓(𝑦|𝑥) =
𝜆2 ∗𝑒 −𝜆∗𝑦
𝜆∗𝑒 − 𝜆∗𝑥
= 𝜆 ∗ 𝑒 − 𝜆∗(𝑦−𝑥) , 𝑦 > 𝑥
𝑦
𝐹(𝑦|𝑥) = 𝜆 ∗ 𝑒 𝜆∗𝑥 ∗ ∫ 𝑒 − 𝜆∗𝑧 ∗ 𝑑𝑧 = 1 − 𝑒 − 𝜆∗(𝑦−𝑥) , 𝑦 > 𝑥
𝑥
To generate a random value of Y given 𝑋 = 𝑥, set
1 − 𝑒 − 𝜆∗(𝑦−𝑥) = 𝑅 where R is a random number. Solving for y/
1
𝑦 = 𝑥 − ∗ 𝑙𝑛(1 − 𝑅)
𝜆
4. Let 𝑣 = 𝑥⁄𝑦 and 𝑤 = 𝑦. Then 𝑥 = 𝑣 ∗ 𝑤 and 𝑦 = 𝑤. The J(acobian) is 𝐽 = 𝑤.
𝑓(𝑣, 𝑤) = 𝜆2 ∗ 𝑒 − 𝜆∗𝑤 ∗ 𝑤 , 𝑤 > 0 , 0 < 𝑣 < 1
∞
2
𝑓(𝑣) = 𝜆 ∗ ∫ 𝑤 ∗ 𝑒 − 𝜆∗𝑤 ∗ 𝑑𝑤 = 1 , 0 < 𝑣 < 1
0
5. 𝐸(𝑉) = 1⁄2
6. Let 𝑤 = 𝑦⁄𝑥 and 𝑢 = 𝑥. Then 𝑦 = 𝑢 ∗ 𝑤 and 𝑥 = 𝑢. The J(acobian) is 𝐽 = 𝑢.
𝑓(𝑤, 𝑢) = 𝜆2 ∗ 𝑒 − 𝜆∗𝑤∗𝑢 ∗ 𝑢 , 𝑤 > 1 , 𝑢 > 0
∞
2
𝑓(𝑤) = 𝜆 ∗ ∫ 𝑢 ∗ 𝑒 − 𝜆∗𝑢∗𝑤 ∗ 𝑑𝑢 = 1⁄𝑤 2 , 𝑤 > 1
0
∞
1
7. 𝐸(𝑊) = ∫1 𝑤 ∗ 𝑤 2 ∗ 𝑑𝑤 = +∞
11
Problem PP168
1. Let 𝐵𝑖 denote the number of balls in box i. Then
𝑋𝑖 = 0 𝑤ℎ𝑒𝑛 𝐵𝑖 = 0
𝑋𝑖 = 𝐵𝑖 − 1 𝑤ℎ𝑒𝑛 𝐵𝑖 > 0
𝑛
𝐸(𝑋𝑖 ) = ∑
𝑛
(𝑏 − 1) ∗ 𝑃(𝐵𝑖 = 𝑏) = ∑
𝑏=1
𝑛
𝑏 ∗ 𝑃(𝐵𝑖 = 𝑏) − ∑
𝑏=1
𝑏=1
𝑃(𝐵𝑖 = 𝑏)
= 𝑛 ∗ 𝑝𝑖 − [1 − 𝑃(𝐵𝑖 = 0)] = 𝑛 ∗ 𝑝𝑖 − 1 + (1 − 𝑝𝑖 )𝑛 as 𝐵𝑖 is binomially distributed
with parameters n and 𝑝𝑖 .
𝑘
𝐸 (∑
𝑘
𝑋𝑖 ) = ∑
𝑖=1
𝑘
𝐸(𝑋𝑖 ) = 𝑛 − 𝑘 + ∑
𝑖=1
(1 − 𝑝𝑖 )𝑛
𝑖=1