8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf ·...

21

Transcript of 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf ·...

Page 1: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y
Page 2: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y

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ai : i = 1, 2, . . .�8���¹���¯���F�8�

�F�0���J©����@�¼�Y�[�2���2�Y�L���L�£¢0���°�£�[�F���Y���[�L�H�·«��L�ai ≥ 0

«������ ∑∞i=1 ai = a < ∞ È@�[�Y��Y�[�2���2�Y���L�L�£¢¤���°�£�[�F���Y���[�L�H�w�°�D�F���N��¸2��ª2�8�7�¬¢j�[�Y�Ã�®�+²<�Y�N���N�

pi = ai

a

§

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2x− x3 ≥ 0

�L�x ≥ 0

�����x2 ≤ 2

«!� § � § «!��� [0,√

2] §ÃÎ �����J� 5

2 >√

2�F�Y�°�É���É�Y�2�É���8�����Y���Y���F����2�

�Y�[�2���2�Y���L�L�£¢¤���N���®�L�£¢ §Î ���*���°���[�L¢H«

2x− x2 > 0�L�

0 ≤ x«W�2���

2 ≥ x§*Ï �Y� 5

2 = 2.5 > 2�����Ð�F�Y�°�É���Y���J�F���2�º�����2���F���Y�2���w�F�Y�¡�R���Y���

�N�2�����°�Y�S�[� §

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P [X > 20] =

∫ ∞

20

f(x)dx =

∫ ∞

20

10

x2dx = lim

K→∞

[

−10

x

]K

20

= 0 +1

2= 0.5

Ö ���Y�Ã�N�����°��¸2��ª2�8�7�¬¢F (x) =

∫ x

−∞ f(t)dt«Y�Y�8���J�

x < 10 : F (x) = 0

x ≥ 10 : F (x) =

∫ x

10

10

t2dt =

[

−10

t

]x

10

= 1− 10

x�¥�Y���[�Q�[���S�

F�����J�H�<�F���<���2�����S�

x = 10 !�Y�N���J�Q�L�)�¯�2±2�+�·�Y�Á�Y�¡Ä!�N�[�N���N�Q¶��Y�N�F�Y�8�

¶D��Y�J©��Y�F = 0

���2�x < 10

�2�x ≥ 10

«4�2���Ð�F�L�¯�����2�F��¢����H���[�Y�*�Y�J©��Y�L�F���2�@���F × ���[�2���F����[�L¸H�<�£Ø § ´]�

F���2�����2�@�7���°�F�N�2�<�F���¬�Y�¡�£¢H«=�J�2�¬ªH�N�<�F���2���°���[�w�Y�J©��Y�

�L�Á�2�ÃÙRE�¾�\:Ú³¾�¿�ES¿¬O»PRCFE�MÛ\�VYK�CÌÚ¡Ü+V�\]ÝÞ Í �S¶f¶D�����&ª2��ߺƷÇÅ�8«D�2���;¶D�w���&ª2�8�·à �j�4�N�8�;�[�2�°�0�2�<¢<�[�Y�L��¸º�2���H���¯�F�Y�8�L�cáNE�Ú³¾�\�Y���®�F�[�L�����F���2� § Ô¹��Y�N�+���F�w�2�[�®���*�Ã�F�����Á�F���N¢����8���&ª2�c�������8���8�����8�H�[�L¢â�¥¶��Y�°�R�

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Page 3: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y

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P [X ≤ 15] = 1 − 10

15=

1

3= p̄

�¥�Y���[�Ð�[���S�7���2���J�2�<�[�L�¬�Y�H���¤Æ·ÇÅ�8«��L�7�j��±H�8���Y�0���LÄ!�N�[�N���N�Ð�¡�öD�@�����N�L���Y�@�2��N½Y�J�������D�F�Y���8�����4�2���H�R�)���4�L�<�F�8�FªS�2���8«��®�����N�Q�[�Y���Y�F�H�������L���¡�£¢c�2�

X���8�L�Y¸��N½��H�Ì�[�L¢

�+²<�������F�;�2�a�N�������H�L�<�w���7åN�8�F�¼ ¯�2�+«��8²<�Y��ªS�����N�<�F��¢2«��4�8�8�����F�2«����H�����<�F�N¸H�[�2�Y����������Ì�[�L�H���

limx→b

∫ x

a

f(t)dt =

∫ b

a

f(t)dt

� § � § «H���<�[�����8�F��ªS�S�[�LªH�8�)���[�D�N�2�<�F���¬�Y�2���[� §�æ �Y�Y��¢<���Y¸É�F�������[¸2�Y�¯�N�<�)¶����¯�J�Å���*�F�Y����F�8ª<���2�����2�[�®��¸2�Y�¯�8�H�+«<�F�Y�c�[�8²<�Y�+�£�[�8�7�Y�F�H�������L���¡�£¢�����¸H�LªH�N�w�<¢

6∑

k=3

(

6k

)

(1 − p̄)kp̄6−k =

= 20 ·(

2

3

)3(1

3

)3

+ 15 ·(

2

3

)4(1

3

)2

+ 6 ·(

2

3

)5(1

3

)

+

(

2

3

)6

=

=23

36

(

20 + 15 · 2 + 6 · 22 + 23)

=8

729(20 + 30 + 24 + 8) =

8 · 82

729= .89986

{Ëçè��8�F�H«�¶������F���F�������2�[���Y�°�

EX =

∫ ∞

−∞xfX(x)dx

¶��Y�°�R��°�»�¯�8�2�Y���Y¸����Y���H�Y��¢��L�xfX

�°�»���¯�L�<�[�N¸2�R�����L�����Y���J�F���2���³���H�Å�L���£�R�����N�2«2�����8���®�L�£¢�����[�Y�7���2�[� c

x2

���2�x ≥ 1

«»¶��¡�[�-�F�Y�L�[�2�Y�L�c«Å¶D�2�Y�°�ã�4�7�������8�8�0�@���8���®�L�£¢2«»�����j�¡�

¶D�2�Y�°�é���&ª2�����â�J½��4�8�Ì�[�8�qªS�2�L�Y� §yê �H�F���[�ã�F���@�4�2���H�+«��F���¼ëD�����R�<¢-���°�£�[�F���Y���[�L�H���¶��L�F�-���N���®�L�£¢ã�����[�Y�����H�F� c

a2+x2

�J«Å¶��Y�°�R�;���Y�4�8�2�[����0�F�Nª2�8�[�2�Q�����Y�L�°�N���F���2���8«Å���°�F����H���Y�¯�J½����+�Ì�[�8�wªS�����Y�+� §Õíì ���R�£�+«��Y�2�F�c�[���S� f ���Á�L�����N�+�Ð�����8���F�¡�£¢7���Y���J�F���2�;�¥�¡�������¤�¯�N����8�Á�2�»�F�Y���¥���¯�L��¢�2�)�F�� µ�N�2�L���8� ×Fî �2�¯�¯�SØÉ���°�£�[�F���Y���[�L�H���R�ÉÈ

∫ ∞

0

1

4xe−

x

2 dx =1

4

(

[

−2xe−x

2

]∞

0+ 4)

= 1

���j�J�H�*�����F�EX

«�¶��c�N�2���N�Y�°�S�F�∫ ∞

−∞xf(x)dx =

1

4

∫ ∞

0

x2e−x

2 dx

ä

Page 4: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y
Page 5: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y
Page 6: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y
Page 7: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y

7 : 45�F�

7 : 50«¬���H����F���R���=�[�L�¯� × ¶��L�����S¶DØ���� ¨�V �¯�L�¬���[�8� § ´]������US�F�Y�c���[�[�LªH�8����¹�F�Y�¯�[�N�j�����Y���Y¸��[�L�¯�¯���H�[�N�[ªS���°�N«!�[�Y� × ¶����Y�Y���Y¸SØ�F�R�����¹¸H�<�+���[� A «W�2���Ð�[���S��°���Sª2�8����F�2�[�2� × ¶��������S¶DØÁ����ö V �*���¬���F�+� § èÁ�N���N�2«¬�[�Y���Y�F�H�������L���¡�£¢¯�2��¸2�H�L��¸Ã�F�

A�°� 40

60 = 23

§ ´µ���J�°���N�<�R�����L¢H«=�F�Y�j���F�H�����Y�����¡�£¢Ð�2�Q�[�Y�j���2�[�F�N�Y¸H�N�c�N���[�R�Y���Y¸w�F�Y��[�[�2�L�7�F���S�����8�&ª2�+���S��W¬È VXV �°��åN�N�[��«Y�F�����J� ∫ 0

0 fp(t)dt = 0§

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[0, 30]�Å�����*���¬���F�+�N«

P [W > 25 |W > 15] =P [W > 25 ∩W > 15]

P [W > 15]=P [W > 25]

P [W > 15]=

5301530

=1

3

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P [W > 30] = 0� § ´]���F�Y�¤�����Ã���[�[�LªS���Å�Y���®�F�[�L�����F���2�¼���H�¼���8�N�

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W�L�@�F�Y�

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15e−

t

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F (10) = 1 − e−10

15 = 1 − e−2

3 = .48658

W

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P [W > 25 |W > 15] =P [W > 25 ∩W > 15]

P [W > 15]=P [W > 25]

P [W > 15]=

=e−

25

15

e−15

15

= e−10

15 = .48658

� § � § «�¢2�2�����Y�[�2�����Y���L�L�£¢¯���W¶����L�F���Y¸ ¦�V �*���¬���F�+���2�D�¯�2�[�����H�Q�������L�¯�Y�[�Sª2�+�¯�2�Y����¡� § ��������¾4ES\D�[�Y���[���¯���F�¡�[���S�[�L�H���H���4�J���2�[�2È

P [W > 30] 6= 0«4�����=«��L���¥�H�Ì�+«

�¬¢��F�Y�Ã�[���¯���N���°�J�������F���2�·«

P [W > 39 |W > 29] = .48658

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Z =X − µ

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P [a < X < b] = P

[

a− µ

σ<X − µ

σ= Z <

b− µ

σ

]

õ·�N�8à �����Y���L¢j�[�Y�°���F�¯�[�Y�ò<�Y�8�®�F���2���c�µ = 10, σ2 = 36, σ =

√36 = 6

�JÈÕ P [X > 5] = P

[

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6

]

= P[

Z > − 56

]

= .79767

ÖP [4 < X < 16] = P

[

4−106 < Z < 16−10

6

]

= P [−1 < Z < 1] = .68269���Y�2�F�

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P [−1 < Z < 1] ' .68, P [−2 < Z < 2] = .95450 ' .955, P [−3 < Z < 3] = .99730 ' .997

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Þ P [X < 8] = P[

Z < 8−106 = − 1

3

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= .36944

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`P [X < 20] = P

[

Z < 20−106 = 5

3

]

= .95221

^P [X > 16] = P

[

Z > 16−106 = 1

]

=

=1

2P [|Z| > 1] =

1

2(1 − P [−1 < Z < 1]) =

1

2(1− .68269) = .15866

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P [X > c] = P

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=

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P [Z > z.1] = .10

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P [.8950 < X < .9050] = P [−.0050 < X − .9000 < .0050] = P

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= P

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j

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P[

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P [Y < 150] = P

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= P

[

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=

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Í ���[��[�Y�¤�J�H�Y�Y�+�Ì�F���2�¹���N�£¶��8�N�a�����@�F���*�N½����+�Ì�F�+�ºª&�2�L���

τ = 1λ

§ ´]�Ã�°�É�F�Y��¯�+�����2�w�F�������°�����³�[�N���N�2�L���8� × �F�Y�����2�¡�)�L�L���ÌØ����)��R�2�Y�L�<�2�Ì�[�LªH�Á�j�S�[�N�[���2�#«�¶����H�F��Y�8�N�&¢��°�&¶y�°�c�2�[�F�Y�¯�8�@�F�w���2���L�S¶ ���º�J½����H�Y�N�<�[���2�)�Y���®�F�[�L�����F���2� § ´]�Ã�°�c�����¯�H�®��[�Y�Ã�[���¯�c�2�D�F���c�J½����+�Ì�[�8�wªS�����Y�2«��Y�w�[�¤���¥�H�Ì�[�2�����

log(2) = .69315§

À �Q�J�H�Y�R�®�H«�¶D�Ã�2�Y�[�������[�Y�*�[���¯�Ã�[�8�F�Y�L���L�Q¶D��¶D�2�[±7�����F�+�Ì�[�L¢�¶��L�F�Ð�[�Y��¸H�LªH�N�@�Y���®�F�[�¡ �Y���[�L�H���8«�¶��¡�[�Y�2�����[�J���8�F�[���Y¸��[�¯�F�Y�c¸H�N�Y�8�[�2�!�F�+�®�Y�L� §Õ Ô¹�c���¬�2±¯���2� a �F�������*���Y���¯�Lå8�8�

ψ(a) =

∫ A

0

|x− a| 1

Adx =

1

A

(

∫ a

0

(a− x) dx+

∫ A

a

(x− a) dx

)

=

=1

A

(

a2 − a2

2+

(

A2 − a2

2

)

− a (A− a)

)

=1

A

(

A2

2− aA+ a2

)

¶������R�·«=�H���¯�������Ì�[�L�H�@���a«!���É�¤�����R���4�2�°�Y«��N�2���8�&ª2�c�Y�W«4¶��L�F�º�j�¯�L���L��Y� �S�

�L�[��ª2�N�F�F�N½=«�� § � § ���a = −−A

2 = A2Ö Ô¹�c���S¶ð���<�H±��S���¯���Y�L�¯��åN���Y¸

∫ ∞

0

λ |x− a| e−λxdx = λ

(∫ a

0

(a− x) e−λxdx+

∫ ∞

a

(x− a) e−λxdx

)

=

= λ

(

a

(∫ a

0

e−λxdx −∫ ∞

a

e−λxdx

)

−(∫ a

0

xe−λxdx+

∫ ∞

a

xe−λxdx

))

ê �Y�¡�[�L���L¢¬���Y¸¯�F�Y�[�2��¸2�Y�H�����F�Y��©��R�®�D�¥�H�Ì�[�2�λ«Y¶���¸H�J�

a− 1

λ

(

1 − 2e−λa)

�¥¶��¯���&ªH����®�+�Ð�[�Y�j�N�¬�D���D�2�¹�N½����H�Y�N�<�F�°���:«1 − e−λx

«W�H�ɶ��8�L�Å�H�����<�F�8¸2�R�S�F�+��¬¢7�����F�[���F�����F�N�[�j�������F�Y�����H�F�

∫ d

c

xe−λxdx =

[

− 1

λxe−λx

]d

c

+1

λ

∫ d

c

e−λxdx =

¦ ÷

Page 16: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y

=1

λ

(

ce−λc − de−λd)

+1

λ2

(

e−λc − e−λd)

�F�Y���®�F�L�F�Y�F���Y¸¤���Y���F�H�Y�F�°�S�[�N��¢Y� §���������������Ì�[�L�H�w�2�

a¶������·���c�¯���Y���*��åN�+���S�

1 + 2e−λa = 1«��H�

a = log 2λ

{ËÊog }Y���Ôº�c�2�Y��¢¤�Y�8�8���F�j�[�N�¯�N����8���F���������2�����w�J½��4�2�Y�8�<�F�°���·Æ·Ç«Y¶��L�F�������R���¯�N�F�N�

λ

P [X > t] = e−λt

Õ Ô¹�Ã�2�F���2�F±2�+�¤�[�¯�8���°�J�Y�°�S�[�

P [X > 2] = e−2λ = e−2· 12 = e−1 = .36788

Ö Ô¹�Ã�2�F���Y�S¶í�2�F±2�+�j���2�

P [X > 10 |X > 9] = P [X > 9 + 1 |X > 9 ] =P [X > 10 ∩X > 9]

P [X > 9]=P [X > 10]

P [X > 9]=e−λ10

e−λ9=

= e−λ = P [X > 1] = e−1

2 = .60653

ab^Qc Õed�f Ôº�É���&ª2���[�J ]�Y�[�Sª2�+���[�Y�É�F�8�¯�2�F±S�2�Y�L�Á�Y�F�H���8�®�£¢*���W�F�Y�É�J½��4�2�Y�8�H�[���2�4�Y���®�F�[�¡ �����F���2�·ÈP [X > s+ t |X > s ] = P [X > t]

ab^Qc Õed�f Í ���[���[���S� λ �°�É�F����[�8�N�L�Y�[���N�2�W���Q�F�Y�*�J½��4�8�Ì�[�8�¹ªS�����Y� § è��8���J�H«=¶��Y�N�¹¶���¯�+�2�F�Y�F�7�F���¯���L�-�Y�H�Y�[�8«λ�°�¯�*�+�2�F�Y�[�8�â���

h−1 «������0¶��Y�N�-¶D�w�¯�+�2�F�Y�F�w����¯���¬���F�+�N«=�L�c���É�¯�8�H�®�Y�[�8�Ð�L�m−1 «=�J�R� § èÁ�N���N�2«=�L�[�ɪS�����Y�2«=���¹�2�¹�����Y�L�°�N���F���2�·«¶������D���8���8���ã�H�¼�F�Y�7�Y�Y�L�[�¶��w���[�j���®���Y¸ § ���Y���4�¬�2±¼�®�Y�H�Y�°�ã���&ª2�¤�F�4�8�J�L©��+�

�[���S���[�Y��ªS�����Y� 12

�¡��¶��2���2�[�®��¸2���L�Y¸��F�λ¶��2�D���

h−1 §ÅÍ ���[�2«��¬¢j�[�Y��¶D�&¢H«H�[���S��[�Y�c�Y�Y�L�[�����®�+�7���2�λ�2�F���F�Y�<�®�c�2�4PRCFK��N¿�KJ¾!ÙJÚ¥KJONÝ

{ËÊcÑÓ}Y���è��8�F�H«��H���J�öD�Ã���j�[�Y���j���F�·«�¶��øH�J���[���F�N�c�[�Y�����LÄ4�8�F�8���J�����N�£¶��8�N�����¡���N�F���*�+���[���S����[�¤�N½¬�4�2���N�<�F�°�����L¢¼���°�£�[�F���Y���[�8�=«������ã���L���J�F���¯�8��F���S�*�2�F�7�Y���+Â@´µ���Y�N�8�·«Å¶��w�����F�+�2��¢±¬�Y�S¶��¥�2�����F�����j�S�[�ж��H�Y�°������F�Y�*�F�2�¯���2�Á�L�@�F�Y��Y�F�8ª¬�L�H�����N½��2�¯�Y�L�&�Ì«��F�����8«4���2��F���c�J½����H�Y�N�<�[���2�·�N�2�F�2«

P [X > t+ s |X > t ] = P [X > s]

�Y�8���J��F���S�

P [X > 10, 000 + 20, 000 |X > 10, 000] = P [X > 20, 000] = e−20000

20000 = e−1 = .36788

¦ ß

Page 17: 8 ¹ ¯ F 8 - University of Washingtonfaculty.washington.edu/fm1/394/Homework/week7/week7.pdf · ¥¶ y ° r ¹ ° c ¯ h ® 2 ¬ª¬ 2 8ÈÁ¢h 2 ¹ 2 f ¶ h 8 f y¸ l q [ y ¯ y

¶��Y�°�R�Ð�°���[�Y�*�[���¯��Y�F�H�������L���¡�£¢w���ŸH�2���Y¸ ¨XV « VXVXV �¯���L�+���F����� Î �¯�L�F�º�2�[�L¸H�L���2�L��¢����2� §´]�£«Y�Y�S¶D�NªH�N�+«2�[�Y�Ã���°�£�[�F���Y���[�L�H�������Y���¡���H�F�Û�2�

[0, 40, 000]«H�F���N�

P [X > 10, 000 + 20, 000 |X > 10, 000] =P [X > 30, 000]

P [X > 10, 000]=

40, 000− 30, 000

40, 000− 10, 000=

10, 000

30, 000=

1

3

À ���N�2�Y�R�F�2«��2�[��¸2���������L¢H«<�F�Y�c�Y�[�2�����Y���L�L�£¢j���)¸2�N�®�[�L�Y¸ ¨XV « VXVXV �¯�L���8�����2�7���8�N� 12

§Z\[6]�^é´µ�ã�[�Y���¯�8�2�F�2«

X�°��¯�+�2�F�Y�F�+�¼�L� ¦�V-VXV �Ã���Á�¯���L�+�à��Y�N���J�

λ�°���¯�+�2�F�Y�F�+�ã���

(1000mi)−1

{ËÊÃçæ ���°�£�[�F���Y���[�L�H�7¶��L�F�����2å8�2�[�����Y���J�F���2�

h(t)�����F���R�7�F�����

P [X > t] = e−h(t)

�Y�8���J�����w�H�Y���N�H�®�H«

Õ P [X > 2] = e−∫

2

0t3dt

= e−24

4 = e−4 = .018316

ÖP [.4 < X < 1.4] = e

−∫

0.4

0t3dt − e

−∫

1.4

0t3dt

= e−.4

4

4 − e−1.4

4

4 = .61088

Þ P [X > 2 |X > 1] = P [X>2]P [X>1] = e

−∫

2

1t3dt

= e−4+ 1

4 = .023518

{ËÊ*ñæ �����®���2�#«��L�N�8à �����Ã�8���[�J���Y�=¶��L�F�������®�H�L���[��ª&�2�L���8� §Õ P

[

|X | > 12

]

= P[

X > 12

]

+ P[

X < − 12

]

= 2 · 14 = 1

2Ö �W�;©����é�[�Y�¹�Y�N���F�¡�£¢H«�¶D�¹�N�2�q�Y�[�¬�N�N�+�q�L� �®�8ª2�N�R���Á¶D�&¢�� § õ·�J�+à �8«����H�������£�R�����N�2«�N�2�¯�Y���[���F�Y�Ã�8�¬�����2�¬ª¬���2���F�L¢H«

FX = 0���2�

x < 0«Y�F�*¶D���F�+�£�[�F�°�Ì���[�

x > 0�JÈ

P [|X | ≤ x] = P [−x ≤ X ≤ x] =

∫ x

−x

1

2dt = x

èÁ�N���N�2« |X | �°���Y�Y�L���2�[�f�H� [0, 1]§

{ÒÑAT }Y���Ôº��¶D�2�[±É­£���£���H��¶D�������7���¤�F���É���F�8ª<���2���Q�Y�[�2���L�8� ���Y�2�F���F�����

ex > 0���2���2�<¢

x«Y�®�

�F���cª&�2�F�°�����L�Y�2�Y��¢¤�R��±2�+�D���<�®�L�F��ª2��ªS�����Y�8�D �¶��É�F�¬�����F�+�£�[�F�°�Ì�D�F�

y > 0�JÈ

P[

eX ≤ y]

= P [X ≤ log y] = log y

¦ W

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as long as logy ∙ 1, i.e., y ∙ e (which is obviously the top range for eX), and logy ¸ 0, i.e. y ¸ 1 (which is obviously the bottom range for eX). The density of Y is the derivative of this function: always equal to 0, except for 1 ∙ y ∙ e , where it is equal to y

1

Theoretical Exercises

#1This is, really, a Calculus exercise. We are enforcingZ

0

1ax2e¡bx2 = 1

The following method relies on the identity ¡³21´= p¼, which you may be familiar with, or

not (it is proved on page 199 in the book), and the property of the Gamma function (see page 215): ¡ (x) = (x¡ 1)¡ (x¡ 1)Taking this identity for granted, we can use substitution, namely x2 = u, to transform the integral into

2aZ

0

1pb

pbu2

1

e¡budu =bpb

a¡¡23¢Z

0

1

¡¡23¢b(bu)23¡1e¡bu

From the definition of the Gamma distribution (on page 237), we know that the integral is equal to 1 (read b as ¸ in the formula). Hence, we have

ab23

¡¡23¢= 1

which, using the fact that ¡³23´=21¡³21´=

2

p¼ results in

a = p¼2b23

= 2bq¼b

#2Following the hint, we have (for absolutely continuous random variables)

E [Y ] =

Z

¡1

1yfY(y)dy =

Z

¡1

0

yfY(y)dy +

Z

0

1yf

Y(y)dy

Integrating by parts both integrals results in

[yF Y (y)]¡11 ¡

Z

¡1

1FY (y)dy

where F Y is the cumulative distribution function of Y. To prove the assertions we only need to observe that the integral is precisely the right hand side of the equation (when a density function exists, the definition of cumulative distribution function in terms of < or ∙ makes no difference), and that, in order for the expected value to exist, the first term in the expression has

to be zero (that is equivalent to the requirement that the integral

Z

¡1

1yf

Y(y)dy exist!)

18

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#3The hint shows a pretty straightforward path. A less rigorous argument (that can, however, be extended to more general situations, once the necessary technicalities are settled) could go as follows.Let X be a continuous random variable. We approximate it by Xn, where the range of X has been “cut up” in small segments

£ak; ak+1

¢, of length n

1, and Xn = ak for all outcomes such that ak ∙ X < ak+1. We now have a discrete random variable, and we have, from Chapter 4

E [g(Xn)] =k

Xg(ak)P [Xn = ak]

By the fact that P [Xn = ak] =

Z

ak

ak+1

f(x)dx, and that the right hand side is now, essentially, a

Riemann sum for our “target" integral, as n! 1, we have our proof.Remark: the argument would work for any random variable X, if we had a rigorous method for passing to the limit in

k

Xg(ak)P

£X 2

£ak; ak+1

¢¤ even when the distribution does not admit a

continuous density. In fact, we would need to extend our definition of integral, in order for this to work, and that is precisely what the Lebesgue integration theory does for functions if a real variables, and general measure theory does for more general situations (recall that our X has an abstract space, the sample space, as its domain).

#5Let’s follow the hint:

E£Xn¤=

Z

0

1P£Xn> xn

¤nxn¡1dx

which proves the statement.

#6This problem has a simple answer: choose Ea as the event when X is not equal to a. It is not such a surprising example, as we are dealing with an uncountable intersection. In fact, if you take complements, the condition requires that the union of the complements of these sets (each having probability zero) have probability 1. This can obviously happen only because we are dealing with an uncountable collection of events.This example reminded me of a different, more surprising, one, which you might feel has some relation to it (or maybe you won’t feel that way - no matter: it is an interesting example, and it has big implications in unexpected areas, like celestial mechanics).Consider the set of rational numbers, say, between 0 and 1. Since they are a countable set, we can write them up as a sequence, say, r1; r2; r3; ... This is a dense set in [0;1]. Now, consider the set made up of the union of intervals of amplitude

2n" around rn, for a small ". The total

probability associated with these intervals is less than or equal to the sum of their amplitudes, which is equal to "

n

X2n1 = ". This is surprising, because " can be as small as you wish, but the

sets we are talking about are all open and dense, so that, in a topological sense, they are “big”. The intersection over a sequence of such sets, as "! 0, will have probability zero. Topologically, though, it will be the intersection of a countable number of open dense sets - this

19

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is called a “second category Baire set”, which, again, topologically is thought to be a “big” set (the intersection of second category Baire sets, is, again, a second category Baire set). The conclusion of this example is that “probabilistic” (or “measure theoretic”) size is very different from “”topological” size.

#10We don’t need to be this lazy, but, thanks to a quick change of variables (it is a standard one: x 7! p

Var(X)x¡ EX ), we might as well prove that f(x) = e¡ 2

x2

has an inflection point at j x j = 1.

Since the second derivative of f (x) is ¡x2 ¡ 1

¢e¡ 2

x2

this is pretty obvious

#15A simple way to show this is to evaluate

P [cX > x] = P [X > cx] = e¡¸cx = e¡c¸x

#16The hazard rate for an absolutely continuous random variable X is defined as the density

formally defined by dx

P [X 2 dx jX > x] , which, with a little care, can be seen to translate, in the

case of absolutely continuous random variables, into 1¡ FX(x)fX(x)

, as discussed on page 234

(section 5.5.1). It is a very intuitive tool in survival and reliability theory. For a uniform variable on [0; a] (notice that, for distributions with a continuous density, it makes no difference if we consider open, closed, semi-open, etc. intervals), this translates to

aa¡xa1= a¡ x1

Not surprisingly, it explodes as x approaches a: after all our gadget has to fail at some point between 0 and a, and as we approach the end of the interval “time for failure is running out”!

#23The density of a Weibull distribution can be defined from

R(t) = P [X > t] = e¡atb

(In reliability/survival analysis we will always have t ¸ 0). While there are approximating argument to justify this choice, the most obvious reason is seen by comparing with the exponential distribution, where

P [X > t] = e¡at

As the question implies, choosing values of b less than, equal to, to greater than 1, corresponds to having hazard rates that are decreasing (a “rejuvenation” effect), constant (“memoryless”), or increasing (“aging” effect). Thus, the Weibull family provides an easy toolbox for a quick and dirty description of the different age-related scenarios for the likelihood of an upcoming failure.The hazard rate calculation is an exercise in derivation:

R(t)

fX(t)=

R(t)¡R0(t)

= ¡dt

dlogR(t)= ¡

dt

d³¡atb

´

= abtb¡1

20

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The conclusion about the different behavior as b > 1; b = 1; b < 1 follows now easily

#27The cumulative distribution function for X is P [X ∙ x] =

b ¡ ax¡ a. Let’s define Y = ®X + ¯.

Since ®a+ ¯ ∙ Y ∙ ®b + ¯ , we need to make sure that ®a+ ¯ = 0, and ®b + ¯ = 1. Thus, we want ® =

b ¡ a1 ; ¯ =

a¡ ba . Checking, for example, the cumulative distribution function

will confirm this is he right choice:

P [Y ∙ y] = P∙X ∙ ®

y ¡ ¯¸= P [X ∙ a¡ (a ¡ b)y] = y

(for 0 ∙ y ∙ 1), by directly substituting.

#29Note that F is a monotone increasing function. Since we are assuming that it is strictly monotone increasing function (and, one could be careful and extend the argument to the non strictly monotone case), it has a proper inverse F ¡1 (in the non strict case, one can use the “right inverse”). Then,

P [F (X)∙ y] = PhX ∙ F ¡1(y)

i= F

³F¡1(y)´= y

This observation is sometimes used to simulate a random variable with a given distribution function, starting from a program that simulates a uniform random variable: starting from a uniform random variable Y, the random variable X = F

¡1(Y ) has distribution function F. This

method is easy, whenever F has an easy inverse e.g., for exponential variables, but not for normal variables), but has its own numerical pitfalls - for example, simulating an exponential variable requires the computation of a logarithm, which is computer intensive, as well as less precise than algebraic calculations.

#31We can compute the probability density function in various ways. A plain route is the following:

P£eX ∙ y

¤= P [X ∙ logy] = p

2¼¾21

Z

¡1

logy

e¡ 2¾2(x¡¹)2

dx

Taking the derivative with respect to y, we obtain

fY(y) =

yp2¼¾2

2¾2(logy¡¹)2

This choice for stock price modeling has the advantage of simplicity, since the logarithm of the random variable has a convenient normal distribution, but it does have a number of drawbacks. One that got a lot of attention in the wake of the recent financial crisis is its lack of “fat tails” (outliers are very unlikely), but there are other concerns as well as to its appropriateness.

21