MATH 20: PROBABILITY Midterm 2 Xingru Chen - - PowerPoint PPT Presentation

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MATH 20: PROBABILITY Midterm 2 Xingru Chen xingru.chen.gr@dartmouth.edu XC 2020 Ex Exam How many hours you spend preparing for the exam? Wrapper Wr How many hours you spend on the exam? Content on the last two weeks.

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MATH 20: PROBABILITY

Midterm 2 Xingru Chen xingru.chen.gr@dartmouth.edu

XC 2020

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Ex Exam Wr Wrapper

for midterm 2

How many hours you spend preparing for the exam? How many hours you spend

  • n

the exam? Content

  • n

the last two weeks. Arrangements for the final. …

XC 2020

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Problem 1: True or False

XC 2020

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?

Can π‘Œ follow a continuous uniform distribution?

XC 2020

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Density Functions of Continuous Random Variable

Β§ Let π‘Œ be a continuous real-valued random

  • variable. A

density function for π‘Œ is a real-valued function 𝑔 that satisfies 𝑄 𝑏 ≀ π‘Œ ≀ 𝑐 = ∫

! " 𝑔 𝑦 𝑒𝑦,

for 𝑏, 𝑐 ∈ ℝ. Β§ If 𝐹 is a subset

  • f

ℝ, then𝑄 𝑦 ∈ 𝐹 = ∫

# 𝑔 𝑦 𝑒𝑦.

Β§ In particular, if 𝐹 is an interval [𝑏, 𝑐], the probability that the

  • utcome
  • f

the experiment falls in 𝐹 is given by 𝑄 [𝑏, 𝑐] = ∫

! " 𝑔 𝑦 𝑒𝑦.

𝒛 π’š 𝑏 𝑐

XC 2020

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SLIDE 6

?

Can π‘Œ follow a continuous uniform distribution?

No

?

Can π‘Œ follow a discrete uniform distribution?

𝑛 𝑦 = β‹―

XC 2020

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SLIDE 7

?

Can π‘Œ follow a discrete uniform distribution?

𝑛 𝑦 = 0 𝑛 𝑦 > 0

XC 2020

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SLIDE 8

?

Can π‘Œ follow a discrete uniform distribution?

𝑛 𝑦 = 0 𝑛 𝑦 > 0 ,

!"# $%

0 = 0 ,

!"# $%

𝑑 = +∞ No

XC 2020

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SLIDE 9

Problem 1: True or False

Discr crete variance ce π‘Š π‘Œ 𝐹 π‘Œ& βˆ’ 𝜈& = 0 = = 𝐹 π‘Œ βˆ’ 𝜈 & = ,

'∈)

(𝑦 βˆ’ 𝜈)&𝑛(𝑦) .

Co Continuous variance π‘Š π‘Œ

0 = 6

*% $%

𝑦 βˆ’ 𝜈 &𝑔 𝑦 𝑒𝑦

π‘Œ = 𝜈

XC 2020

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Problem 2: Computation

XC 2020

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?

𝐹 π‘Œ&#&# = 𝐹&#&#(π‘Œ)

XC 2020

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The Product of Two Random Variables

Β§ Let π‘Œ and 𝑍 be independent real-valued continuous random variables with finite expected

  • values. Then

we have 𝐹(π‘Œπ‘) = 𝐹(π‘Œ)𝐹(𝑍). Β§ More generally, for π‘œ mutually independent random variables π‘Œ!, we have 𝐹 π‘Œ+π‘Œ& β‹― π‘Œ, = 𝐹 π‘Œ+ 𝐹 π‘Œ& β‹― 𝐹(π‘Œ,).

?

𝐹 π‘Œ&#&# = 𝐹&#&#(π‘Œ)

XC 2020

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?

Are π‘Œ and π‘Œ independent? Β§ If π‘Œ is any random variable and 𝑑 is any constant, then π‘Š π‘‘π‘Œ = 𝑑&π‘Š(π‘Œ), π‘Š π‘Œ + 𝑑 = π‘Š(π‘Œ). Β§ Let π‘Œ and 𝑍 be two independent random

  • variables. Then

π‘Š(π‘Œ + 𝑍) = π‘Š(π‘Œ) + π‘Š(𝑍). π‘Š 2π‘Œ = 4π‘Š π‘Œ ? π‘Š π‘Œ + π‘Œ = 2π‘Š π‘Œ ?

XC 2020

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Problem 2: Computation

XC 2020

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!

Moment: 𝐹 π‘Œ, , where π‘œ = 1, 2, 3, β‹―

XC 2020

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Problem 3: Proof

𝑦! βˆ’ Μ… 𝑦 = (𝑦! βˆ’ 𝜈) + (𝜈 βˆ’ Μ… 𝑦)

XC 2020

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𝐹 𝑑& = π‘œ βˆ’ 1 π‘œ 𝜏& How to redefine 𝑑&, so that 𝐹 𝑑& = 𝜏&?

XC 2020

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SLIDE 18

XC 2020

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!

Extreme: 𝑔- 𝑦 = 0

!

Maximum: 𝑔-- 𝑦 < 0

!

Minimum: 𝑔-- 𝑦 > 0

XC 2020

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Problem 4: Manipulation

XC 2020

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SLIDE 21

XC 2020

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SLIDE 22

XC 2020

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v

Ca Calvin At Atkeson, , Max ax Te Telemaque

𝑉 = π‘π‘Œ + 𝑐, 𝐹 𝑉 = 𝑏𝐹 π‘Œ + 𝑐, π‘Š 𝑉 = 𝑏&π‘Š(π‘Œ) 𝑉: uniform

  • n

[0, 1], 𝐹 𝑉 = +

&,

π‘Š 𝑉 = +

+&

π‘Œ: uniform

  • n

[𝑑, 𝑒], 𝐹 π‘Œ = .$/

& ,

π‘Š π‘Œ = (/*.)!

+&

XC 2020

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Problem 5: Educational Attainment

XC 2020

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Bin Binomia ial d dist istrib ibutio ion

𝑐 π‘œ, π‘ž, 𝑙 = π‘œ 𝑙 π‘ž!π‘Ÿ"#!

Po Poisson Distribution

𝑄 π‘Œ = 𝑙 = πœ‡! 𝑙! 𝑓#$

Which one to use? !

tw two parameter eters

!

  • n
  • ne

parameter

𝒒 𝒐 𝝁

XC 2020

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SLIDE 26

Bin Binomia ial d dist istrib ibutio ion

𝑐 π‘œ, π‘ž, 𝑙 = π‘œ 𝑙 π‘ž!π‘Ÿ"#!

Po Poisson Distribution

𝑄 π‘Œ = 𝑙 = πœ‡! 𝑙! 𝑓#$

Which one to use? !

𝒐 < +∞

!

𝒐 β†’ +∞

XC 2020

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Bin Binomia ial d dist istrib ibutio ion

𝑐 π‘œ, π‘ž, 𝑙 = π‘œ 𝑙 π‘ž!π‘Ÿ"#!

Po Poisson Distribution

𝑄 π‘Œ = 𝑙 = πœ‡! 𝑙! 𝑓#$

Which one to use? !

𝒐 = πŸ‘, πŸ”, β‹―

!

𝒐 = πŸ”πŸ, 𝟐𝟏𝟏, β‹―

=

𝒐𝒒 = 𝝁

XC 2020

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MATH 20 BABY PROBABILISTS

When you cannot explain something: use Po Poisson di distribu bution (Poisson process)!

XC 2020

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SLIDE 29

XC 2020

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Problem 6: Cupid’s Arrow

XC 2020

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SLIDE 31

XC 2020

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𝑄(𝑍 > π‘Œ) 6

'"# $%

6

2"' $%

𝑒𝑧𝑒𝑦

XC 2020

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SLIDE 33

XC 2020

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𝑄(π‘Œ > 𝑍) 6

'"# $%

6

2"# '

𝑒𝑧𝑒𝑦

XC 2020

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SLIDE 35

𝑄(𝑍 β‰₯ π‘Œ + 300) 6

'"# $%

6

2"'$3## $%

𝑒𝑧𝑒𝑦

XC 2020

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?

Daphne will break free from the enchantment first? 𝑄 π‘Œ > 𝑍 = 2 7 golden arrow: 200 years lead arrow: 500 years

πœ‡. = 1 200 πœ‡/ = 1 500

1 500 1 200 + 1 500 = 2 7 𝑄 π‘Œ > 𝑍 = πœ‡4 πœ‡5 + πœ‡4

XC 2020

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?

Apollo will break free from the enchantment first and Daphne has to wait another 300 years

  • r

more before her arrow wears

  • ff?

𝑄 𝑍 > π‘Œ ∩ 𝑍 β‰₯ π‘Œ + 300 𝑄 𝑍 β‰₯ π‘Œ + 300 = 5 7 𝑓*3/7

XC 2020

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SLIDE 38

𝑄 𝑍 β‰₯ π‘Œ + 300 = 5 7 𝑓*3/7

πœ‡. = 1 200 πœ‡/ = 1 500

𝑄 π‘Œ > 𝑍 = πœ‡4 πœ‡5 + πœ‡4 = 2 7 𝑄 𝑍 > π‘Œ = πœ‡5 πœ‡5 + πœ‡4 = 5 7 𝑄 𝑍 > 𝑧 = 𝑓*8"2 𝑄 𝑍 > 300 = 𝑓*3/7

XC 2020

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𝑄 𝑍 β‰₯ π‘Œ + 300 = 5 7 𝑓*3/7 𝑄 𝑍 > π‘Œ = πœ‡5 πœ‡5 + πœ‡4 = 5 7 𝑄 𝑍 > 300 = 𝑓*3/7 𝑄 𝑍 β‰₯ π‘Œ + 300 = 𝑄 𝑍 > 300 𝑄(𝑍 > π‘Œ) 𝑄 𝑍 > π‘Œ + 300|𝑍 > π‘Œ 𝑄(𝑍 > π‘Œ)

𝑄 π‘Œ > 𝑠 + 𝑑|π‘Œ > 𝑠 = 𝑄 π‘Œ > 𝑑

Memoryless Property

XC 2020

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Problem 7: Man with No Name: A fistful of Nuts

XC 2020

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Conditional Expectation

Β§ If 𝐺 is any event and π‘Œ is a random variable with sample space Ξ© = {𝑦+, 𝑦&, β‹― }, then the conditional expectation given 𝐺 is defined by 𝐹 π‘Œ 𝐺 = βˆ‘9 𝑦9𝑄(π‘Œ = 𝑦9|𝐺). Β§ Let π‘Œ be a random variable with sample space Ξ©. If 𝐺

+,

𝐺&, β‹―, 𝐺

: are

events such that 𝐺! ∩ 𝐺

9 = βˆ… for

𝑗 β‰  π‘˜ and Ξ© =βˆͺ9 𝐺

9,

then 𝐹 π‘Œ = βˆ‘9 𝐹 π‘Œ 𝐺

9 𝑄(𝐺 9).

XC 2020

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Conditional Expectation

Β§ Conditional density 𝑔

.|/ 𝑦 𝑧 = 5$,&(7, 8) 5&(8) .

Β§ Conditional expected value 𝐹 π‘Œ 𝑍 = 𝑧 = ∫ 𝑦𝑔

.|/ 𝑦 𝑧 𝑒𝑦.

Β§ Expected value 𝐹 π‘Œ = ∫ 𝐹 π‘Œ 𝑍 = 𝑧 𝑔

/ 𝑧 𝑒𝑧. marginal density joint density

XC 2020

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EXAMPLE

Farming Sim

XC 2020

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Example

Β§ A point 𝑍 is chosen at random from [0, 1] uniformly. A second point π‘Œ is then uniformly and randomly chosen from the interval [0, 𝑍]. Find the expected value for π‘Œ. 𝑔

5|4 𝑦 𝑧 = 𝑔 5,4(𝑦, 𝑧)

𝑔

4(𝑧)

𝐹 π‘Œ 𝑍 = 𝑧 = 6 𝑦𝑔

5|4 𝑦 𝑧 𝑒𝑦

1

𝑧

XC 2020

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!

sa same me

!

in independent

Ge Geometric Di Distribu bution 𝑄 π‘ˆ = π‘œ = π‘Ÿ,*+π‘ž

tr trial 1

tr trial 2

!

tw two

!

fi first

XC 2020

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SLIDE 46

𝐹 π‘Œ = 1 π‘ž Ge Geometric 𝑄 π‘ˆ = π‘œ = π‘Ÿ,*+π‘ž Ge Geometric Di Distribu bution 𝑄 π‘ˆ = π‘œ = π‘Ÿ,*+π‘ž 𝑄 𝑂 = π‘œ = 𝑦,(1 βˆ’ 𝑦)

XC 2020

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SLIDE 47

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v

…

𝑄 𝑂 = π‘œ π‘Œ = 𝑦 = 𝑦,(1 βˆ’ 𝑦) 𝐹 𝑂 = π‘œ π‘Œ = 𝑦 = 6

# +

𝑦𝑄 𝑂 = π‘œ π‘Œ = 𝑦 𝑒𝑦 𝐹 𝑂 = π‘œ π‘Œ = 𝑦 = ,

,"# $%

π‘œπ‘„ 𝑂 = π‘œ π‘Œ = 𝑦

XC 2020

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SLIDE 49

XC 2020

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Problem 8: Man with No Name: Out of San Pecan

XC 2020

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Expectation of Functions of Random Variables

Β§ If π‘Œ is a real-valued random variable and if 𝜚: 𝑆 β†’ 𝑆 is a continuous real-valued function with domain [𝑏, 𝑐], then 𝐹 𝜚(π‘Œ) = ∫

*% $%𝜚(𝑦)𝑔 𝑦 𝑒𝑦,

provided the integral exists. Discr crete expect cted value 𝑭 𝝔(𝒀) ,

'∈)

𝜚(𝑦)𝑛(𝑦) Con

  • ntinuou
  • us

expect cted value 𝑭 𝝔(𝒀) 6

*% $%

𝜚(𝑦)𝑔 𝑦 𝑒𝑦

XC 2020

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SLIDE 52

XC 2020