MATH 20: PROBABILITY Expected Value of Discrete Random - - PowerPoint PPT Presentation

MATH 20: PROBABILITY

Expected Value

• f

Discrete Random Variables Xingru Chen xingru.chen.gr@dartmouth.edu

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Important Distributions

𝑛 𝜕 = 1 𝑜

Discrete Uniform Distribution

𝑄 𝑈 = 𝑜 = 𝑟!"#𝑞

Geometric Distribution

𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞$𝑟!"$

Binomial Distribution

ℎ 𝑂, 𝑙, 𝑜, 𝑦 =

$ % &"$ !"% & !

Hypergeometric Distribution

𝑣 𝑦, 𝑙, 𝑞 = 𝑦 − 1 𝑙 − 1 𝑞$𝑟%"$

Negative Binomial Distribution

𝑄 𝑌 = 𝑙 = 𝜇$ 𝑙! 𝑓"'

Poisson Distribution

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Average Score of a Pre-employment Psychometric Test

1 2 3 4 5 6 7 8 9 10 10

poor good

Sc Scot

• tt

Su Summers Ra Raven Darkh kholme Ki Kitty Pr Pryde He Henry Ha Hank McCoy Bobby Drake ke Er Eric Leh Lehns nser err Je Jean Gr Grey Em Emma Fr Frost Charles Xa Xavier Ja James Logan Howlett

Av Average

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1 2 3 4 5 6 7 8 9 10 10 Sc Scot

• tt

Su Summers Ra Raven Darkh kholme Ki Kitty Pr Pryde He Henry Ha Hank McC McCoy Bobby Drake ke Er Eric Leh Lehns nser err Je Jean Gr Grey Em Emma Fr Frost Charles Xa Xavier Ja James Logan Ho Howlett

Average 𝜈

5 + 7 + 6 + 2 + 4 + 7 + 9 + 6 + 10 + 2 10 = 5.8 2×2 + 4 + 5 + 2×6 + 2×7 + 9 + 10 10 = 5.8 1 5 ×2 + 1 10 ×4 + 1 5 ×6 + 1 5 ×7 + 1 10 ×9 + 1 10 ×10 = 5.8

Av Average 𝝂 #

!∈#

frequency×value

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Expected Value

§ Let 𝑌 be a numerica cally-val valued discrete random variable with sample space Ω and distribution function 𝑛(𝑦). The expected value 𝐹(𝑌) is defined by 𝐹 𝑌 = ∑!∈$ 𝑦𝑛(𝑦), provided this sum con converges abs bsol

• lutely.

§ The expected value 𝐹(𝑌) is

• ften

referred to as the mean, and can be denoted by 𝜈 for short. § If the above sum does not converge absolutely, then we say 𝑌 does not have an expected value.

!

| #

!∈$

𝑦𝑛 𝑦 | < ∞

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Law of Large Numbers

The Law

• f

Large Numbers can help us justify frequency concept

• f

probability and the interpretation

• f

expected value as the average value to be expected in experiments repeated a large number

• f

times.

Av Average 𝝂 #

!∈#

frequency×value Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦)

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Example 1

Tos

• ss

a coi coin head

• r

tail 𝑛 𝑦 = 1 2 𝐹 𝑌 = ⋯

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Example 2

Rol

• ll

a dice ce 1, 2, 3, 4, 5,

• r

6 𝑛 𝑦 = 1 6 𝐹 𝑌 = ⋯

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Example 3

Suppose that we flip a coin until a head first appears, and if the number

• f

tosses equals 𝑜, then we are paid 𝑜 dollars. What is the expected value

• f

the payment? Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) #

%&' ()

𝑜𝑟%*'𝑞 = #

%&' ()

𝑜 1 2

%*'

(1 2) = #

%&' ()

𝑜 1 2

%

= 2 (1/2 + 1/4 + 1/8 + ⋯ ) + (1/4 + 1/8 + 1/16 + ⋯ ) + (1/8 + 1/16 + ⋯ ) + ⋯ = 1 + 1/2 + 1/4 + 1/8 + ⋯ = 2

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Example 3 (continued)

Suppose that we flip a coin until a head first appears, and if the number

• f

tosses equals 𝑜, then we are paid 2% dollars. What is the expected value

• f

the payment? Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) #

%&' ()

2%𝑟%*'𝑞 = #

%&' ()

2% 1 2

%

= #

+&' ()

1 = +∞

!

| #

!∈$

𝑦𝑛 𝑦 | < ∞

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Example 4

Consider the general Bernoulli trial

• process. As

usual, we let 𝑌 = 1 if the

• utcome

is a success and 0 if it is a failure. Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) 1×𝑞 + 0× 1 − 𝑞 = 𝑞 Ber Bernoulli t tria ial 𝑛 𝑦 = K 𝑞, 𝑌 = 1 1 − 𝑞, 𝑌 = 0

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Bin Binomia ial 𝐹 𝑌 = 𝑜𝑞 𝐹 𝑌 = 1 𝑞 𝐹 𝑌 = 𝜇 𝐹 𝑌 = 𝑙 𝑟 𝑞 Ge Geometric Po Poisson Ne Negative binomi

• mial

𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞,𝑟%*, 𝑄 𝑈 = 𝑜 = 𝑟%*'𝑞 𝑄 𝑌 = 𝑙 = 𝜇, 𝑙! 𝑓*- 𝑣 𝑦, 𝑙, 𝑞 = 𝑦 − 1 𝑙 − 1 𝑞,𝑟!*, 𝐹 𝑌 = 𝑜 𝑙 𝑂 Hy Hyper ergeo eomet etric ic ℎ 𝑂, 𝑙, 𝑜, 𝑦 =

, ! .*, %*! . %

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Binomial Distribution and Poisson Distribution

Po Poisson Distribution 𝑄 𝑌 = 𝑙 = 𝜇$ 𝑙! 𝑓"' Bi Binomial Dist stribution 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞$𝑟!"$

=

𝑞 = -/

%,

𝑢 = 1, 𝑜 → ∞

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Bin Binomia ial 𝐹 𝑌 = 𝑜𝑞 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞,𝑟%*, Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) #

,&0 %

𝑙 𝑜 𝑙 𝑞,𝑟%*, = #

,&' %

𝑙 𝑜 𝑙 𝑞,𝑟%*, = #

,&' %

𝑙 𝑜! 𝑙! 𝑜 − 𝑙 ! 𝑞,𝑟%*, = #

,&' %

𝑜𝑞 (𝑜 − 1)! (𝑙 − 1)! 𝑜 − 𝑙 ! 𝑞,*'𝑟%*, = 𝑜𝑞 #

,&' %

𝑜 − 1 𝑙 − 1 𝑞,*'𝑟%*, = 𝑜𝑞 #

1&0 %*' 𝑜 − 1

𝑚 𝑞1𝑟%*'*1 = 𝑜𝑞

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𝐹 𝑌 = 𝜇 Po Poisson 𝑄 𝑌 = 𝑙 = 𝜇, 𝑙! 𝑓*- Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) #

,&0 ()

𝑙 𝜇, 𝑙! 𝑓*- = #

,&' ()

𝑙 𝜇, 𝑙! 𝑓*- = #

,&' ()

𝜇 𝜇,*' (𝑙 − 1)! 𝑓*- = 𝜇 #

1&0 () 𝜇1

𝑚! 𝑓*- = 𝜇

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Expectation of Functions of Random Variables

§ If 𝑌 is a discrete random variable with sample space Ω and distribution function 𝑛(𝑦), and if 𝜚: Ω → 𝑆 is a function, then 𝐹 𝜚(𝑌) = ∑!∈$ 𝜚(𝑦)𝑛(𝑦), provided the series converges absolutely. Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) Expect cted value 𝑭 𝝔(𝒀) #

!∈$

𝜚(𝑦)𝑛(𝑦)

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Example 1

Tos

• ss

a coi coin head

• r

tail 𝑛 𝑦 = 1 2 𝜚 𝑌 = K1, 𝑌 = Head 0, 𝑌 = Tail 𝐹 𝜚 𝑌 = ⋯

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Fixed Points

§ Since a permutation is a

• ne-to-one

mapping

• f

the set

• nto

itself, it is

• f

interest to ask how many points (elements) are mapped

• nto
• themselves. Such

points are called fi fixed po points of the mapping.

1 2 3 1 2 3 2 1 3 1 2 3 2 3 1 1 2 3

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Example 2

Let 𝑍 be the number

• f

fixed points in a random permutation

• f

the set {𝑏, 𝑐, 𝑑}. To find the expected value

• f

𝑍, it is helpful to consider the basic random variable associated with this experiment, namely the random variable 𝑌 which represents the random permutation.

1 2 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1

Ra Random pe perm rmutation 3! = 6 possible

• utcomes

𝑛 𝑦 = 1 6 𝐹 𝑍 = ⋯

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Example 2

Let 𝑍 be the number

• f

fixed points in a random permutation

• f

the set {𝑏, 𝑐, 𝑑}. To find the expected value

• f

𝑍, it is helpful to consider the basic random variable associated with this experiment, namely the random variable 𝑌 which represents the random permutation.

1 2 3 1 2 3 1 3 2 2 1 3 2 3 1 3 1 2 3 2 1

Ra Random pe perm rmutation 3! = 6 possible

• utcomes

𝑛 𝑦 = 1 6 𝐹 𝑍 = 3 1 6 + 1 1 6 + 1 1 6 + 0 1 6 + 0 1 6 + 1 1 6 = 1

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The Sum of Two Random Variables

§ Let 𝑌 and 𝑍 be random variables with finite expected

• values. Then

𝐹(𝑌 + 𝑍) = 𝐹(𝑌) + 𝐹(𝑍), and if 𝑑 is any constant, then 𝐹(𝑑𝑌) = 𝑑𝐹(𝑌). § It can be shown that the expected value

• f

the sum

• f

any finite number

• f

random variables is the sum

• f

the expected values

• f

the individual random variables. 𝐹 𝑌' + 𝑌2 + ⋯ + 𝑌% = 𝐹 𝑌' + 𝐹 𝑌2 + ⋯ + 𝐹(𝑌%).

!

Mu Mutual independence ce

• f
• f

the summands is not

• t

ne needed.

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Linearity

𝐹(𝑌 + 𝑍) = 𝐹(𝑌) + 𝐹(𝑍) 𝐹(𝑑𝑌) = 𝑑𝐹(𝑌). 𝐹(𝑏𝑌 + 𝑐) = 𝑏𝐹(𝑌) + 𝑐

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Let the sample spaces

• f

𝑌 and 𝑍 be denoted by Ω( and Ω), and suppose that Consider the random variable 𝑌 + 𝑍 to be the result

• f

applying the function 𝜚 𝑦, 𝑧 = 𝑦 + 𝑧 to the joint random variable (𝑌, 𝑍). Then according to the Expectation

• f

Functions

• f

Random Variables, we have Ω( = 𝑦#, 𝑦*, ⋯ and Ω) = 𝑧#, 𝑧*, ⋯ .

.

!

𝑄(𝑌 = 𝑦", 𝑍 = 𝑧!) = 𝑄(𝑌 = 𝑦")

𝐹 𝑌 + 𝑍 = ?

+

?

$

𝑦+ + 𝑧$ 𝑄(𝑌 = 𝑦+, 𝑍 = 𝑧$) = ?

+

?

$

𝑦+𝑄(𝑌 = 𝑦+, 𝑍 = 𝑧$) + ?

+

?

$

𝑧$𝑄(𝑌 = 𝑦+, 𝑍 = 𝑧$) = ?

+

𝑦+𝑄(𝑌 = 𝑦+) + ?

$

𝑧$𝑄(𝑍 = 𝑧$) = 𝐹 𝑌 + 𝐹(𝑍).

Proof

.

"

𝑄(𝑌 = 𝑦", 𝑍 = 𝑧!) = 𝑄(𝑍 = 𝑧!)

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Bin Binomia ial 𝐹 𝑌 = 𝑜𝑞 𝑐 𝑜, 𝑞, 𝑙 = 𝑜 𝑙 𝑞,𝑟%*, Expect cted value 𝑭(𝒀) #

!∈$

𝑦𝑛(𝑦) A single Bernoulli trial 𝑌+ = K1, success 0, failure 𝐹 𝑌+ = 1×𝑞 + 0× 1 − 𝑞 = 𝑞 𝑜 Bernoulli trials 𝐹 𝑌 = 𝐹 𝑌' + 𝑌2 + ⋯ + 𝑌% = 𝑜𝑞

1 2 3 4 5 ✗ ✓ ✗ ✗ ✓ 2 ✓

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Number of fixed points

We now give a very quick way to calculate the average number

• f

fixed points in a random permutation

• f

the set {1, 2, 3, . . . , 𝑜}.

3 5 4 2 1 1 2 3 4 5

Let 𝑎 denote the random permutation. For each 𝑗, 1 ≤ 𝑗 ≤ 𝑜, let 𝑌+ equal 1 if 𝑎 fixes 𝑗, and 0 otherwise. 𝑌+ = K 1, 𝑎 mixes 𝑗 0,

• therwise

let 𝐺 denote the number

• f

fixed points in 𝑎.

?

𝐺 = ⋯

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Number of fixed points

3 5 4 2 1 1 2 3 4 5

Let 𝑎 denote the random permutation. For each 𝑗, 1 ≤ 𝑗 ≤ 𝑜, let 𝑌+ equal 1 if 𝑎 fixes 𝑗, and 0 otherwise. 𝑌+ = K 1, 𝑎 mixes 𝑗 0,

• therwise

let 𝐺 denote the number

• f

fixed points in 𝑎.

!

𝐺 = 𝑌@ + 𝑌A + ⋯ + 𝑌B

!

𝐹(𝐺) = 𝐹(𝑌') + 𝐹(𝑌2) + ⋯ + 𝐹(𝑌%)

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Number of fixed points

3 5 4 2 1 1 2 3 4 5

Let 𝑎 denote the random permutation. For each 𝑗, 1 ≤ 𝑗 ≤ 𝑜, let 𝑌+ equal 1 if 𝑎 fixes 𝑗, and 0 otherwise. 𝑌+ = K 1, 𝑎 mixes 𝑗 0,

• therwise

let 𝐺 denote the number

• f

fixed points in 𝑎.

!

𝐺 = 𝑌@ + 𝑌A + ⋯ + 𝑌B

!

𝐹(𝐺) = 𝐹(𝑌') + 𝐹(𝑌2) + ⋯ + 𝐹(𝑌%)

𝐹 𝑌C = 1 𝑜

!

𝐹 𝐺 = 𝑜× 1 𝑜 = 1

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The Product of Two Random Variables

§ Let 𝑌 and 𝑍 be random variables with finite expected

• values. If

𝑌 and 𝑍 are independent random variables, then 𝐹(𝑌𝑍) = 𝐹(𝑌)𝐹(𝑍).

!

Mu Mutual independence ce

• f
• f

the summands is ne needed.

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Let the sample spaces

• f

𝑌 and 𝑍 be denoted by Ω( and Ω), and suppose that Consider the random variable 𝑌𝑍 to be the result

• f

applying the function 𝜚 𝑦, 𝑧 = 𝑦𝑧 to the joint random variable (𝑌, 𝑍). Then according to the Expectation

• f

Functions

• f

Random Variables, we have Ω( = 𝑦#, 𝑦*, ⋯ and Ω) = 𝑧#, 𝑧*, ⋯ .

𝑄 𝑌 = 𝑦3, 𝑍 = 𝑧, = (𝑌 = 𝑦3)𝑄(𝑍 = 𝑧,)

𝐹 𝑌𝑍 = ?

+

?

$

𝑦+𝑧$ 𝑄(𝑌 = 𝑦+, 𝑍 = 𝑧$) = ?

+

?

$

𝑦+𝑧$ 𝑄(𝑌 = 𝑦+)𝑄(𝑍 = 𝑧$) = (?

+

𝑦+𝑄(𝑌 = 𝑦+))(?

$

𝑧$𝑄 𝑍 = 𝑧$ ) = 𝐹 𝑌 𝐹(𝑍).

Proof

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Open book Scope: Chapter 4, 5, 6 § conditional probability § distributions and densities § expected value and variance Materials: Slides, homework, quizzes, textbook Date & Time: August 10, 3 hours, 24 hours Office hours: August 10, August 11 Homework due: 11:00 pm August 14

Au August 2020

Mid Midter erm 2

26 26 27 27 28 28 29 29 30 30 31 31 01 01 02 02 03 03 04 04 05 05 06 06 07 07 08 08 09 09 10 10 11 11 12 12 13 13 14 14 15 15 16 16 17 17 18 18 19 19 20 20 21 21 22 22 23 23 24 24 25 25 26 26 28 28 27 27 30 30 05 05 31 31 01 01 02 02 04 04 03 03 29 29

Sun Mon Tue Wed Thu Fri Sat

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Conditional Expectation

§ If 𝐺 is any event and 𝑌 is a random variable with sample space Ω = {𝑦', 𝑦2, ⋯ }, then the conditional expectation given 𝐺 is defined by 𝐹 𝑌 𝐺 = ∑3 𝑦3𝑄(𝑌 = 𝑦3|𝐺). § Let 𝑌 be a random variable with sample space Ω. If 𝐺

',

𝐺2, ⋯, 𝐺

4 are

events such that 𝐺+ ∩ 𝐺

3 = ∅ for

𝑗 ≠ 𝑘 and Ω =∪3 𝐺

3,

then 𝐹 𝑌 = ∑3 𝐹 𝑌 𝐺

3 𝑄(𝐺 3).

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We have

𝐹 𝑌 𝐺 = #

3

𝑦3𝑄(𝑌 = 𝑦3|𝐺) #

3

𝑄(𝑌 = 𝑦, ∩ 𝐺

3)

= 𝑄(𝑌 = 𝑦,)

?

+

𝐹 𝑌 𝐺

+ 𝑄(𝐺 +) = ? +

?

$

𝑦$𝑄 𝑌 = 𝑦$ 𝐺

+ 𝑄(𝐺 +)

= ?

+

?

$

𝑦$𝑄(𝑌 = 𝑦$ ∩ 𝐺

+)

= ?

$

𝑦$ ?

+

𝑄(𝑌 = 𝑦$ ∩ 𝐺

+)

= ?

$

𝑦$𝑄(𝑌 = 𝑦$) = 𝐹(𝑌)

Proof

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Why We Need Conditional Expectation?

𝑭 𝒀

𝑭 𝒀 𝑮𝟐 𝑸(𝑮𝟐) 𝑭 𝒀 𝑮𝟕 𝑸(𝑮𝟕) 𝑭 𝒀 𝑮𝟔 𝑸(𝑮𝟔) 𝑭 𝒀 𝑮𝟑 𝑸(𝑮𝟑) 𝑭 𝒀 𝑮𝟓 𝑸(𝑮𝟓) 𝑭 𝒀 𝑮𝟒 𝑸(𝑮𝟒)

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EXAMPLE

Farming Sim

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Ra Rain Dr Droughts Sn Snow Spring 100% Summer 20% 80% Fall 100% Winter 20% 80%

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Spec Special wea eather er Rain Droughts Snow Spring 100% Summer 20% 80% Fall 100% Winter 20% 80% Al All weath ther Normal Rain Droughts Snow Spr Spring 75% 25% Su Summer er 75% 5% 20% Fa Fall 75% 25% Wi Wint nter 75% 5% 20%

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Al All weath ther (a season) Normal Rain Droughts Snow Spr Spring 3/4 1/4 Su Summer er 3/4 1/20 1/5 Fa Fall 3/4 1/4 Wi Wint nter 3/4 1/20 1/5 Al All weath ther (a year) r) Normal Rain Droughts Snow Spr Spring 3/16 1/16 Su Summer er 3/16 1/80 1/20 Fa Fall 3/16 1/16 Wi Wint nter 3/16 1/80 1/5

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No Norm rmal Ra Rain Dr Droughts Sn Snow income 1000 500 200 100 Al All weath ther (a year) r) Normal Rain Droughts Snow Spr Spring 3/16 1/16 Su Summer er 3/16 1/80 1/20 Fa Fall 3/16 1/16 Wi Wint nter 3/16 1/80 1/20

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No Norm rmal Ra Rain Dr Droughts Sn Snow income 1000 500 200 100 No Norm rmal Ra Rain Dr Droughts Sn Snow To Total Spr Spring 3/16 1/16 1/4 Su Summer er 3/16 1/80 1/20 1/4 Fa Fall 3/16 1/16 1/4 Wi Wint nter 3/16 1/80 1/5 1/4 To Total 3/4 3/20 1/20 1/20 1/4

§ 𝑌: income § 𝐺

',

𝐺2, 𝐺;, 𝐺

<:

seasons spring, summer, fall, winter 𝐹 𝑌|𝐺+ = ⋯ 𝐹 𝑌 = ⋯

!

𝐹 𝑌 𝐺 = ∑3 𝑦3𝑄(𝑌 = 𝑦3|𝐺).

!

𝐹 𝑌 = ∑3 𝐹 𝑌 𝐺

3 𝑄(𝐺 3).

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No Normal Ra Rain in Dr Droughts Sn Snow income 1000 500 200 100

No Norm rmal Ra Rain Dr Droughts Sn Snow To Total Spr Spring 3/16 1/16 1/4 Su Summer er 3/16 1/80 1/20 1/4 Fa Fall 3/16 1/16 1/4 Wi Wint nter 3/16 1/80 1/5 1/4 To Total 3/4 3/20 1/20 1/20 1/4

§ 𝑌: income § 𝐺

',

𝐺2, 𝐺;, 𝐺

<:

seasons spring, summer, fall, winter

𝐹 𝑌|spring = 1000× 3 4 + 500× 1 4 = 3500 4 𝐹 𝑌|summer = 1000× 3 4 + 500× 1 20 + 200× 1 5 = 16300 20 𝐹 𝑌|fall = 1000× 3 4 + 500× 1 4 = 3500 4 𝐹 𝑌|winter = 1000× 3 4 + 500× 1 20 + 100× 1 5 = 15900 20 𝐹 𝑌 = 𝐹 𝑌|spring 𝑄 spring + 𝐹 𝑌|summer 𝑄 summer + 𝐹 𝑌|fall 𝑄 fall + 𝐹 𝑌|winter 𝑄 winter = 3500 4 + 16300 20 + 3500 4 + 15900 20 × 1 4 = 840

!

𝐹 𝑌 𝐺 = ∑3 𝑦3𝑄(𝑌 = 𝑦3|𝐺).

!

𝐹 𝑌 = ∑3 𝐹 𝑌 𝐺

3 𝑄(𝐺 3).

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No Norm rmal Ra Rain Dr Droughts Sn Snow income 1000 500 200 100 No Norm rmal Ra Rain Dr Droughts Sn Snow To Total Spr Spring 3/16 1/16 1/4 Su Summer er 3/16 1/80 1/20 1/4 Fa Fall 3/16 1/16 1/4 Wi Wint nter 3/16 1/80 1/5 1/4 To Total 3/4 3/20 1/20 1/20 1/4

§ 𝑌: income § 𝐺

',

𝐺2, 𝐺;, 𝐺

<:

weather normal, rain, droughts, snow 𝐹 𝑌|𝐺+ = ⋯ 𝐹 𝑌 = ⋯

!

𝐹 𝑌 𝐺 = ∑3 𝑦3𝑄(𝑌 = 𝑦3|𝐺).

!

𝐹 𝑌 = ∑3 𝐹 𝑌 𝐺

3 𝑄(𝐺 3).

XC 2020

No Normal Ra Rain in Dr Droughts Sn Snow income 1000 500 200 100

No Norm rmal Ra Rain Dr Droughts Sn Snow To Total Spr Spring 3/16 1/16 1/4 Su Summer er 3/16 1/80 1/20 1/4 Fa Fall 3/16 1/16 1/4 Wi Wint nter 3/16 1/80 1/5 1/4 To Total 3/4 3/20 1/20 1/20 1/4

§ 𝑌: income § 𝐺

',

𝐺2, 𝐺;, 𝐺

<:

weather normal, rain, droughts, snow

𝐹 𝑌|normal = 1000, 𝐹 𝑌|rain = 500, 𝐹 𝑌|droughts = 200, 𝐹 𝑌|snow = 100 𝑄 normal =

, #- ×4 = , .,

𝑄 rain =

, */

𝑄 droughts =

# */,

𝑄 snow =

# */

𝐹 𝑌 = 𝐹 𝑌|normal 𝑄 normal + 𝐹 𝑌|rain 𝑄 rain + 𝐹 𝑌|droughts 𝑄 droughts + 𝐹 𝑌|snow 𝑄 snow = 1000× 3 4 + 500× 3 20 + 200× 1 20 + 100× 1 20 = 840

!

𝐹 𝑌 𝐺 = ∑3 𝑦3𝑄(𝑌 = 𝑦3|𝐺).

!

𝐹 𝑌 = ∑3 𝐹 𝑌 𝐺

3 𝑄(𝐺 3).

XC 2020

MARTINGALES

Fairness to a Player

XC 2020

Martingales

§ Let 𝑇', 𝑇2, ⋯ , 𝑇% be the sequential

• utcomes
• f

a repeated experiment (such as fair games), we have 𝑭 𝑻𝒐 𝑻𝒐*𝟐 = 𝒃, ⋯ , 𝑻𝟑 = 𝒖, 𝑻𝟐 = 𝒔 = 𝑻𝒐*𝟐. Samuelson suggested in 1965 that the stock prices follow a martingale: Proof

• of

That Prop

• perly

Antici cipated Price ces Fluct ctuate Ra Randomly ly.

Mathematical Finance

T T

“Given all I know today, expected price tomorrow is the price today.”

XC 2020

Martingales in Finance

Many asset prices are believed to behave approximately like martingales, at least in the short term.

!

𝐹 𝑇% 𝑇%*' = 𝑏, ⋯ , 𝑇2 = 𝑢, 𝑇' = 𝑠 = 𝑇%*'.

XC 2020

Martingales in Finance

§ New information is instantly absorbed into the stock value, so expected value

• f

the stock tomorrow should be the value today. If it were higher, statistical arbitrageurs would bid up today’s price until this was not the case. § But there are some caveats: interest, risk premium, etc.

Effici cient marke ket hypot

• thesis

!

𝐹 𝑇! 𝑇!"# = 𝑏, ⋯ , 𝑇* = 𝑢, 𝑇# = 𝑠 = 𝑇!"#.

XC 2020

Martingales in Finance

§ According to the fu fundamental theorem

• f
• f

asset prici cing, the discounted price

>(%) A(%),

where 𝐵 is a risk-free asset, is a martingale with respected to ri risk ne neutr tral probability ty. Ri Risk neutral pr probabili lity

!

𝐹 𝑇! 𝑇!"# = 𝑏, ⋯ , 𝑇* = 𝑢, 𝑇# = 𝑠 = 𝑇!"#.

XC 2020