06: Random Variables - Stanford University...The probability of the event that a random variable takes on the value 𝑘! For discrete random variables, this is a probability mass

06: Random VariablesLisa Yan

April 17, 2020

1

Random Variables

2

06b_random_variables

Lisa Yan, CS109, 2020

Conditional independence review

3

Next Episode Playing in 5 seconds


Random variables are like typed

variables (with uncertainty)

int a = 5;

double b = 4.2;

bit c = 1;

𝐴 is the number of Pokemon we bring to our future battle.

𝐴 ∈ 1, 2, … , 6

𝐵 is the amount of money we get after we win a battle.

𝐵 ∈ ℝ+

𝐶 is 1 if we successfully beat the Elite Four. 0 otherwise.

𝐶 ∈ {0,1}

4

Random variables are like typed variables

name

Random

variablesCS variables


Random Variable

A random variable is a real-valued function defined on a sample space.

5

Experiment 𝑋 = 𝑘

2. What is the event (set of outcomes) where 𝑋 = 2?

Outcome

Example:

3 coins are flipped.

Let 𝑋 = # of heads.

𝑋 is a random variable.

1. What is the value of 𝑋 for the outcomes:

• (T,T,T)?

• (H,H,T)?

3. What is 𝑃 𝑋 = 2 ? 🤔


Random Variable

A random variable is a real-valued function defined on a sample space.

6

Experiment 𝑋 = 𝑘

2. What is the event (set of outcomes) where 𝑋 = 2?

Outcome

1. What is the value of 𝑋 for the outcomes:

• (T,T,T)?

• (H,H,T)?

3. What is 𝑃 𝑋 = 2 ?

Example:





Random variables are NOT events!

It is confusing that random variables and events use the same notation.

• Random variables ≠ events.

• We can define an event to be a particular assignmentof a random variable.

7

event

𝑋 = 2probability

(number b/t 0 and 1)

𝑃(𝑋 = 2)

Example:





Random variables are NOT events!

It is confusing that random variables and events use the same notation.

• Random variables ≠ events.

• We can define an event to be a particular assignmentof a random variable.

8

Example:




𝑋 = 𝟏 {(H, T, T), (T, H, T),

(T, T, H)}

3/8

𝑋 = 𝟐 {(H, H, T), (H, T, H),

(T, H, H)}

3/8

𝑋 = 𝟑 {(H, H, H)} 1/8

𝑋 ≥ 4 { } 0

𝑋 = 𝑥 Set of outcomes 𝑃 𝑋 = 𝑘

𝑋 = 𝟎 {(T, T, T)} 1/8

PMF/CDF

9

06c_pmf_cdf


So far

3 coins are flipped. Let 𝑋 = # of heads. 𝑋 is a random variable.

10

Experiment 𝑋 = ___Outcome

(flip __ heads)𝑃 𝑋 = ___

Can we get a “shorthand” for

this last step?

Seems like it might be useful!


Probability Mass Function


11

𝑃(𝑋 = 𝑘)return value/output

number between

0 and 1

parameter/input 𝑘

A function on 𝑘with range [0,1]

What would be a useful function to define?The probability of the event that a random variable 𝑋 takes on the value 𝑘!

For discrete random variables, this is a probability mass function.


Probability Mass Function


12

𝑃(𝑋 = 𝑘)

2

0.375return value/output:

probability of the event

𝑋 = 𝑘

parameter/input 𝑘:

a value of 𝑋

N = 3P = 0.5

def prob_event_y_equals(k):n_ways = scipy.special.binom(N, k)p_way = np.power(P, k) * np.power(1 - P, N-k)return n_ways * p_way

A function on 𝑘with range [0,1]

probability mass function


A random variable 𝑋 is discrete if it can take on countably many values.

• 𝑋 = 𝑥, where 𝑥 ∈ 𝑥1, 𝑥2, 𝑥3, …

The probability mass function (PMF) of a discrete random variable is

𝑃 𝑋 = 𝑥

• Probabilities must sum to 1:

This last point is a

good way to verify

any PMF you create.

Discrete RVs and Probability Mass Functions

13

𝑖=1

∞

𝑝 𝑥𝑖 = 1

shorthand notation

= 𝑝 𝑥 = 𝑝𝑋(𝑥)


PMF for a single 6-sided die

Let 𝑋 be a random variable that represents the result of a singledice roll.

• Support of 𝑋 : 1, 2, 3, 4, 5, 6

• Therefore 𝑋 is a discreterandom variable.

• PMF of X:

𝑝 𝑥 = ቊ1/6 𝑥 ∈ {1, … , 6}0 otherwise

14

1/6

1 2 3 4 5 60

𝑋 = 𝑥

𝑃𝑋=𝑥


Cumulative Distribution Functions

For a random variable 𝑋, the cumulative distribution function (CDF) is defined as

𝐹 𝑎 = 𝐹𝑋 𝑎 = 𝑃 𝑋 ≤ 𝑎 ,where −∞ < 𝑎 < ∞

For a discrete RV 𝑋, the CDF is:

𝐹 𝑎 = 𝑃 𝑋 ≤ 𝑎 =

all 𝑥≤𝑎

𝑝(𝑥)

15


Let 𝑋 be a random variable that represents the result of a single dice roll.

16

1/6

1 2 3 4 5 60

𝑋 = 𝑥

𝑃𝑋=𝑥

CDFs as graphsCDF (cumulative

distribution function)𝐹 𝑎 = 𝑃 𝑋 ≤ 𝑎

𝐹𝑎

𝑋 = 𝑥

5/6

4/6

3/6

2/6

1/6

PMF of 𝑋

CDF of 𝑋

𝑃 𝑋 ≤ 0 = 0

𝑃 𝑋 ≤ 6 = 1

Expectation

17

06d_expectation


Discrete random variables

18

Discrete

Random

Variable, 𝑋

Experiment

outcomes

PMF

𝑃 𝑋 = 𝑥 = 𝑝(𝑥)

Definition

Properties

Support

CDF 𝐹 𝑥

Without performing the experiment:

• The support gives us a ballpark of what values our RV will take on

• Next up: How do we report the “average” value?


Expectation

The expectation of a discrete random variable 𝑋 is defined as:

𝐸 𝑋 =

𝑥:𝑝 𝑥 >0

𝑝 𝑥 ⋅ 𝑥

• Note: sum over all values of 𝑋 = 𝑥 that have non-zero probability.

• Other names: mean, expected value, weighted average,center of mass, first moment

19


Expectation of a die roll

What is the expected value of a 6-sided die roll?

20

Expectation

of 𝑋𝐸 𝑋 =

𝑥:𝑝 𝑥 >0

𝑝 𝑥 ⋅ 𝑥

1. Define random variables

2. Solve

𝑃 𝑋 = 𝑥 = ቊ1/6 𝑥 ∈ {1, … , 6}0 otherwise

𝑋 = RV for value of roll

𝐸 𝑋 = 11

6+ 2

1

6+ 3

1

6+ 4

1

6+ 5

1

6+ 6

1

6=7

2


Important properties of expectation

1. Linearity:

𝐸 𝑎𝑋 + 𝑏 = 𝑎𝐸 𝑋 + 𝑏

2. Expectation of a sum = sum of expectation:

𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌

3. Unconscious statistician:

𝐸 𝑔 𝑋 =

𝑥

𝑔 𝑥 𝑝(𝑥)

21

• Let 𝑋 = 6-sided dice roll,

𝑌 = 2𝑋 − 1.

• 𝐸 𝑋 = 3.5• 𝐸 𝑌 = 6

Sum of two dice rolls:

• Let 𝑋 = roll of die 1

𝑌 = roll of die 2

• 𝐸 𝑋 + 𝑌 = 3.5 + 3.5 = 7

These properties let you avoid

defining difficult PMFs.

Proofs (OK to stop here)

22



1. Linearity:



𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌


𝐸 𝑔 𝑋 =

𝑥


23

Review


Linearity of Expectation proof


Proof:

𝐸 𝑎𝑋 + 𝑏 =

𝑥

𝑎𝑥 + 𝑏 𝑝 𝑥 =

𝑥

𝑎𝑥𝑝 𝑥 + 𝑏𝑝 𝑥

= 𝑎

𝑥

𝑥𝑝(𝑥) + 𝑏

𝑥

𝑝 𝑥

= 𝑎 𝐸 𝑋 + 𝑏 ⋅ 1

24

𝐸 𝑋 =

𝑥:𝑝 𝑥 >0

𝑝 𝑥 ⋅ 𝑥


Expectation of Sum intuition

𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌

25

(we’ll prove this

in two weeks)

𝑋 𝑌 𝑋 + 𝑌

3 6 9

2 4 6

6 12 18

10 20 30

-1 -2 -3

0 0 0

8 16 24

1

7(28)

Average:1

𝑛

𝑖=1

𝑛

𝑥𝑖1

𝑛

𝑖=1

𝑛

𝑦𝑖1

𝑛

𝑖=1

𝑛

𝑥𝑖 + 𝑦𝑖

Intuition

for now:

+ =

+ =1

7(56)

1

7(84)

𝐸 𝑋 =

𝑥:𝑝 𝑥 >0

𝑝 𝑥 ⋅ 𝑥


LOTUS proof

26

Let 𝑌 = 𝑔(𝑋), where 𝑔 is a real-valued function.

𝐸 𝑔 𝑋 = 𝐸 𝑌 =

𝑗

𝑦𝑗𝑝(𝑦𝑗)

=

𝑗

𝑦𝑗

𝑖:𝑔 𝑥𝑖 = 𝑦𝑗

𝑝(𝑥𝑖)

=

𝑗


𝑦𝑗 𝑝(𝑥𝑖)

=

𝑗


𝑔(𝑥𝑖) 𝑝(𝑥𝑖)

=

𝑖

𝑔(𝑥𝑖) 𝑝(𝑥𝑖)

Expectation

of 𝑔 𝑋𝐸 𝑔 𝑋 =

𝑥


For you to review

so that you can

sleep at night

Variance

27

07a_variance_i


Stanford, CA

𝐸 high = 68°F

𝐸 low = 52°F

Washington, DC

𝐸 high = 67°F

𝐸 low = 51°F

28

Average annual weather

Is 𝐸 𝑋 enough?


Stanford, CA

𝐸 high = 68°F

𝐸 low = 52°F

Washington, DC

𝐸 high = 67°F

𝐸 low = 51°F

29

Average annual weather

0

0.1

0.2

0.3

0.4

0

0.1

0.2

0.3

0.4

𝑃𝑋=𝑥

𝑃𝑋=𝑥

Stanford high temps Washington high temps

35 50 65 80 90 35 50 65 80 90

68°F 67°F

Normalized histograms are approximations of PMFs.


Variance = “spread”

Consider the following three distributions (PMFs):

• Expectation: 𝐸 𝑋 = 3 for all distributions

• But the “spread” in the distributions is different!

• Variance, Var 𝑋 : a formal quantification of “spread”

30


Variance

The variance of a random variable 𝑋 with mean 𝐸 𝑋 = 𝜇 is

Var 𝑋 = 𝐸 𝑋 − 𝜇 2

• Also written as: 𝐸 𝑋 − 𝐸 𝑋 2

• Note: Var(X) ≥ 0

• Other names: 2nd central moment, or square of the standard deviation

Var 𝑋

def standard deviation SD 𝑋 = Var 𝑋

31

Units of 𝑋2

Units of 𝑋


Variance of Stanford weather

32

Variance

of 𝑋Var 𝑋 = 𝐸 𝑋 − 𝐸 𝑋 2

Stanford, CA

𝐸 high = 68°F

𝐸 low = 52°F

0

0.1

0.2

0.3

0.4

𝑃𝑋=𝑥

Stanford high temps

35 50 65 80 90

𝑋 𝑋 − 𝜇 2

57°F 124 (°F)2

𝐸 𝑋 = 𝜇 = 68

71°F 9 (°F)2

75°F 49 (°F)2

69°F 1 (°F)2

… …

Variance 𝐸 𝑋 − 𝜇 2 = 39 (°F)2

Standard deviation = 6.2°F


Stanford, CA

𝐸 high = 68°F

Washington, DC

𝐸 high = 67°F

33

Comparing variance

0

0.1

0.2

0.3

0.4

0

0.1

0.2

0.3

0.4

𝑃𝑋=𝑥

𝑃𝑋=𝑥

Stanford high temps Washington high temps

35 50 65 80 90 35 50 65 80 90

Var 𝑋 = 39 °F 2 Var 𝑋 = 248 °F 2

Variance


68°F 67°F

Properties of Variance

34

07b_variance_ii


Properties of variance

Definition Var 𝑋 = 𝐸 𝑋 − 𝐸 𝑋 2


Property 1 Var 𝑋 = 𝐸 𝑋2 − 𝐸 𝑋 2

Property 2 Var 𝑎𝑋 + 𝑏 = 𝑎2Var 𝑋

35

Units of 𝑋2

Units of 𝑋

• Property 1 is often easier to compute than the definition

• Unlike expectation, variance is not linear







36

Units of 𝑋2

Units of 𝑋




Computing variance, a proof

37

Var 𝑋 = 𝐸 𝑋 − 𝐸 𝑋 2

= 𝐸 𝑋2 − 𝐸 𝑋 2

= 𝐸 𝑋2 − 𝜇2= 𝐸 𝑋2 − 2𝜇2 + 𝜇2= 𝐸 𝑋2 − 2𝜇𝐸 𝑋 + 𝜇2

= 𝐸 𝑋 − 𝜇 2Let 𝐸 𝑋 = 𝜇

Everyone,

please

welcome the

second

moment!

Variance


= 𝐸 𝑋2 − 𝐸 𝑋 2

=

𝑥

𝑥 − 𝜇 2𝑝 𝑥

=

𝑥

𝑥2 − 2𝜇𝑥 + 𝜇2 𝑝 𝑥

=

𝑥

𝑥2𝑝 𝑥 − 2𝜇

𝑥

𝑥𝑝 𝑥 + 𝜇2

𝑥

𝑝 𝑥

⋅ 1


Let Y = outcome of a single die roll. Recall 𝐸 𝑌 = 7/2 .

Calculate the variance of Y.

38

Variance of a 6-sided dieVariance


= 𝐸 𝑋2 − 𝐸 𝑋 2

1. Approach #1: Definition

Var 𝑌 =1

61 −

7

2

2

+1

62 −

7

2

2

+1

63 −

7

2

2

+1

64 −

7

2

2

+1

65 −

7

2

2

+1

66 −

7

2

2

2. Approach #2: A property

𝐸 𝑌2 =1

612 + 22 + 32 + 42 + 52 + 62

= 91/6

Var 𝑌 = 91/6 − 7/2 2

= 35/12 = 35/12







39

Units of 𝑋2

Units of 𝑋




Property 2: A proof


40

Var 𝑎𝑋 + 𝑏

= 𝐸 𝑎𝑋 + 𝑏 2 − 𝐸 𝑎𝑋 + 𝑏 2 Property 1

= 𝐸 𝑎2𝑋2 + 2𝑎𝑏𝑋 + 𝑏2 − 𝑎𝐸 𝑋 + 𝑏 2

= 𝑎2𝐸 𝑋2 + 2𝑎𝑏𝐸 𝑋 + 𝑏2 − 𝑎2 𝐸[𝑋] 2 + 2𝑎𝑏𝐸[𝑋] + 𝑏2

= 𝑎2𝐸 𝑋2 − 𝑎2 𝐸[𝑋] 2

= 𝑎2 𝐸 𝑋2 − 𝐸[𝑋] 2

= 𝑎2Var 𝑋 Property 1

Proof:

Factoring/

Linearity of

Expectation

Bernoulli RV

41

07c_bernoulli


Jacob Bernoulli

42

Jacob Bernoulli (1654-1705), also known as “James”, was a Swiss mathematician

One of many mathematicians in Bernoulli family

The Bernoulli Random Variable is named for him

My academic great14 grandfather


Consider an experiment with two outcomes: “success” and “failure.”

def A Bernoulli random variable 𝑋 maps “success” to 1 and “failure” to 0.

Other names: indicator random variable, boolean random variable

Examples:• Coin flip

• Random binary digit

• Whether a disk drive crashed

Bernoulli Random Variable

43

𝑃 𝑋 = 1 = 𝑝 1 = 𝑝𝑃 𝑋 = 0 = 𝑝 0 = 1 − 𝑝𝑋~Ber(𝑝)

Support: {0,1} Variance

Expectation

PMF

𝐸 𝑋 = 𝑝

Var 𝑋 = 𝑝(1 − 𝑝)

Remember this nice property of

expectation. It will come back!


Defining Bernoulli RVs

44

Serve an ad.• User clicks w.p. 0.2• Ignores otherwise

Let 𝑋: 1 if clicked

𝑋~Ber(___)

𝑃 𝑋 = 1 = ___

𝑃 𝑋 = 0 = ___

Roll two dice.• Success: roll two 6’s

• Failure: anything else

Let 𝑋 : 1 if success

𝑋~Ber(___)

𝐸 𝑋 = ___

𝑋~Ber(𝑝) 𝑝𝑋 1 = 𝑝𝑝𝑋 0 = 1 − 𝑝𝐸 𝑋 = 𝑝

Run a program• Crashes w.p. 𝑝• Works w.p. 1 − 𝑝

Let 𝑋: 1 if crash

𝑋~Ber(𝑝)

𝑃 𝑋 = 1 = 𝑝

𝑃 𝑋 = 0 = 1 − 𝑝 🤔


Defining Bernoulli RVs

45

Serve an ad.• User clicks w.p. 0.2• Ignores otherwise

Let 𝑋: 1 if clicked

𝑋~Ber(___)

𝑃 𝑋 = 1 = ___

𝑃 𝑋 = 0 = ___

Roll two dice.• Success: roll two 6’s

• Failure: anything else

Let 𝑋 : 1 if success

𝑋~Ber(___)

𝐸 𝑋 = ___

𝑋~Ber(𝑝) 𝑝𝑋 1 = 𝑝𝑝𝑋 0 = 1 − 𝑝𝐸 𝑋 = 𝑝

Run a program• Crashes w.p. 𝑝• Works w.p. 1 − 𝑝

Let 𝑋: 1 if crash

𝑋~Ber(𝑝)

𝑃 𝑋 = 1 = 𝑝

𝑃 𝑋 = 0 = 1 − 𝑝

(live)06: Random Variables ISlides by Lisa Yan

July 6, 2020

46

Today’s discussion thread: https://us.edstem.org/courses/667/discussion/84211

https://us.edstem.org/courses/667/discussion/84211


Discrete random variables

47

Discrete

Random

Variable, 𝑋

Experiment

outcomes𝑃 𝑋 = 𝑥 = 𝑝(𝑥)

𝐸 𝑋

Definition

Properties

Review

Note: Random Variables

also called distributions

Support


Event-driven probability

• Relate only binary events◦ Either happens (𝐸)

◦ or doesn’t happen (𝐸𝐶)

• Can only report probability

• Lots of combinatorics

Random Variables

• Link multiple similar events together (𝑋 = 1, 𝑋 = 2,… , 𝑋 = 6)

• Can compute statistics: report the “average” outcome

• Once we have the PMF (discrete RVs), we can do regular math

48

A Whole New World with Random Variables


PMF for the sum of two dice

Let 𝑌 be a random variable that represents the sum oftwo independent dice rolls.

Support of 𝑌: 2, 3, … , 11, 12

𝑝 𝑦 =

𝑦−1

36𝑦 ∈ ℤ, 2 ≤ 𝑦 ≤ 6

13−𝑦

36𝑦 ∈ ℤ, 7 ≤ 𝑦 ≤ 12

0 otherwise

Sanity check:

49

6/36

0

𝑌 = 𝑦

5/36

2 3 4 5 6 7 8 9 10 11 12

4/36

3/36

2/36

1/36

𝑃𝑌=𝑦

𝑦=2

12

𝑝 𝑦 = 1

Think, then Breakout Rooms

Then check out the question on the next slide. Post any clarifications here!


Think by yourself: 1 min

Breakout rooms: 5 min.

50

🤔



Example random variable

Consider 5 flips of a coin which comes up heads with probability 𝑝.Each coin flip is an independent trial. Let 𝒀 = # of heads on 5 flips.

1. What is the support of 𝑌? In other words, what are the values that 𝑌 can take on with non-zero probability?

2. Define the event 𝑌 = 2. What is 𝑃 𝑌 = 2 ?

3. What is the PMF of 𝑌? In other words, whatis 𝑃 𝑌 = 𝑘 , for 𝑘 in the support of 𝑌?

51

🤔


Example random variable

Consider 5 flips of a coin which comes up heads with probability 𝑝.Each coin flip is an independent trial. Let 𝒀 = # of heads on 5 flips.

1. What is the support of 𝑌? In other words, what are the values that 𝑌 can take on with non-zero probability?

2. Define the event 𝑌 = 2. What is 𝑃 𝑌 = 2 ?

3. What is the PMF of 𝑌? In other words, whatis 𝑃 𝑌 = 𝑘 , for 𝑘 in the support of 𝑌?

52

0, 1, 2, 3, 4, 5

𝑃 𝑌 = 𝑘 =52

𝑝2 1 − 𝑝 3

𝑃 𝑌 = 𝑘 =5𝑘

𝑝𝑘 1 − 𝑝 5−𝑘


Expectation

Remember that the expectation of a die roll is 3.5.

53

𝑋 = RV for value of roll

𝐸 𝑋 = 11

6+ 2

1

6+ 3

1

6+ 4

1

6+ 5

1

6+ 6

1

6=7

2

Review

𝐸 𝑋 =

𝑥:𝑝 𝑥 >0

𝑝 𝑥 ⋅ 𝑥 Expectation: The average value

of a random variable


Lying with statistics

54

“There are three kinds of lies:

lies, damned lies, and statistics”

–popularized by Mark Twain, 1906


Lying with statistics

A school has 3 classes with 5, 10, and 150 students.

What is the average class size?

55

1. Interpretation #1

• Randomly choose a classwith equal probability.

• 𝑋 = size of chosen class

𝐸 𝑋 = 51

3+ 10

1

3+ 150

1

3

=165

3= 55

2. Interpretation #2

• Randomly choose a studentwith equal probability.

• 𝑌 = size of chosen class

𝐸 𝑌

= 55

165+ 10

10

165+ 150

150

165

=22635

165≈ 137

Average student perception of class sizeWhat universities usually report

Interlude for fun/announcements

56


The pedagogy behind concept checks

• Spaced practice (vs. ”practice makes perfect”): better memory retention

• Low-stakes testing: better concept retrieval, actively connect concepts

It is okay if you don’t understand material off-the-bat. In fact,

learning research suggests that you will learn more in the long run.

Announcements

57

Problem Set 2

Out: last Wed.

Due: Friday

Covers: through today(ish)

Problem Set 1

Due: last Wed.

On-time grades: this Wed.

Solutions: next Wed.

https://www.dartmouth.edu/~cogedlab/pubs/Kang(2016,PIBBS).pdf

http://learninglab.psych.purdue.edu/downloads/2006_Roediger_Karpicke_PsychSci.pdfJOKE HERE

https://www.dartmouth.edu/~cogedlab/pubs/Kang(2016,PIBBS).pdf

http://learninglab.psych.purdue.edu/downloads/2006_Roediger_Karpicke_PsychSci.pdf


Ethics in Probability: Misleading Averages

“Averages can be misleading when a distribution is heavily skewed at one end, with a small number of unrepresentative outliers pulling the average in their direction. “

“In 2011, for example, the average income of the 7,878 households in Steubenville, Ohio, was $46,341. But if just two people, Warren Buffett and Oprah Winfrey, relocated to that city, the average household income in Steubenville would rise 62 percent overnight, to $75,263 per household.”

58

https://www.nytimes.com/2013/05/26/opinion/sunday/wh

en-numbers-mislead.html

https://www.nytimes.com/2013/05/26/opinion/sunday/when-numbers-mislead.html



1. Linearity:



𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌


𝐸 𝑔 𝑋 =

𝑥

𝑔 𝑥 𝑝(𝑥)59

Roll a die, outcome is 𝑋. You win $2𝑋 − 1.

What are your expected winnings?

Review

Let 𝑋 = 6-sided dice roll.

𝐸 2𝑋 − 1 = 2 3.5 − 1 = 6



1. Linearity:



𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌


𝐸 𝑔 𝑋 =

𝑥




Review

What is the expectation of the sum of

two dice rolls?


𝐸 2𝑋 − 1 = 2 3.5 − 1 = 6

Let 𝑋 = roll of die 1, 𝑌 = roll of die 2.

𝐸 𝑋 + 𝑌 = 3.5 + 3.5 = 7



1. Linearity:



𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌


𝐸 𝑔 𝑋 =

𝑥




Review

What is the expectation of the sum of

two dice rolls?


𝐸 2𝑋 − 1 = 2 3.5 − 1 = 6

(next up)

Let 𝑋 = roll of die 1, 𝑌 = roll of die 2.

𝐸 𝑋 + 𝑌 = 3.5 + 3.5 = 7

Think, then Breakout Rooms




Breakout rooms: 5 min. Introduce yourself!

62

🤔



Being a statistician unconsciously

Let 𝑋 be a discrete random variable.

• 𝑃 𝑋 = 𝑥 =1

3for 𝑥 ∈ {−1, 0, 1}

Let 𝑌 = |𝑋|. What is 𝐸 𝑌 ?

63

Expectation


𝑥


A.1

3⋅ 1 +

1

3⋅ 0 +

1

3⋅ −1 = 0

B. 𝐸 𝑌 = 𝐸 0 = 0

C.1

3⋅ 0 +

2

3⋅ 1 =

2

3

D.1

3⋅ −1 +

1

3⋅ 0 +

1

31 =

2

3

E. C and D 🤔


A.1

3⋅ 1 +

1

3⋅ 0 +

1

3⋅ −1 = 0

B. 𝐸 𝑌 = 𝐸 0 = 0

C.1

3⋅ 0 +

2

3⋅ 1 =

2

3

D.1

3⋅ −1 +

1

3⋅ 0 +

1

31 =

2

3

E. C and D

Being a statistician unconsciously

Let 𝑋 be a discrete random variable.

• 𝑃 𝑋 = 𝑥 =1

3for 𝑥 ∈ {−1, 0, 1}

Let 𝑌 = |𝑋|. What is 𝐸 𝑌 ?

64

Expectation


𝑥


𝐸 𝑋

1. Find PMF of 𝑌: 𝑝𝑌 0 =1

3, 𝑝𝑌 1 =

2

3

2. Compute 𝐸[𝑌]

Use LOTUS by using PMF of X:

1. 𝑃 𝑋 = 𝑥 ⋅ 𝑥2. Sum up

𝐸 𝐸 𝑋

❌

❌

Think




65

🤔(by yourself)



St. Petersburg Paradox

• A fair coin (comes up “heads” with 𝑝 = 0.5)

• Define 𝑌 = number of coin flips (“heads”) before first “tails”

• You win $2𝑌

How much would you pay to play? (How much you expect to win?)

66

Expectation

of 𝑔 𝑋

A. $10000

B. $∞C. $1

D. $0.50

E. $0 but let me play

F. I will not play

𝐸 𝑔 𝑥 =

𝑥


🤔(by yourself)


• A fair coin (comes up “heads” with 𝑝 = 0.5)

• Define 𝑌 = number of coin flips (“heads”) before first “tails”

• You win $2𝑌

How much would you pay to play? (How much you expect to win?)

67

St. Petersburg Paradox


2. Solve

For 𝑖 ≥ 0:

Let 𝑊 = your winnings, 2𝑌.

𝐸 𝑊 = 𝐸 2𝑌 =1

2

1

20 +1

2

2

21 +1

2

3

22 +⋯

𝑃 𝑌 = 𝑖 =1

2

𝑖+1

=

𝑖=0

∞1

2

𝑖+1

2𝑖 =

𝑖=0

∞1

2= ∞

Expectation

of 𝑔 𝑋𝐸 𝑔 𝑥 =

𝑥


Lisa Yan, CS109, 2020 68

St. Petersburg + Reality


2. Solve 𝐸 𝑊 =1

2

1

20 +1

2

2

21 +1

2

3

22 +⋯

=

𝑖=0

𝑘1

2

𝑖+1

2𝑖 =

𝑖=0

161

2= 8.5

Expectation

of 𝑔 𝑋𝐸 𝑔 𝑥 =

𝑥


What if dealer has only $65,536?

• Same game

• If you win over $65,536, dealer leaves the country

• Define 𝑌 = # heads before first tails

• You win 𝑊 = $2𝑌

𝑘 = log2 65,536

= 16

For 𝑖 ≥ 0:

Let 𝑊 = your winnings, 2𝑌.

𝑃 𝑌 = 𝑖 =1

2

𝑖+1

Breakout Rooms

Check out the questions on the next slide.Post any clarifications here!


Breakout rooms: 5 min. Introduce yourself!

69

🤔



Statistics: Expectation and variance

1. a. Let 𝑋 = the outcome of a 4-sideddie roll. What is 𝐸 𝑋 ?

b. Let 𝑌 = the sum of three rolls of a4-sided die. What is 𝐸 𝑌 ?

2. Compare the variances of𝐵1~Ber 0.1 and 𝐵2~Ber(0.5).

3. a. Let 𝑍 = # of tails on 10 flips of abiased coin (w.p. 0.4 of heads). What is 𝐸 𝑍 ?

b. What is Var(𝑍)?

70

🤔


Statistics: Expectation and variance

1. a. Let 𝑋 = the outcome of a 4-sideddie roll. What is 𝐸 𝑋 ?

b. Let 𝑌 = the sum of three rolls of a4-sided die. What is 𝐸 𝑌 ?

2. Compare the variances of𝐵1~Ber 0.1 and 𝐵2~Ber(0.5).

3. a. Let 𝑍 = # of tails on 10 flips of abiased coin (w.p. 0.4 of heads). What is 𝐸 𝑍 ?

b. What is Var(𝑍)?

71

If you can identify common RVs, just look up statistics instead of re-deriving from definitions.


Our first common RVs

73

𝑋 ~ Ber(𝑝)

𝑌 ~ Bin(𝑛, 𝑝)

1. The random

variable

2. is distributed

as a3. Bernoulli 4. with parameter

Example: Heads in one coin flip,

P(heads) = 0.8 = p

Example: # heads in 40 coin flips,

P(heads) = 0.8 = p

Review

Next Time: Binomial, Poisson, and more ☺

74

Documents

06: Random Variables - Stanford University...The probability of the event that a random variable takes on the value 𝑘! For discrete random variables, this is a probability mass