18: central limit theorem - stanford university...•perform an experiment •based on experiment...

18: Central Limit TheoremLisa YanNovember 1, 2019

1

Lisa Yan, CS109, 2019

Beta random variabledef An Beta random variable 𝑋 is defined as follows:

2

𝑓 𝑥 =1

𝐵 𝑎, 𝑏𝑥!"# 1 − 𝑥 $"#𝑋~Beta(𝑎, 𝑏)

VarianceExpectation

PDF

𝐸 𝑋 =𝑎

𝑎 + 𝑏 Var 𝑋 =𝑎𝑏

𝑎 + 𝑏 - 𝑎 + 𝑏 + 1

Support of 𝑋: 0, 1

𝑎 > 0, 𝑏 > 0

Beta is a distribution for probabilities.

where 𝐵 𝑎, 𝑏 = ∫/# 𝑥!"# 1 − 𝑥 $"#𝑑𝑥, normalizing constant

Review


🤔

0.0

1.0

2.0

3.0

4.0

0.0 0.2 0.4 0.6 0.8 1.0

3

CS109 focus: Beta where 𝑎, 𝑏 both positive integers

If 𝑎, 𝑏 are positive integers,Beta parameters 𝑎, 𝑏 couldcome from an experiment:

𝑎 = “successes” + 1𝑏 = “failures” + 1

𝑋~Beta(𝑎, 𝑏)

Beta(5,5)

Beta(2,8) Beta(8,2)

Review


0.0

1.0

2.0

3.0

4.0

0.0 0.2 0.4 0.6 0.8 1.0

Back to flipping coins• Start with a 𝑋~Uni 0,1 over probability• Observe 𝑛 = 7 successes and 𝑚 = 1 failures• Your new belief about the probability of 𝑋 is:

𝑓5|7 𝑥 𝑛 =1𝑐𝑥8 1 − 𝑥 #,where 𝑐 = 5

/

#𝑥8 1 − 𝑥 #𝑑𝑥

4𝑓 7|9𝑥|𝑛

Posterior belief, 𝑋|𝑁

𝑥

Posterior belief, 𝑋|𝑁:Beta(𝑎 = 8, 𝑏 = 2)

Beta(𝑎 = 𝑛 + 1, 𝑏 = 𝑚 + 1)

𝑓5|7 𝑥 𝑛 =1𝑐𝑥9"# 1 − 𝑥 :"#


Understanding Beta• Start with a 𝑋~Uni 0,1 over probability• Observe 𝑛 successes and 𝑚 failures• Your new belief about the probability of 𝑋 is:

𝑋|𝑁~Beta(𝑎 = 𝑛 + 1, 𝑏 = 𝑚 + 1)

5


Understanding Beta• Start with a 𝑋~Uni 0,1 over probability• Observe 𝑛 successes and 𝑚 failures• Your new belief about the probability of 𝑋 is:

𝑋|𝑁~Beta(𝑎 = 𝑛 + 1, 𝑏 = 𝑚 + 1)Check this out:

Beta 𝑎 = 1, 𝑏 = 1 has PDF:

So our prior 𝑋~Beta 𝑎 = 1, 𝑏 = 1 !

6

𝑓 𝑥 =1

𝐵 𝑎, 𝑏𝑥?@A 1 − 𝑥 B@A

where 0 < 𝑥 < 1

=1

𝐵 𝑎, 𝑏𝑥D 1 − 𝑥 D =

1

∫DA 1𝑑𝑥


If the prior is a Beta…Let 𝑋 be our random variable for probability of success and 𝑁• If our prior belief about 𝑋 is beta:

• …and if we observe 𝑛 successes and 𝑚 failures:

• …then our posterior belief about 𝑋is also beta.

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𝑁|𝑋~Bin(𝑛 + 𝑚, 𝑥)

𝑋|𝑁~Beta(𝑎 + 𝑛, 𝑏 + 𝑚)

This is the main takeaway of Beta.👉

likelihood





8


Proof:

𝑓5|7 𝑥 𝑛 =𝑝7|5 𝑛|𝑥 𝑓5 𝑥

𝑝7 𝑛 =

𝑛 +𝑚𝑛 𝑥G 1 − 𝑥 H ⋅ 1

𝐵 𝑎, 𝑏 𝑥?@A 1 − 𝑥 B@A

𝑝9 𝑛

= 𝐶 ⋅ 𝑥> 1 − 𝑥 ? ⋅ 𝑥!"# 1 − 𝑥 $"#constants that don’t depend on 𝑥

= 𝐶 ⋅ 𝑥>@!"# 1 − 𝑥 ?@$"# ✅



likelihood





9



Beta is a conjugate distribution.• Prior and posterior parametric forms are the same• Practically, conjugate means easy update:

Add number of “heads” and “tails” seento Beta parameter.

𝑁|𝑋~Bin(𝑛 + 𝑚, 𝑥)likelihood





10



You can set the prior to reflect how biased you think the coin is apriori.• This is a subjective probability!• 𝑋~Beta(𝑎, 𝑏): have seen 𝑎 + 𝑏 − 2 imaginary trials, where

𝑎 − 1 are heads, 𝑏 − 1 tails• Then Beta 1, 1 = Uni(0, 1) means we haven’t seen any imaginary trials

𝑁|𝑋~Bin(𝑛 + 𝑚, 𝑥)likelihood





11



This is the main takeaway of Beta.👉

likelihood

Beta(𝑎 = 𝑛A?!B + 1, 𝑏 = 𝑚A?!B + 1)Prior

Beta 𝑎 = 𝑛A?!B + 𝑛 + 1, 𝑏 = 𝑚A?!B +𝑚 + 1Posterior



The enchanted dieLet 𝑋 be the probability of rolling a 6 on Lisa’s die.• Prior: Imagine 5 die rolls where only 6 showed up• Observation: roll it a few times…

What is the updated distribution of 𝑋 after our observation?

Check out the demo!

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Beta(𝑎 = 𝑛A?!B + 𝑛 + 1, 𝑏 = 𝑚A?!B + 𝑚 + 1)Posterior

http://web.stanford.edu/class/cs109/demos/beta.html


Medicinal Beta• Before being tested, a medicine is believed to “work” 80% of the time.• The medicine is tried on 20 patients.• It “works” for 12, “doesn’t work” for 8.

What is your new belief that the drug “works”?

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Frequentist

Let 𝑝 be the probabilityyour drug works.

𝑝 ≈1220

= 0.6

Bayesian

A frequentist view will not incorporateprior/expert belief about probability.👉




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Frequentist

Let 𝑝 be the probabilityyour drug works.

𝑝 ≈1420

= 0.7

Bayesian

Let 𝑋 be the probabilityyour drug works.

𝑋 is a random variable.


🤔15



What is the prior distribution of 𝑋? (select all that apply)

A. 𝑋~Beta 1, 1 = Uni 0, 1B. 𝑋~Beta 81, 101C. 𝑋~Beta 80, 20D. 𝑋~Beta 81, 21E. 𝑋~Beta 5, 2



(Bayesian interpretation)


🤔16



What is the prior distribution of 𝑋? (select all that apply)

A. 𝑋~Beta 1, 1 = Uni 0, 1B. 𝑋~Beta 81, 101C. 𝑋~Beta 80, 20D. 𝑋~Beta 81, 21E. 𝑋~Beta 5, 2



Interpretation: 80 successes / 100 imaginary trials

Interpretation: 4 successes / 5 imaginary trials


(you can choose either; we choose E on next slide)


🤔17



Prior: 𝑋~Beta 𝑎 = 5, 𝑏 = 2Posterior: 𝑋~Beta 𝑎 = 5 + 12, 𝑏 = 2 + 8

~Beta 𝑎 = 17, 𝑏 = 10




0.01.02.03.04.05.0

0.0 0.2 0.4 0.6 0.8 1.0

Posterior

Prior

𝑥


🤔18




~Beta 𝑎 = 17, 𝑏 = 10




What do you report to pharmacists?A. Expectation of posteriorB. Mode of posteriorC. Distribution of posteriorD. Nothing

0.01.02.03.04.05.0

0.0 0.2 0.4 0.6 0.8 1.0

Posterior

Prior

𝑥

mode


🤔19




~Beta 𝑎 = 17, 𝑏 = 10




What do you report to pharmacists?A. Expectation of posteriorB. Mode of posteriorC. Distribution of posteriorD. Nothing

0.01.02.03.04.05.0

0.0 0.2 0.4 0.6 0.8 1.0

Posterior

Prior

𝑥

𝐸 𝑋 =𝑎

𝑎 + 𝑏=

1717 + 10

≈ 0.63

mode 𝑋 =𝑎 − 1

𝑎 + 𝑏 − 2=

1616 + 9

= 0.64

mode


Food for thought

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𝑌~Ber 𝑝

In this lecture: If we don’t know the parameter 𝑝,Bayesian statisticians will:• Treat the parameter as a random variable 𝑋

with a Beta prior distribution• Perform an experiment• Based on experiment outcomes, update the

posterior distribution of 𝑋Food for thought:

Any parameter for a “parameterized” random variable can be thought of as

a random variable.𝑌~𝒩 𝜇, 𝜎:

🤔

👉


Today’s plan

Finish Beta

Central Limit Theorem (CLT)

CLT exercises

21

(silent drumroll)

22


Central Limit TheoremConsider 𝑛 independent and identically distributed (i.i.d.) variables 𝑋#, 𝑋:, … , 𝑋>with 𝐸 𝑋A = 𝜇 and Var 𝑋A = 𝜎:.

JAK#

>

𝑋A ~𝒩(𝑛𝜇, 𝑛𝜎:)

The sum of 𝑛 i.i.d. random variables is normally distributed with mean 𝑛𝜇and variance 𝑛𝜎:.

23

As 𝑛 → ∞


True happiness

24



JAK#

>



25

As 𝑛 → ∞


🤔26

Quick checkWhat dimensions are the following random variables?1. 𝑋#

2. 𝑋#, 𝑋:, … , 𝑋>

3.

4.

A. 1-D random variableB. 𝑛 -D random variable (a vector)C. not a random variable

WXYA

G

𝑋X

1𝑛WXYA

G

𝑋X


🤔27

Quick checkWhat dimensions are the following random variables?1. 𝑋#

2. 𝑋#, 𝑋:, … , 𝑋>

3.

4.

A. 1-D random variableB. 𝑛 -D random variable (a vector)C. not a random variable

WXYA

G

𝑋X

1𝑛WXYA

G

𝑋X

(A)

(B)

(A)

(A) (aka the sample mean)

(aka a sample)



JAK#

>



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As 𝑛 → ∞


i.i.d. random variablesConsider 𝑛 variables 𝑋#, 𝑋:, … , 𝑋>.𝑋#, 𝑋:, … , 𝑋> are independent and identically distributed if• 𝑋#, 𝑋:, … , 𝑋> are independent, and• All have the same PMF (if discrete) or PDF (if continuous).⇒ 𝐸 𝑋A = 𝜇 for 𝑖 = 1,… , 𝑛⇒ Var 𝑋A = 𝜎: for 𝑖 = 1,… , 𝑛

Side note: Multiple random variables 𝑋A, 𝑋-, … , 𝑋G are independent iff

for all subsets 𝑋A,… , 𝑋\29

𝑃 𝑋# ≤ 𝑥#, 𝑋: ≤ 𝑥:, … , 𝑋N ≤ 𝑥N =OAK#

N

𝑃 𝑋A ≤ 𝑥A

Same thing: i.i.d. iid IID


🤔30

Quick check

Are 𝑋#, 𝑋:, … , 𝑋> i.i.d. with the following distributions?

1. 𝑋A~Exp 𝜆 , 𝑋A independent

2. 𝑋A~Exp 𝜆A , 𝑋A independent

3. 𝑋A~Exp 𝜆 , 𝑋# = 𝑋: = ⋯ = 𝑋>

4. 𝑋A~Bin 𝑛A , 𝑝 , 𝑋A independent


🤔31

Quick check

Are 𝑋#, 𝑋:, … , 𝑋> i.i.d. with the following distributions?

1. 𝑋A~Exp 𝜆 , 𝑋A independent

2. 𝑋A~Exp 𝜆A , 𝑋A independent

3. 𝑋A~Exp 𝜆 , 𝑋# = 𝑋: = ⋯ = 𝑋>

4. 𝑋A~Bin 𝑛A , 𝑝 , 𝑋A independent

✅

(unless 𝜆X equal)

dependent: 𝑋A = 𝑋- = ⋯ = 𝑋G

(unless 𝑛X equal)Note underlying Bernoulli RVs are i.i.d.!



JAK#

>



32

As 𝑛 → ∞

(demo)


Sum of dice rolls

Roll 𝑛 independent dice. Let 𝑋A be the outcome of roll 𝑖. 𝑋P are i.i.d.

33

WXYA

A

𝑋X

2 4 6 8 10 120.0

0.1

0.2

1 2 3 4 5 60.0

0.1

0.2

3 5 7 9 11131517

WXYA

-

𝑋X WXYA

_

𝑋XSum of 1die roll

Sum of 2die rolls

Sum of 3die rolls

How many wayscan you roll a totalof 3 vs 11?


CLT explains a lot

34

WXYA

G

𝑋X ~𝒩(𝑛𝜇, 𝑛𝜎-)

Normal approximation of BinomialSum of i.i.d. Bernoulli RVs ≈ Normal

𝑋~Bin(𝑛, 𝑝)𝑋 =WXYA

G

𝑋X

𝑋~𝒩 𝑛𝑝, 𝑛𝑝(1 − 𝑝)

Proof:

𝑋~𝒩 𝑛𝜇, 𝑛𝜎: CLT, as 𝑛 → ∞

(substitute mean,variance of Bernoulli)

Let 𝑋X~Ber(𝑝) for 𝑖 = 1,… , 𝑛, where 𝑋X are i.i.d.𝐸 𝑋X = 𝑝, Var 𝑋X = 𝑝(1 − 𝑝)

As 𝑛 → ∞The sum of 𝑛 i.i.d. random variables is normally distributed with mean 𝑛𝜇 and variance 𝑛𝜎-.


CLT explains a lot

35

WXYA

G


Sample ofsize 15,

sum values

Distribution of 𝑋X Distribution of ∑XYAAb 𝑋X



CLT explains a lot

36

WXYA

G


Sample ofsize 15,

average values

(sample mean) Distribution of AAb∑XYAAb 𝑋XDistribution of 𝑋X



CLT explains a lot

37

WXYA

G


0 1 2 3 4 5

𝑛 = 5

Galton Board, by Sir Francis Galton(1822-1911)


Break for Friday/ announcements

38

BAYESIANS


Midterm exam

It’s done!Grades: Friday 11/1Solutions: Friday 11/1

Announcements

39

Problem Set 4

Due: Wednesday 11/6Covers: Up to Law of Total Expectation

Late day reminder: No late days permitted past last day of the quarter, 12/7


Proof of CLT

40

The sum of 𝑛 i.i.d. random variables is normally distributed with mean 𝑛𝜇 and variance 𝑛𝜎-.W

XYA

G


• The Fourier Transform of a PDF is called a characteristic function.• Take the characteristic function of the probability mass of the sample

distance from the mean, divided by standard deviation• Show that this approaches an

exponential function in the limit as 𝑛 → ∞: • This function is in turn the characteristic function of the Standard

Normal, 𝑍~ 𝒩(0,1).

Proof:

𝑓 𝑥 = 𝑒"ST:

(this proof is beyond the scope of CS109)

As 𝑛 → ∞


Implications of CLT

Anything that is a sum/average of independent random variables is normal…meaning in real life, many things are normally distributed:• Movie ratings: averages of independent viewer scores• Polling:◦ Ask 100 people if they will vote for candidate 1◦ 𝑝# = # “yes”/100◦ Sum of Bernoulli RVs (each person independently says “yes” w.p. 𝑝)

◦ Repeat this process with different groups to get 𝑝A,… , 𝑝G (different sample statistics)

◦ Normal distribution over sample means 𝑝\◦ Confidence interval: “How likely is it that an estimate for true 𝑝 is close?”

41


What about other functions?

42

WAK#

>


?

Sum of i.i.d. RVs

Average of i.i.d. RVs

Max of i.i.d. RVs

(sample mean)?

Let 𝑋#, 𝑋:, … , 𝑋> i.i.d., where 𝐸 𝑋A = 𝜇, Var 𝑋A = 𝜎:. As 𝑛 → ∞:



43

WAK#

>


?


Max of i.i.d. RVs

(sample mean)?


Sum of i.i.d. RVs


🤔44

Distribution of sample meanLet 𝑋#, 𝑋:, … , 𝑋> i.i.d., where 𝐸 𝑋A = 𝜇, Var 𝑋A = 𝜎:. As 𝑛 → ∞:

Define:

𝑌~𝒩 𝑛𝜇, 𝑛𝜎:

f𝑋 = #>𝑌

f𝑋~𝒩( ? , ? )

U𝑋 =1𝑛JAK#

>

𝑋A (sample mean) 𝑌 =JAK#

>

𝑋A (sum)

(Linear transform of a Normal)

A. f𝑋~𝒩(𝑛𝜇, 𝑛𝜎-)B. f𝑋~𝒩(hG ,

iT

G )C. f𝑋~𝒩(𝜇, 𝜎-)D. f𝑋~𝒩(𝜇, i

T

G)

(CLT, as 𝑛 → ∞)


🤔45


Define:


f𝑋 = #>𝑌

f𝑋~𝒩( ? , ? )

U𝑋 =1𝑛JAK#

>


>

𝑋A (sum)


A. f𝑋~𝒩(𝑛𝜇, 𝑛𝜎-)B. f𝑋~𝒩(hG ,

iT

G )C. f𝑋~𝒩(𝜇, 𝜎-)D. f𝑋~𝒩(𝜇, i

T

G)



🤔46


Define:


f𝑋 = #>𝑌

f𝑋~𝒩 𝜇, VT

>

U𝑋 =1𝑛JAK#

>


>

𝑋A (sum)



The average of i.i.d. random variables (i.e., sample mean) is normally distributed with mean 𝜇 and variance 𝜎-/𝑛.

1𝑛WXYA

G

𝑋X ~𝒩(𝜇,𝜎-

𝑛)

Demo: http://onlinestatbook.com/stat_sim/sampling_dist/

http://onlinestatbook.com/stat_sim/sampling_dist/


?


47

WAK#

>



Max of i.i.d. RVs

(sample mean)


Sum of i.i.d. RVs

1𝑛WAK#

>

𝑋A ~𝒩(𝜇,𝜎:

𝑛)

Gumbel(see Fisher-Tippett Gnedenko Theorem)


Once upon a time…

48

Abraham de MoivreCLT for 𝑋~Ber 1/21733

Aubrey Drake Graham(Drake)


A short history of the CLT

49

1700

1800

1900

2000

1733: CLT for 𝑋~Ber 1/2postulated by Abraham de Moivre

1823: Pierre-Simon Laplace extends de Moivre’swork to approximating Bin 𝑛, 𝑝 with Normal

1901: Alexandr Lyapunov provides precisedefinition and rigorous proof of CLT

2018: Drake releases Scorpion• It was his 5th studio album, bringing his total # of songs to 190• Mean quality of subsamples of songs is normally distributed (thanks

to the Central Limit Theorem)


Today’s plan

Central Limit Theorem (CLT)

CLT exercises

50


Working with the CLT

51

⚠If 𝑋X is discrete:

Use the continuity correction on 𝑌!

WAK#

>


1𝑛WAK#

>


𝑛)

Sum of i.i.d. RVs

Average of i.i.d. RVs(sample mean)



🤔52

Dice gameYou will roll 10 6-sided dice 𝑋#, 𝑋:, … , 𝑋#/ .• Let 𝑋 = 𝑋# + 𝑋: +⋯+ 𝑋#/, the total value of all 10 rolls.• You win if 𝑋 ≤ 25 or 𝑋 ≥ 45.

And now the truth (according to the CLT)…

WAK#

>

𝑋A ~𝒩(𝑛𝜇, 𝑛𝜎:)As 𝑛 → ∞:

1. Define RVs andstate goal.

2. Solve.

𝐸 𝑋X = 3.5,Var 𝑋X = 35/12𝑋 ≈ 𝑌~𝒩(10 3.5 , 10 35/12 )

Want:𝑃 𝑋 ≤ 25 or 𝑋 ≥ 45

A. 𝑃 25 ≤ 𝑌 ≤ 45B. 𝑃 𝑌 ≤ 25.5 + 𝑃 𝑌 ≥ 44.5C. 1 − 𝑃 25 ≤ 𝑌 ≤ 45D. 1 − 𝑃 25.5 ≤ 𝑌 ≤ 44.5


🤔53



WAK#

>



2. Solve.

𝐸 𝑋X = 3.5,Var 𝑋X = 35/12𝑋 ≈ 𝑌~𝒩(10 3.5 , 10 35/12 )

Want:𝑃 𝑋 ≤ 25 or 𝑋 ≥ 45≈ 𝑃 𝑌 ≤ 25.5 + 𝑃 𝑌 ≥ 44.5

𝑃 𝑌 ≤ 25.5 + 𝑃 𝑌 ≥ 44.5

= Φ25.5 − 3510 35/12

− 1 − Φ44.5 − 3510 35/12

≈ Φ −1.76 − 1 − Φ 1.76≈ 1 − 0.9608 − 1 − 0.9608= 0.0784


🤔54



WAK#

>


0

0.02

0.04

0.06

0.08

10 20 30 40 50 60

≈ 𝑃 𝑌 ≤ 25.5 + 𝑃 𝑌 ≥ 44.5

≈ 0.0784

𝑃 𝑋 ≤ 25 or 𝑋 ≥ 45 ≈ 0.0780

(by computer)

(by CLT)


As 𝑛 → ∞:Clock running time

55

f𝑋~𝒩 𝑡,4𝑛

𝑃 𝑡 − 0.5 ≤ f𝑋 ≤ 𝑡 + 0.5 = 0.95Want:

𝑃 0.5 ≤ f𝑋 − 𝑡 ≤ 0.5 = 0.95f𝑋 − 𝑡~𝒩 0,4𝑛

(linear transform of

a normal)

(CLT)

Want to find the mean (clock)runtime of an algorithm, 𝜇 = 𝑡 sec.• Suppose variance of

runtime is 𝜎: = 4 sec2.How many trials do we need s.t. estimated time = 𝑡 ± 0.5 with 95% certainty?

Run algorithm repeatedly (i.i.d. trials):• 𝑋X = runtime of 𝑖-th run (for 1 ≤ 𝑖 ≤ 𝑛)• Estimate runtime to be

average of 𝑛 trials, f𝑋


2. Solve.

1𝑛WAK#

>


𝑛 )


Clock running timeWant to find the mean (clock)runtime of an algorithm, 𝜇 = 𝑡 sec.• Suppose variance of


56




0.95 =𝑃 0.5 ≤ f𝑋 − 𝑡 ≤ 0.5

f𝑋 − 𝑡~𝒩 0,4𝑛

2. Solve.

0.95 = 𝐹 U5"] 0.5 − 𝐹 U5"] −0.5

= Φ0.5 − 04/𝑛

− Φ−0.5 − 0

4/𝑛= 2Φ

𝑛4

− 1

As 𝑛 → ∞:1𝑛WAK#

>


𝑛 )


Clock running timeWant to find the mean (clock)runtime of an algorithm, 𝜇 = 𝑡 sec.• Suppose variance of


57




0.95 =𝑃 0.5 ≤ f𝑋 − 𝑡 ≤ 0.5

f𝑋 − 𝑡~𝒩 0,4𝑛

2. Solve.

0.95 = 𝐹 U5"] 0.5 − 𝐹 U5"] −0.5

= Φ0.5 − 04/𝑛

− Φ−0.5 − 0

4/𝑛= 2Φ

𝑛4

− 1

0.975 = Φ 𝑛/4𝑛/4 = Φ"# 0.975 ≈ 1.96 𝑛 ≈ 62

As 𝑛 → ∞:1𝑛WAK#

>


𝑛 )


Wonderful form of cosmic order

I know of scarcely anything so apt to impress the imagination as the wonderful form of cosmic order expressed by the ”[Central limit theorem]". The law would have been personified by the Greeks and deified, if they had known of it. It reigns with serenity and in complete self-effacement, amidst

the wildest confusion. The huger the mob, and the greater the apparent anarchy, the more perfect is its sway.

– Sir Francis Galton

58

(of the Galton Board)

It reigns with serenity and in complete self-effacement, amidst the wildest confusion. The huger the mob, and the greater the apparent anarchy, the more perfect is its sway. It is the supreme law of Unreason.

Whenever a large sample of chaotic elements are taken in hand and marshalled in the order of their magnitude, an unsuspected and most

beautiful form of regularity proves to have been latent all along.


Extra slides

Extra CLT exercises

59


Crashing website• Let 𝑋 = number of visitors to a website, where 𝑋~Poi 100 .• The server crashes if there are ≥ 120 requests/minute.

What is 𝑃 server crashes in next minute ?

60

Strategy: Poisson (exact) Strategy: CLT (approximate)State goal

𝑃 𝑋 ≥ 120

Want: 𝑃 𝑋 ≥ 120Solve

= JNK#:/

^120 N𝑒"#//

𝑘!

≈ 0.0282


Crashing website• Let 𝑋 = number of visitors to a website, where 𝑋~Poi 100 .• The server crashes if there are ≥ 120 requests/minute.

What is 𝑃 server crashes in next minute ?

61

Strategy: Poisson (exact) Strategy: CLT (approximate)State goal

𝑃 𝑋 ≥ 120

Want: 𝑃 𝑋 ≥ 120Solve

= JNK#:/

^120 N𝑒"#//

𝑘!

≈ 0.0282

State approx.goal

Poi 100 ~WXYA

G

Poi 100/𝑛 (sum of IIDPoisson = Poisson)

𝑋 ≈ 𝑌~𝒩(𝑛𝜇, 𝑛𝜎-) 𝜇 =100𝑛

= 𝜎-

Want: 𝑃 𝑋 ≥ 120 ≈ 𝑃 𝑌 ≥ 119.5Solve𝑃 𝑌 ≥ 119.5 = 1 − Φ

119.5 − 100100

= 1 − Φ 1.95≈ 0.0256

18: central limit theorem - stanford university...•perform an experiment •based on experiment...

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