limit theorems Consider random variables X 1 + X 2 + . . . + X n - - PowerPoint PPT Presentation

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Nov 23, 2022 406 likes •470 views

Limit Theorems Consider i.i.d. (independent, identically distributed) random vars X 1 , X 2 , X 3 , X i has = E[X i ] and 2 = Var[X i ] limit theorems Consider random variables X 1 + X 2 + . . . + X n and n 1 X X i n i =1 Law of

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limit theorems

Limit Theorems Consider i.i.d. (independent, identically distributed) random vars X1, X2, X3, … Xi has μ = E[Xi] and σ2 = Var[Xi] Consider random variables and X1 + X2 + . . . + Xn 1 n

i=1

Xi Law of Large Numbers If we observe a random variable X many times (independently) and take the average, this average will converge to a real number which is E(X). Formally, let be independent, identically distributed random variables with mean . Define Then for any we have Proof: Use Chebychev’s inequality.

X1, . . . , Xn µ An = 1 n

i=1

Xi α > 0 Pr(|An − µ| > α) → 0 as n → ∞ the central limit theorem (CLT) Consider i.i.d. (independent, identically distributed) random vars X1, X2, X3, … Xi has μ = E[Xi] and σ2 = Var[Xi] As n → ∞, Restated: As n → ∞,

Mn = 1 n

X

i=1

Xi → N ✓ µ, σ2 n ◆

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CLT applies even to even wacky distributions

CLT in the real world

CLT is the reason many things appear normally distributed Many quantities = sums of (roughly) independent random vars Exam scores: sums of individual problems People’s heights: sum of many genetic & environmental factors Measurements: sums of various small instrument errors ...

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in the real world…

the central limit theorem (CLT) Consider i.i.d. (independent, identically distributed) random vars X1, X2, X3, … Xi has μ = E[Xi] and σ2 = Var[Xi] As n → ∞, Restated: As n → ∞,

Mn = 1 n

X

i=1

Xi → N ✓ µ, σ2 n ◆

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Example 1 Number of students who enroll in a class is Poisson (100). Professor will teach course in two sections if more than 120 students enroll. What is the probability of two sections? When applying approximation, use continuity correction: Think of Pr (X=i) = Pr (i-0.5 < X < i + 0.5) Recall Poisson (100) is sum of 100 independent Poisson(1) random variables, so we can apply CLT.

Pr(X > 119.5) = Pr ✓X − 100 √ 100 ≥ 119.5 − 100 √ 100 ◆

≈ 1 − Φ(1.95) ≈ 0.0256. Example 2 If 10 fair die are rolled, find the approximate probability that the sum obtained is between 30 and 40, inclusive, using CLT. Xi is the value of die # i. E(Xi) = 7/2 and Var(Xi) = 35/12

X = X1 + X2 + ...X10

Pr(29.5 ≤ X ≤ 40.5) = Pr 29.5 − 35 p 350/12 ≤ X − 35 p 350/12 ≤ 40.5 − 35 p 350/12 !

≈ 2 · Φ(1.0184) − 1 ≈ 0.692. E(X) = 35 V ar(X) = 350/12