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Infinity 郭俊利 2010/05/11. 09. Limit. 4.6 ~ 5.4. Outline. Estimation error Inequalities Large numbers Convergence Central limit theorem. Linear Estimator. ^ X = E[X| Y] ≒ a Y + b
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Infinity 郭俊利 2010/05/11 09. Limit
4.6 ~ 5.4 Outline • Estimation error • Inequalities • Large numbers • Convergence • Central limit theorem
Linear Estimator • ^X = E[X| Y] ≒ a Y + b • If the measurement Y = X2, X is uniformly in [0, 2], what is the optimal estimate after a measured value = 2.25 ? ^X = Y½ ≒ 1/3 Y + 2/3 ^X = 1.5 ≒ 1.416
Linear Least Estimator • ^X = E[X| Y] ≒ a Y + b a = cov(X, Y) / var(Y) = ρ(σX / σY) b = E[X] – a E[Y] E[~X] = (1 –ρ2) σX2 σ is standard deviations, ρ is correlation coefficient
X = aY + b X – aY = b E[b] = E[X – aY] ↓ b = E[X] – a E[Y] ↓ ~X = X –^X Liner = X – aY – b E[~X] = E[X – aY – b] ↓ = var(X – aY) ↓ a = cov(X, Y) / var(Y) = ρ(σX / σY) E[~X] = (1 – ρ2) σX2 E[(X –^X Liner) Y] = 0 Linear Least Squares Estimator
Linear Estimated Example (1/2) X = Y½^X Liner = 1/2 Y +1/3 a = ρ(σX / σY) ρ = cov(X, Y) / σXσY σ = √var = √E[2] – E[]2, cov(X, Y) = E[XY] – E[X] E[Y] E[X] = 1, E[Y] = E[X2] = 4/3, E[X3] = 2, E[X4] = 16/5 cov(X, Y) = 2 – 1*4/3 = 2/3 var(X) = 1/3, var(Y) ≒ 1.42 ρ ≒ 0.97 a ≒ 1/2 b = E[X] – a E[Y] = 1/3 ^X Liner = 1/2 Y +1/3 E[~X] = (1 – ρ2) σX2≒ 0.02
Linear Estimated Example (2/2) Y X ~X 1/3 Y + 2/3 1/2 Y +1/3
E[X] a σ2 c2 Inequalities (1/4) • Markov • Chebyshev • When 0 ≤ E[X] ≤ a, Markov is informative; When 0 ≤ σ ≤ c, Chebyshev is informative. • If Chebyshev is informative, Chebyshev is exacter than Markov P(X ≥ a) ≤ P(|X – μ| ≥ c) ≤
X is uniformly distributed in [0, 4] Exact probabilities P(X ≥ 2) = 0.5 P(X ≥ 3) = 0.25 P(X ≥ 4) = 0 Markov P(X ≥ 2) ≤ 1 P(X ≥ 3) ≤ 2/3 P(X ≥ 4) ≤ 1/2 Chebyshev P(X ≥ 3) ≤ 4/3 ! P(X ≥ 4) ≤ 1/3 Inequalities (2/4)
Inequalities (3/4) • Xiao-Ming can type X words/minute and X is random variable. The average is 50 words/minute. (a) P(X ≥ 75) ≤ Real probability < 1/4 (b) If σX = 5, P(40 ≤ X ≤ 60) ≥
Inequalities (4/4) • Both Markov and Chebyshev are suitable for exponential distribution, but not for normal distribution. • X is exponentially distributed with λ = 1. For c > 1, to find P(X > c). • Exact probability = e–c • Markov ≤ 1 / c • Chebyshev ≤ 1 / (c – 1)2
σ2 nε2 Convergence (1/4) • Sequence Xn Constant c n ∞ lim Xn = c • Mn = (ΣXi) / n = Sn / n • E[Mn] = μ; var(Mn) = σ2/n • From Chebyshev P(| Mn– μ| ≥ε) ≤ 0 for all ε > 0, Mn converges to μ
Convergence (2/4) lim P(| Yn – a | ≥ε) = 0 • Xiao-Bao calculate Yn = min{X1, …, Xn}, Xi is uniformly in [0, 1], find the converged value of Yn. n ∞ ∵ P(|Yn – 0| ≥ ε) = P(X1 ≥ ε, …, Xn ≥ ε) = P(X1 ≥ ε) …P(Xn ≥ ε) = (1 – ε)n For all ε,ε = 0 ~ 1 ∴ lim P(| Yn – 0 | ≥ε) = lim (1 – ε)n = 0
Convergence (3/4) • If Xiao-Bao guess wrong, he gets the converged value = ½, what happened after finishing calculating? ∵ P(|Yn – ½| ≥ ε) = P(Yn – ½ ≥ ε∪ ½ – Yn ≥ ε) = P(Yn ≥ ε+ ½) + P(Yn ≤ ½ – ε) = (½ – ε)n+ (½ – ε)n ε = 0 ~ ½ ∴ lim P(| Yn – 0 | ≥ε) when ε = ½ ~ 1 cannot support.
Convergence (4/4) • Yn = max{X1, …, Xn}, Xi is uniformlyin [-1, 1], find the converged value of Yn. P(|Yn – 1| ≥ ε) = P(Yn ≥ ε + 1) + P(Yn ≤ 1 – ε) = P(max(Xi) ≥ ε + 1) + P(max(Xi) ≤ 1 – ε) = P(max(Xi) ≤ 1 – ε) = [ P(X ≤ 1 – ε) ]n = (1 – ½ ε)n ε = 0 ~ 1 lim (1 – ½ ε)n = 0
N(z) =1/ √2π ∫e –x/2dx 2 X1 + … + Xn – nμ σ√n Zn = Central Limit • Let X1, X2, … be a sequence of independent random variables with mean μ and variance σ2 CDF Zn converges to the standard normal CDF lim∞ P(Zn ≤ z) = N(z)
k – np – 1/2 √np(1 – p) l – np +1/2 √np(1 – p) De Moivre - Laplace • P(k ≤ Sn ≤ l) ≒ N( ) – N( ) • Xiao-Hua answers 100 questions and the probability of his correctness is 0.8. • P(X ≥ 70) ≒ P(S ≥ 69.5) ≒ N(–2.875) • P(X > 70) ≒ P(S ≥ 70.5) • P(X ≤ 85) ≒ N(1.375) • P(X = 88) ≒ N(8.5/4) – N(7.5/4)
