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Material Type: Notes; Professor: Abayomi; Class: Regression Analysis; Subject: Industrial & Systems Engr; University: Georgia Institute of Technology-Main Campus; Term: Spring 2009;

Typology: Study notes

Pre 2010

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Download Solutions to 12 Problems - Regression Analysis | ISYE 6414 and more Study notes Systems Engineering in PDF only on Docsity! ISYE 6414 - Spring 2009 Solution Lecture 1 - Prerequisites Problem 1. ∏n i=1 c · exi = cne ( ∑n i=1 xi) Problem 2. • The most likely sum of two rolled six-sided dice is 7 • The most likely sum of three rolled seven-sided dice is 12 Problem 3. Cov(aX + b, Y ) = E [(aX + b− E(aX + b)) (Y − E(Y ))] = = E [(aX + b− aE(X)− b) (Y − E(Y ))] = E [(a (X − E(X))) (Y − E(Y ))] = = E [(X − E(X)) (Y − E(Y ))] = aCov(aX + b, Y ) Problem 4. Cov n∑ i=1 Xi, n∑ j=1 Yj = E ( n∑ i=1 Xi − E ( n∑ i=1 Xi )) n∑ j=1 Yj − E n∑ j=1 Yj = = E ( n∑ i=1 Xi − n∑ i=1 E (Xi) ) n∑ j=1 Yj − n∑ j=1 E (Yj) = = E n∑ i=1 (Xi − E (Xi)) n∑ j=1 (Yj − E (Yj)) = = n∑ i=1 E (Xi − E (Xi)) n∑ j=1 (Yj − E (Yj)) = = n∑ i=1 n∑ j=1 E [(Xi − E (Xi)) (Yj − E (Yj))] Problem 5. By definition we know that λ(x) ≥ 0; x ∈ <. Also: ∫∞ −∞ λ(x)dx = ∫∞ x0 λ(x)dx = ∫∞ x0 f(x)dx 1−F (x0) = 1−F (x0) 1−F (x0) = 1. Now suppose X ∼ f(x) = e−x · 1(0,∞). Then: P (X < x|X ≥ x0) = P (x0 < X ∩X ≤ x) P (X ≥ x0) = P (x0 ≤ X ≤ x) P (X ≥ x0) = e−x0 − e−x e−x0 = 1− e−(x−x0) ⇒ f (x|X > x0) = d [ 1− e−(x−x0) ] dx · 1(x0,∞) = e −(x−x0) · 1(x0,∞) = λ(x) 1 Problem 6. In this case x0 = a, then, the probability that a nonsmoker with age a will survive ro age b is: P (nonsmoker) = ∫ ∞ b λn(x)dx = 1− ∫ b a λn(x)dx And the same probability for a smoker is: P (smoker) = ∫ ∞ b λs(x)dx = ∫ ∞ b 1 2 λn(x)dx = 1 2 · P (nonsmoker) Problem 7. We know that if X ∼ f(x), then Z = aX + b, where a ≥ 0 and b ∈ <, has density function: fZ(z) = fX ( z − b a ) · a−1 In this case a = σ−1 and b = −µσ , then: fZ(z) = fX (σz − µ) · σ = (2π)− 1 2 e z2 2 ∼ Normal(0, 1) Problem 8. We know that the moment generating function of Xi is: MXi(t) = exp ( µit+ t2σ2i 2 ) . Then, assuming that the random variables X ′is are independent, we have: MSn(t) = n∏ i=1 MXi(t) = n∏ i=1 exp ( µit+ t2σ2i 2 ) = exp ( t n∑ i=1 µi + t2 2 n∑ i=1 σ2i ) Then, Sn ∼ Normal (∑n i=1 µi, ∑n i=1 σ 2 i ) . Problem 9. The maximum likelihood estimator for σ2 from N ( µ, σ2 ) is σ̂2 = ∑n i=1(Xi−X̄) 2 n . Then: E ( σ̂2 ) = E (∑n i=1 ( Xi − X̄ )2 n ) However: n∑ i=1 ( Xi − X̄ )2 = n∑ i=1 X2i − nX̄2 Then: E ( σ̂2 ) = E (∑n i=1X 2 i − nX̄2 n ) = = ∑n i=1E ( X2i ) − nE ( X̄2 ) n = = ∑n i=1E ( µ2 + σ2 ) − nE ( µ2 + σ 2 n ) n = σ2 n− 1 n That is, σ̂2 is an biased estimator for σ2. Problem 10. Both estimators are the same. 2

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