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hw6_solns - μ 1 = m 1 = ⇒ = 1 n n ± i =1 X i...

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ORIE 4580/5580/5581 Solutions for Homework Assignment 6 Question 1. a) Choose m = 2 20 , a = 37 and c = 1. Following Theorem 1 in the course packet, m and c are relatively prime; the only prime divisor of m is 2 and 2 divides a - 1; since m is divisible by 4, a - 1 is also divisible by 4. b) See the accompanying spreadsheet. The samples look like they are coming from the stan- dard normal distribution. c) See the accompanying spreadsheet. The pairs of samples lie on a spiral. Question 2. a) Let σ ij be the covariance between X i and X j . Note that ρ = σ 12 / σ 11 σ 22 = 3 . 6 / (2 . 0 × 3 . 0) = 0 . 6. Thus, setting L = [ σ 11 0 ρ σ 22 σ 22 1 - ρ 2 ] = [ 2 . 0 0 0 . 6 × 3 . 0 3 . 0 × 1 - 0 . 36 ] would do the job. b,c,d) See the accompanying spreadsheet. The covariance between X 1 and X 2 is positive. These random variables are positively correlated. In fact, the correlation coefficient between X 1 and X 2 is 0.6. Since X 1 and X 2 are positively correlated, whenever one of them takes a large value, the other one takes a large value as well. Question 3. If X Poisson( λ ), then E { X } = λ . Set m 1 = 1 n n i =1 X i , μ 1 = λ . All we need to do is to set m 1 = μ 1 and solve for the unknown parameter.
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Unformatted text preview: μ 1 = m 1 = ⇒ λ = 1 n n ± i =1 X i . Therefore, ˆ λ , our estimator of λ , is ˆ λ = 1 n n ± i =1 X i . Question 4. If X ∼ Binomial(7 , p ), then E { X } = 7 p . Set m 1 = 1 n ∑ n i =1 X i , μ 1 = 7 p . All we need to do is to set m 1 = μ 1 and solve for the unknown parameter. μ 1 = m 1 = ⇒ 7 p = 1 n n ± i =1 X i . 1 Therefore, ˆ p , our estimator of p , is ˆ p = 1 7 n n ∑ i =1 X i . Question 5. Let m 1 = 1 n n ∑ i =1 X i and m 2 = 1 n n ∑ i =1 X 2 i . Set m 1 = ˆ n ˆ p and m 2 = m 2 1 + ˆ n ˆ p (1-ˆ p ). Solving this system of equations yields: m 1 = ˆ n ˆ p = ⇒ ˆ n = m 1 ˆ p m 2 = m 2 1 + ˆ n ˆ p (1-ˆ p ) = m 2 1 + m 1 ˆ p ˆ p (1-ˆ p ) = ⇒ ˆ p = 1-m 2-m 2 1 m 1 = m 1-( m 2-m 2 1 ) m 1 Therefore, ˆ n = m 1 ˆ p = m 2 1 m 1-( m 2-m 2 1 ) (Note that whenever the sample mean is smaller than the sample variance, ˆ n and ˆ p both become negative.) 2...
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