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To simplify the calculations, we make use of the following random variables for the empirical distribution P: for i = {1, 2,...,n}, with Pn(i) = 1/n. (i) Verify that Ŷ = a + BX + ê. Using this, show that Conclude that X (i) = x₁, Ŷ (i) = Yi, ê(i) = Ei, (ii) Show that Vy := Varp (Ý) = = 3²¹vx + Varp (ê) + 2/6 Cov, (X, ê). Evy = ß²v₂ + E Var (ê) + 2ß E Covp₁ (‚ ê). n - 1 n Pn (Hint: recall that if v = Var (X) is the variance of the empirical distribution for i.i.d sample outcomes x1, x2,., n of a random variable X, then we know Ev(X₁, X2,..., Xn) n-1 Var X.4) E Var (2) = ·0²

To simplify the calculations, we make use of the following random variables for the empirical distribution P: for i = {1, 2,...,n}, with Pn(i) = 1/n. (i) Verify that Ŷ = a + BX + ê. Using this, show that Conclude that X (i) = x₁, Ŷ (i) = Yi, ê(i) = Ei, (ii) Show that Vy := Varp (Ý) = = 3²¹vx + Varp (ê) + 2/6 Cov, (X, ê). Evy = ß²v₂ + E Var (ê) + 2ß E Covp₁ (‚ ê). n - 1 n Pn (Hint: recall that if v = Var (X) is the variance of the empirical distribution for i.i.d sample outcomes x1, x2,., n of a random variable X, then we know Ev(X₁, X2,..., Xn) n-1 Var X.4) E Var (2) = ·0²

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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Can you show me the answer to (i) and (ii)? Thank you so much for your help

In this exercise, we fill in a key missing step in our computation of Eô² = "=²0². Namely, we will show that n Conclude that E vy (1) n To simplify the calculations, we make use of the following random variables for the empirical distribution Pn: (ii) Show that n 1 = for i = {1, 2,...,n}, with Pn(i) = 1/n. (i) Verify that Ŷ = a + ß + ĉ. Using this, show that = Â(i) = xi, Ŷ (i) = Yi, ê(i) = ɛi, -0² + B²vx. Vy := Varp (Ý) 3²v₂ + Varp (ê) + 2/3 Cov, (Â, ê). Evy = 3²v + E Var (2) + 2/3 E Covin E Var în (â) = η - 1 - -0². (Â,Ê). n (Hint: recall that if v = Var (X) is the variance of the empirical distribution for n i.i.d sample outcomes 1, 2,...,n of a random variable X, then we know Ev(X₁, X2, ..., Xn) = n-1 Var X.4) n
Transcribed Image Text:In this exercise, we fill in a key missing step in our computation of Eô² = "=²0². Namely, we will show that n Conclude that E vy (1) n To simplify the calculations, we make use of the following random variables for the empirical distribution Pn: (ii) Show that n 1 = for i = {1, 2,...,n}, with Pn(i) = 1/n. (i) Verify that Ŷ = a + ß + ĉ. Using this, show that = Â(i) = xi, Ŷ (i) = Yi, ê(i) = ɛi, -0² + B²vx. Vy := Varp (Ý) 3²v₂ + Varp (ê) + 2/3 Cov, (Â, ê). Evy = 3²v + E Var (2) + 2/3 E Covin E Var în (â) = η - 1 - -0². (Â,Ê). n (Hint: recall that if v = Var (X) is the variance of the empirical distribution for n i.i.d sample outcomes 1, 2,...,n of a random variable X, then we know Ev(X₁, X2, ..., Xn) = n-1 Var X.4) n
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