(iii) Verify that5 Using this, show that n Covp, (X, ê) = ¹ (α; — ä₁. - x) ei. n i=1 E Covê (Â, ê) = 0. (iv) By combining the previous parts, conclude that the identity (1) holds.
(iii) Verify that5 Using this, show that n Covp, (X, ê) = ¹ (α; — ä₁. - x) ei. n i=1 E Covê (Â, ê) = 0. (iv) By combining the previous parts, conclude that the identity (1) holds.
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 question 3 and 4? Thank you so much for your help

Transcribed Image Text:(iii) Verify that5
Using this, show that
Cov p. (Â, ê) = ¹
n
n
i=1
(x¡ — T)e į.
E Covp, (X, ê) = 0.
(iv) By combining the previous parts, conclude that the identity (1) holds.

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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