5. In the regression model, Y₁ = ẞ0+ ẞ₁xi1 + B₂xi2 + ... + ßpxip +Єi, n = 1, . . ., n. Suppose we correct Y₁,..., Yn for their arithmetic average Ỹ, to obtain Y₁ = Y₁ – Ỹ. Then V₁,, have average zero, i.e., they are centred on zero. Similary, correct the values of the jth explanatory variable, x1,...,xnj, by subtracting their average, á: Xij = Xij − Ñj. Define the n× 1 vector, y = (1, №2...Vn), and the n xp matrix (no intercept column), X11 Xip -(1-5) x = We also define the vector of slop parameters (no intercept), ß₁:p = (ß₁, . . ., ßp). (a). (b). (c). (d). written as Please show that the least squares estimate for the slope parameters can be = (xTx)¯¹xy. B1P = SS(B₁,..., Bp Bo) in the ANOVA Table (with the intercept ignored) can be written as Σ(Ŷ - Ÿ)². 3) SS (B₁, ..., ßp|ẞ0) can also be written as Tŷ, and hence as Show that E[SS(B1, ..., ßp |ẞo)] = po² +ß{ :pXT Xß₁:p· T
5. In the regression model, Y₁ = ẞ0+ ẞ₁xi1 + B₂xi2 + ... + ßpxip +Єi, n = 1, . . ., n. Suppose we correct Y₁,..., Yn for their arithmetic average Ỹ, to obtain Y₁ = Y₁ – Ỹ. Then V₁,, have average zero, i.e., they are centred on zero. Similary, correct the values of the jth explanatory variable, x1,...,xnj, by subtracting their average, á: Xij = Xij − Ñj. Define the n× 1 vector, y = (1, №2...Vn), and the n xp matrix (no intercept column), X11 Xip -(1-5) x = We also define the vector of slop parameters (no intercept), ß₁:p = (ß₁, . . ., ßp). (a). (b). (c). (d). written as Please show that the least squares estimate for the slope parameters can be = (xTx)¯¹xy. B1P = SS(B₁,..., Bp Bo) in the ANOVA Table (with the intercept ignored) can be written as Σ(Ŷ - Ÿ)². 3) SS (B₁, ..., ßp|ẞ0) can also be written as Tŷ, and hence as Show that E[SS(B1, ..., ßp |ẞo)] = po² +ß{ :pXT Xß₁:p· T
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter7: Analytic Trigonometry
Section7.6: The Inverse Trigonometric Functions
Problem 92E
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![5.
In the regression model, Y₁ = ẞ0+ ẞ₁xi1 + B₂xi2 + ... + ßpxip +Єi, n = 1, . . ., n.
Suppose we correct Y₁,..., Yn for their arithmetic average Ỹ, to obtain Y₁ = Y₁ – Ỹ. Then
V₁,, have average zero, i.e., they are centred on zero. Similary, correct the values of
the jth explanatory variable, x1,...,xnj, by subtracting their average, á: Xij = Xij − Ñj.
Define the n× 1 vector, y = (1, №2...Vn), and the n xp matrix (no intercept column),
X11
Xip
-(1-5)
x =
We also define the vector of slop parameters (no intercept), ß₁:p = (ß₁, . . ., ßp).
(a).
(b).
(c).
(d).
written as
Please show that the least squares estimate for the slope parameters can be
= (xTx)¯¹xy.
B1P
=
SS(B₁,..., Bp Bo) in the ANOVA Table (with the intercept ignored) can be
written as Σ(Ŷ - Ÿ)².
3) SS (B₁, ..., ßp|ẞ0) can also be written as Tŷ, and hence as
Show that E[SS(B1, ..., ßp |ẞo)] = po² +ß{ :pXT Xß₁:p·
T](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F643c9073-bfc9-41d3-b50c-3b14187fea75%2F633cbdf1-1c77-45a4-acf9-c627f14d8c7b%2Fds8ggbb_processed.png&w=3840&q=75)
Transcribed Image Text:5.
In the regression model, Y₁ = ẞ0+ ẞ₁xi1 + B₂xi2 + ... + ßpxip +Єi, n = 1, . . ., n.
Suppose we correct Y₁,..., Yn for their arithmetic average Ỹ, to obtain Y₁ = Y₁ – Ỹ. Then
V₁,, have average zero, i.e., they are centred on zero. Similary, correct the values of
the jth explanatory variable, x1,...,xnj, by subtracting their average, á: Xij = Xij − Ñj.
Define the n× 1 vector, y = (1, №2...Vn), and the n xp matrix (no intercept column),
X11
Xip
-(1-5)
x =
We also define the vector of slop parameters (no intercept), ß₁:p = (ß₁, . . ., ßp).
(a).
(b).
(c).
(d).
written as
Please show that the least squares estimate for the slope parameters can be
= (xTx)¯¹xy.
B1P
=
SS(B₁,..., Bp Bo) in the ANOVA Table (with the intercept ignored) can be
written as Σ(Ŷ - Ÿ)².
3) SS (B₁, ..., ßp|ẞ0) can also be written as Tŷ, and hence as
Show that E[SS(B1, ..., ßp |ẞo)] = po² +ß{ :pXT Xß₁:p·
T
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