Finding the regression function
$begingroup$
We have $$y(t) = B_0 +B_1(t) + ε(t)$$ $$z(t) = y^2(t) - y^2(t-1),$$ $$u(t) = x(t) - x(t-1),$$ $$v(t)= x(t)+x(t-1).$$
Find the regression function $mathbb {E}(z(t) vert u(t), v(t)).$
Use:
$mathbb {E}ε(t) = 0. $
$mathbb {E}ε(i)ε(j) = 0, $ if $ i≠j$, and $mathbb {E}ε(i)ε(j) = δ^2,$ if $ i=j.$
Solution:
$mathbb {E}(z(t) vert u(t), v(t)) = mathbb {E}(( y^2(t) - y^2(t-1))vert u(t), v(t)) = mathbb {E}((( B_0+B_1x(t) + ε(t))^2 -(B_0+B_1x(t-1) + ε(t-1))^2) )vert u(t), v(t)) $
And I do not know what to do next. Can someone give me some advises?
conditional-expectation regression expected-value
asked Dec 19 '18 at 19:36
PhilipPhilip
897
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add a comment |
$begingroup$
We have $$y(t) = B_0 +B_1(t) + ε(t)$$ $$z(t) = y^2(t) - y^2(t-1),$$ $$u(t) = x(t) - x(t-1),$$ $$v(t)= x(t)+x(t-1).$$
Find the regression function $mathbb {E}(z(t) vert u(t), v(t)).$
Use:
$mathbb {E}ε(t) = 0. $
$mathbb {E}ε(i)ε(j) = 0, $ if $ i≠j$, and $mathbb {E}ε(i)ε(j) = δ^2,$ if $ i=j.$
Solution:
$mathbb {E}(z(t) vert u(t), v(t)) = mathbb {E}(( y^2(t) - y^2(t-1))vert u(t), v(t)) = mathbb {E}((( B_0+B_1x(t) + ε(t))^2 -(B_0+B_1x(t-1) + ε(t-1))^2) )vert u(t), v(t)) $
And I do not know what to do next. Can someone give me some advises?
conditional-expectation regression expected-value
asked Dec 19 '18 at 19:36
PhilipPhilip
897
$endgroup$
$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35
add a comment |
$begingroup$
We have $$y(t) = B_0 +B_1(t) + ε(t)$$ $$z(t) = y^2(t) - y^2(t-1),$$ $$u(t) = x(t) - x(t-1),$$ $$v(t)= x(t)+x(t-1).$$
Find the regression function $mathbb {E}(z(t) vert u(t), v(t)).$
Use:
$mathbb {E}ε(t) = 0. $
$mathbb {E}ε(i)ε(j) = 0, $ if $ i≠j$, and $mathbb {E}ε(i)ε(j) = δ^2,$ if $ i=j.$
Solution:
$mathbb {E}(z(t) vert u(t), v(t)) = mathbb {E}(( y^2(t) - y^2(t-1))vert u(t), v(t)) = mathbb {E}((( B_0+B_1x(t) + ε(t))^2 -(B_0+B_1x(t-1) + ε(t-1))^2) )vert u(t), v(t)) $
And I do not know what to do next. Can someone give me some advises?
conditional-expectation regression expected-value
asked Dec 19 '18 at 19:36
PhilipPhilip
897
$endgroup$
We have $$y(t) = B_0 +B_1(t) + ε(t)$$ $$z(t) = y^2(t) - y^2(t-1),$$ $$u(t) = x(t) - x(t-1),$$ $$v(t)= x(t)+x(t-1).$$
Find the regression function $mathbb {E}(z(t) vert u(t), v(t)).$
Use:
$mathbb {E}ε(t) = 0. $
$mathbb {E}ε(i)ε(j) = 0, $ if $ i≠j$, and $mathbb {E}ε(i)ε(j) = δ^2,$ if $ i=j.$
Solution:
$mathbb {E}(z(t) vert u(t), v(t)) = mathbb {E}(( y^2(t) - y^2(t-1))vert u(t), v(t)) = mathbb {E}((( B_0+B_1x(t) + ε(t))^2 -(B_0+B_1x(t-1) + ε(t-1))^2) )vert u(t), v(t)) $
And I do not know what to do next. Can someone give me some advises?
conditional-expectation regression expected-value
conditional-expectation regression expected-value
asked Dec 19 '18 at 19:36
PhilipPhilip
897
asked Dec 19 '18 at 19:36
PhilipPhilip
897
asked Dec 19 '18 at 19:36
PhilipPhilip
897
asked Dec 19 '18 at 19:36
PhilipPhilip
897
asked Dec 19 '18 at 19:36
PhilipPhilip
897
897
$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35
add a comment |
$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35
$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35
$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35
add a comment |
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$begingroup$
You have missed a x out of the first equation. Just keep terms with no $epsilon$s in and those with either $epsilon(t)^2$ or $epsilon(t-1)^2$
$endgroup$
– user121049
Dec 19 '18 at 20:35