Elliptical Confidence Set Calculation
$begingroup$
I got stuck in a homework question:
In Linear regression model with assumption $varepsilon_{i} sim cal{N}(0, sigma^{2})$, iid.
$$Y_{i} = X_{i}^{intercal}theta^{*} + varepsilon_{i}, ~ i = 1, cdots, n, ~ mathbb{E}varepsilon_{i} sim cal{N}(0,sigma^{2})$$
I am asked to find an elliptical confidence set for the mean response $mathbb{E}(Y)$ in this model, with $sigma^{2} > 0$ unknown.
Below is what I tried. I got stuck since my answer contains $sigma^{2}$ and not sure if that is the right path.
Typically the mean response confidence interval is like $hat{y}_{h} pm t_{frac{alpha}{2}, n-p} se(hat{y}_{h})$. Compare to that, I claim $hat{theta}$ is the estimated parameter, and $hat{Y} = X^{intercal}hat{theta}$ is the estimated value. Remaining is to find variance.
$$Var(hat{Y}) = Var(X^{intercal}hat{theta}) = Var(X^{intercal}(XX^{intercal})^{-1}XY)\
=X^{intercal}(XX^{intercal})^{-1}X Var(Y) X^{intercal} (XX^{intercal})^{-1}X = sigma^{2} X^{intercal}(XX^{intercal})^{-1}X$$
since $Var(Y) = Var(varepsilon) = sigma^{2} I_{n}$
Now that $(hat{y}/se(hat{y})^{2} leqslant F_{alpha/2, 1, n-p}$, I used the idea that square of a t-distributed r.v. follows a 1,n-p F-distribution. In fact the right hand side with misuse of notation I mean the F-value with 1,n-p df and $alpha/2$ cutoff.
My major question is about left hand side, if I similarly plug in the expression, this is what I got eventually:
$$hat{theta}^{intercal}XX^{intercal}hat{theta} frac{1}{sigma^{2}} (X^{intercal}(XX^{intercal})^{-1}X)^{-1} \
= frac{1}{sigma^{2}} Y^{intercal}(X^{intercal}(XX^{intercal})^{-1}X) Y (X^{intercal}(XX^{intercal})^{-1}X)^{-1}$$
Of course the confidence set is the set where this expression is bounded by that F-value.
Am I totally wrong from the beginning? Or there is something I can do to fix my calculation? It just doesn't look right at this moment.
(Kind of have idea now. Please allow me some time to update it when I finish my homework).
I think it should be:
$$mathbb{E}(Y) = X^{intercal}theta^{*} in { X^{intercal} theta in mathbb{R} ^ {n}: || (XX^{intercal}) ^{ frac{1}{2}} (theta - hat{theta}||^{2} leqslant frac{p}{n-p} S(hat{theta}) Q(1-alpha, F(p,n-p) ) }$$
where $S(hat{theta}) = ||Y - X^{intercal}hat{theta}||^{2}$ and $hat{theta} = (XX^{intercal})^{-1}XY$. $Q$ is the quantile inverse function.
linear-regression
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
$endgroup$
add a comment |
$begingroup$
I got stuck in a homework question:
In Linear regression model with assumption $varepsilon_{i} sim cal{N}(0, sigma^{2})$, iid.
$$Y_{i} = X_{i}^{intercal}theta^{*} + varepsilon_{i}, ~ i = 1, cdots, n, ~ mathbb{E}varepsilon_{i} sim cal{N}(0,sigma^{2})$$
I am asked to find an elliptical confidence set for the mean response $mathbb{E}(Y)$ in this model, with $sigma^{2} > 0$ unknown.
Below is what I tried. I got stuck since my answer contains $sigma^{2}$ and not sure if that is the right path.
Typically the mean response confidence interval is like $hat{y}_{h} pm t_{frac{alpha}{2}, n-p} se(hat{y}_{h})$. Compare to that, I claim $hat{theta}$ is the estimated parameter, and $hat{Y} = X^{intercal}hat{theta}$ is the estimated value. Remaining is to find variance.
$$Var(hat{Y}) = Var(X^{intercal}hat{theta}) = Var(X^{intercal}(XX^{intercal})^{-1}XY)\
=X^{intercal}(XX^{intercal})^{-1}X Var(Y) X^{intercal} (XX^{intercal})^{-1}X = sigma^{2} X^{intercal}(XX^{intercal})^{-1}X$$
since $Var(Y) = Var(varepsilon) = sigma^{2} I_{n}$
Now that $(hat{y}/se(hat{y})^{2} leqslant F_{alpha/2, 1, n-p}$, I used the idea that square of a t-distributed r.v. follows a 1,n-p F-distribution. In fact the right hand side with misuse of notation I mean the F-value with 1,n-p df and $alpha/2$ cutoff.
My major question is about left hand side, if I similarly plug in the expression, this is what I got eventually:
$$hat{theta}^{intercal}XX^{intercal}hat{theta} frac{1}{sigma^{2}} (X^{intercal}(XX^{intercal})^{-1}X)^{-1} \
= frac{1}{sigma^{2}} Y^{intercal}(X^{intercal}(XX^{intercal})^{-1}X) Y (X^{intercal}(XX^{intercal})^{-1}X)^{-1}$$
Of course the confidence set is the set where this expression is bounded by that F-value.
Am I totally wrong from the beginning? Or there is something I can do to fix my calculation? It just doesn't look right at this moment.
(Kind of have idea now. Please allow me some time to update it when I finish my homework).
I think it should be:
$$mathbb{E}(Y) = X^{intercal}theta^{*} in { X^{intercal} theta in mathbb{R} ^ {n}: || (XX^{intercal}) ^{ frac{1}{2}} (theta - hat{theta}||^{2} leqslant frac{p}{n-p} S(hat{theta}) Q(1-alpha, F(p,n-p) ) }$$
where $S(hat{theta}) = ||Y - X^{intercal}hat{theta}||^{2}$ and $hat{theta} = (XX^{intercal})^{-1}XY$. $Q$ is the quantile inverse function.
linear-regression
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
$endgroup$
add a comment |
$begingroup$
I got stuck in a homework question:
In Linear regression model with assumption $varepsilon_{i} sim cal{N}(0, sigma^{2})$, iid.
$$Y_{i} = X_{i}^{intercal}theta^{*} + varepsilon_{i}, ~ i = 1, cdots, n, ~ mathbb{E}varepsilon_{i} sim cal{N}(0,sigma^{2})$$
I am asked to find an elliptical confidence set for the mean response $mathbb{E}(Y)$ in this model, with $sigma^{2} > 0$ unknown.
Below is what I tried. I got stuck since my answer contains $sigma^{2}$ and not sure if that is the right path.
Typically the mean response confidence interval is like $hat{y}_{h} pm t_{frac{alpha}{2}, n-p} se(hat{y}_{h})$. Compare to that, I claim $hat{theta}$ is the estimated parameter, and $hat{Y} = X^{intercal}hat{theta}$ is the estimated value. Remaining is to find variance.
$$Var(hat{Y}) = Var(X^{intercal}hat{theta}) = Var(X^{intercal}(XX^{intercal})^{-1}XY)\
=X^{intercal}(XX^{intercal})^{-1}X Var(Y) X^{intercal} (XX^{intercal})^{-1}X = sigma^{2} X^{intercal}(XX^{intercal})^{-1}X$$
since $Var(Y) = Var(varepsilon) = sigma^{2} I_{n}$
Now that $(hat{y}/se(hat{y})^{2} leqslant F_{alpha/2, 1, n-p}$, I used the idea that square of a t-distributed r.v. follows a 1,n-p F-distribution. In fact the right hand side with misuse of notation I mean the F-value with 1,n-p df and $alpha/2$ cutoff.
My major question is about left hand side, if I similarly plug in the expression, this is what I got eventually:
$$hat{theta}^{intercal}XX^{intercal}hat{theta} frac{1}{sigma^{2}} (X^{intercal}(XX^{intercal})^{-1}X)^{-1} \
= frac{1}{sigma^{2}} Y^{intercal}(X^{intercal}(XX^{intercal})^{-1}X) Y (X^{intercal}(XX^{intercal})^{-1}X)^{-1}$$
Of course the confidence set is the set where this expression is bounded by that F-value.
Am I totally wrong from the beginning? Or there is something I can do to fix my calculation? It just doesn't look right at this moment.
(Kind of have idea now. Please allow me some time to update it when I finish my homework).
I think it should be:
$$mathbb{E}(Y) = X^{intercal}theta^{*} in { X^{intercal} theta in mathbb{R} ^ {n}: || (XX^{intercal}) ^{ frac{1}{2}} (theta - hat{theta}||^{2} leqslant frac{p}{n-p} S(hat{theta}) Q(1-alpha, F(p,n-p) ) }$$
where $S(hat{theta}) = ||Y - X^{intercal}hat{theta}||^{2}$ and $hat{theta} = (XX^{intercal})^{-1}XY$. $Q$ is the quantile inverse function.
linear-regression
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
$endgroup$
I got stuck in a homework question:
In Linear regression model with assumption $varepsilon_{i} sim cal{N}(0, sigma^{2})$, iid.
$$Y_{i} = X_{i}^{intercal}theta^{*} + varepsilon_{i}, ~ i = 1, cdots, n, ~ mathbb{E}varepsilon_{i} sim cal{N}(0,sigma^{2})$$
I am asked to find an elliptical confidence set for the mean response $mathbb{E}(Y)$ in this model, with $sigma^{2} > 0$ unknown.
Below is what I tried. I got stuck since my answer contains $sigma^{2}$ and not sure if that is the right path.
Typically the mean response confidence interval is like $hat{y}_{h} pm t_{frac{alpha}{2}, n-p} se(hat{y}_{h})$. Compare to that, I claim $hat{theta}$ is the estimated parameter, and $hat{Y} = X^{intercal}hat{theta}$ is the estimated value. Remaining is to find variance.
$$Var(hat{Y}) = Var(X^{intercal}hat{theta}) = Var(X^{intercal}(XX^{intercal})^{-1}XY)\
=X^{intercal}(XX^{intercal})^{-1}X Var(Y) X^{intercal} (XX^{intercal})^{-1}X = sigma^{2} X^{intercal}(XX^{intercal})^{-1}X$$
since $Var(Y) = Var(varepsilon) = sigma^{2} I_{n}$
Now that $(hat{y}/se(hat{y})^{2} leqslant F_{alpha/2, 1, n-p}$, I used the idea that square of a t-distributed r.v. follows a 1,n-p F-distribution. In fact the right hand side with misuse of notation I mean the F-value with 1,n-p df and $alpha/2$ cutoff.
My major question is about left hand side, if I similarly plug in the expression, this is what I got eventually:
$$hat{theta}^{intercal}XX^{intercal}hat{theta} frac{1}{sigma^{2}} (X^{intercal}(XX^{intercal})^{-1}X)^{-1} \
= frac{1}{sigma^{2}} Y^{intercal}(X^{intercal}(XX^{intercal})^{-1}X) Y (X^{intercal}(XX^{intercal})^{-1}X)^{-1}$$
Of course the confidence set is the set where this expression is bounded by that F-value.
Am I totally wrong from the beginning? Or there is something I can do to fix my calculation? It just doesn't look right at this moment.
(Kind of have idea now. Please allow me some time to update it when I finish my homework).
I think it should be:
$$mathbb{E}(Y) = X^{intercal}theta^{*} in { X^{intercal} theta in mathbb{R} ^ {n}: || (XX^{intercal}) ^{ frac{1}{2}} (theta - hat{theta}||^{2} leqslant frac{p}{n-p} S(hat{theta}) Q(1-alpha, F(p,n-p) ) }$$
where $S(hat{theta}) = ||Y - X^{intercal}hat{theta}||^{2}$ and $hat{theta} = (XX^{intercal})^{-1}XY$. $Q$ is the quantile inverse function.
linear-regression
linear-regression
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
edited Dec 7 '18 at 6:31
adosdeci
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
asked Dec 7 '18 at 4:13
adosdeciadosdeci
376
376
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