Elliptical Confidence Set Calculation












1












$begingroup$


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







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    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







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      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







      share|cite|improve this question











      $endgroup$




      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






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 7 '18 at 6:31







      adosdeci

















      asked Dec 7 '18 at 4:13









      adosdeciadosdeci

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