Expected prediction error
$begingroup$
In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
begin{equation}
mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
end{equation}
i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.
Thank you for any help.
statistics
asked Dec 24 '18 at 23:05
yjntyjnt
113
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add a comment |
$begingroup$
In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
begin{equation}
mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
end{equation}
i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.
Thank you for any help.
statistics
asked Dec 24 '18 at 23:05
yjntyjnt
113
$endgroup$
add a comment |
$begingroup$
In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
begin{equation}
mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
end{equation}
i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.
Thank you for any help.
statistics
asked Dec 24 '18 at 23:05
yjntyjnt
113
$endgroup$
In the case of linear regression where $x_0$ is an input vector, and $Y_0=f(x_0)+epsilon_0$, i don't understand this formulation of the expected prediction error, with : $hat{f}(x_0)=x_0 ^That{beta}$ and $f(x_0)=x_0 ^Tbeta$
begin{equation}
mathbb{E}big[(Y_0 - hat{f}(x_0))^2big]=sigma^2 + mathbb{E}big[(x_0 ^That{beta} - f(x_0))^2big]
end{equation}
i don't get what is random in $(x_0 ^That{beta} - f(x_0))^2$, hence why we're taking its expectation.
Thank you for any help.
statistics
statistics
asked Dec 24 '18 at 23:05
yjntyjnt
113
asked Dec 24 '18 at 23:05
yjntyjnt
113
asked Dec 24 '18 at 23:05
yjntyjnt
113
asked Dec 24 '18 at 23:05
yjntyjnt
113
asked Dec 24 '18 at 23:05
yjntyjnt
113
113
add a comment |
add a comment |
1 Answer
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$begingroup$
$hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.
Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
$endgroup$
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
add a comment |
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1 Answer
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1 Answer
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active
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active
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$begingroup$
$hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.
Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
$endgroup$
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
add a comment |
$begingroup$
$hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.
Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
$endgroup$
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
add a comment |
$begingroup$
$hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.
Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
$endgroup$
$hat beta$ is random in $( x_0^T hat beta - f(x_0 ))^2$.
Remember, $hat beta$ is the estimator of $beta$, calculated from your training data. Since your training data is random, $hat beta$ is random too.
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
answered Dec 24 '18 at 23:23
Kenny WongKenny Wong
19.3k21441
19.3k21441
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
add a comment |
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
$begingroup$
oh ok it's clearer now. Thank you
$endgroup$
– yjnt
Dec 24 '18 at 23:24
add a comment |
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