Mixing
Deriving the least squares estimate of $beta_{k-1}$
$begingroup$
Let $y_i=Sigma^k_{j=0} x_{ij} beta_j+epsilon_i$
$epsilon_i$ is $NID(0,sigma^2)$ and $x_{ij}, i=1,...,n, j=0,...,k$ is the $(i,j)^{th}$ elelement of the $n times (k+1)$ matrix $X$, which is of full rank, and $beta_o,...,beta_k$ are constants. Also $g_{ij}$ is the $(i,j)^{th}$ element of the matrix $(X'X)$. The first column of $X$ corresponds to $j=0$
Specifying the value of $M_{kj}$, show that the least squares estimate of $beta_{k-1}$ is given by $Sigma^n_{i=1}y_iSigma^k_{j=o}x_{ij}M_{kj}$
I have no idea how to approach this one. Please can someone give me some pointers? I am familiar with getting estimates of $beta$ where there are only two (ie $beta_0, beta_1$) but not when there are lots. Presumably matrix algebra is involved? I really don't know where to start.
statistics self-learning estimation least-squares
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
$endgroup$
add a comment |
$begingroup$
Let $y_i=Sigma^k_{j=0} x_{ij} beta_j+epsilon_i$
$epsilon_i$ is $NID(0,sigma^2)$ and $x_{ij}, i=1,...,n, j=0,...,k$ is the $(i,j)^{th}$ elelement of the $n times (k+1)$ matrix $X$, which is of full rank, and $beta_o,...,beta_k$ are constants. Also $g_{ij}$ is the $(i,j)^{th}$ element of the matrix $(X'X)$. The first column of $X$ corresponds to $j=0$
Specifying the value of $M_{kj}$, show that the least squares estimate of $beta_{k-1}$ is given by $Sigma^n_{i=1}y_iSigma^k_{j=o}x_{ij}M_{kj}$
I have no idea how to approach this one. Please can someone give me some pointers? I am familiar with getting estimates of $beta$ where there are only two (ie $beta_0, beta_1$) but not when there are lots. Presumably matrix algebra is involved? I really don't know where to start.
statistics self-learning estimation least-squares
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
$endgroup$
1
$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02
add a comment |
$begingroup$
Let $y_i=Sigma^k_{j=0} x_{ij} beta_j+epsilon_i$
$epsilon_i$ is $NID(0,sigma^2)$ and $x_{ij}, i=1,...,n, j=0,...,k$ is the $(i,j)^{th}$ elelement of the $n times (k+1)$ matrix $X$, which is of full rank, and $beta_o,...,beta_k$ are constants. Also $g_{ij}$ is the $(i,j)^{th}$ element of the matrix $(X'X)$. The first column of $X$ corresponds to $j=0$
Specifying the value of $M_{kj}$, show that the least squares estimate of $beta_{k-1}$ is given by $Sigma^n_{i=1}y_iSigma^k_{j=o}x_{ij}M_{kj}$
I have no idea how to approach this one. Please can someone give me some pointers? I am familiar with getting estimates of $beta$ where there are only two (ie $beta_0, beta_1$) but not when there are lots. Presumably matrix algebra is involved? I really don't know where to start.
statistics self-learning estimation least-squares
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
$endgroup$
Let $y_i=Sigma^k_{j=0} x_{ij} beta_j+epsilon_i$
$epsilon_i$ is $NID(0,sigma^2)$ and $x_{ij}, i=1,...,n, j=0,...,k$ is the $(i,j)^{th}$ elelement of the $n times (k+1)$ matrix $X$, which is of full rank, and $beta_o,...,beta_k$ are constants. Also $g_{ij}$ is the $(i,j)^{th}$ element of the matrix $(X'X)$. The first column of $X$ corresponds to $j=0$
Specifying the value of $M_{kj}$, show that the least squares estimate of $beta_{k-1}$ is given by $Sigma^n_{i=1}y_iSigma^k_{j=o}x_{ij}M_{kj}$
I have no idea how to approach this one. Please can someone give me some pointers? I am familiar with getting estimates of $beta$ where there are only two (ie $beta_0, beta_1$) but not when there are lots. Presumably matrix algebra is involved? I really don't know where to start.
statistics self-learning estimation least-squares
statistics self-learning estimation least-squares
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
edited Jan 21 at 20:20
Maths Barry
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
asked Jan 19 at 12:06
Maths BarryMaths Barry
438
438
1
$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02
add a comment |
1
$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02
1
1
$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02
add a comment |
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$begingroup$
What is $M_{kj}$?
$endgroup$
– V. Vancak
Jan 19 at 12:36
$begingroup$
Great question - I don't know,
$endgroup$
– Maths Barry
Jan 19 at 13:02