Deriving the least squares estimate of $beta_{k-1}$












0












$begingroup$


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







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$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 21 at 20:20







Maths Barry

















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














  • 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










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