Suppose that you have simple two variable model:
Y=b0+b1X1+e
The least squares estimator for the slope coefficient, b1 can be obtained with
b1=cov(X1,Y)/var(X1)
the intercept term can be calculated from the means of X1 and Y
b0=mean(Y)-b1*mean(X1)
In a larger model, Y=b0+b1X1+b2X2+e
the estimator for b1 can be found with
b1=(cov(X1,Y)var(X2)-cov(X2,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2)
to find b2, simply swap the X1 and X2 terms in the above to get
b2=(cov(X2,Y)var(X1)-cov(X1,Y)cov(X1,X2))/(var(X1)var(X2)-cov(X1,X2)2)
Find the intercept with
b0=mean(Y)-b1*mean(X1)-b2*mean(X2)
Beyond two regressors, it just gets ugly.
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consider a matrix Aobtain a transformation which will diagonalize the matrix.Whatare the coordinates of an arbitrary vectora=traspos(x,y,z)with respect to the basis set which diagonalizes A?
Assume two matrices are to be multiplied (the generalization to any number is discussed below). Arithmetic process of multiplying numbers (solid lines) in row i in matrix A and column j in matrix B, then adding the terms (dashed lines) to obtain entry ij in the final matrix.If A is an n × m matrix and B is an m × p matrix, the matrix product ABwhere each i, j entry is given by multiplying the entries Aik (across row i of A) by the entries Bkj (down column j of B), for k = 1, 2, ..., m, and summing the results over k:Thus the product AB is defined only if the number of columns in A is equal to the number of rows in B, in this case m. Each entry may be computed one at a time.Refer to link below for more details.
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The graphical method is often approximate but can be applied to any function. If done on a computer, the region surrounding the solution can be enlarged to obtain more accurate estimates. A numerical method will give an exact result is an analytical solution is possible. If not, the solution will depend on the numerical method used and, sometimes, the starting "guesstimate".