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Multiple Linear Regression


Date: 01/23/97 at 08:21:22
From: dulud
Subject: Multiple linear regression

Bonjour from Paris,

I am looking for the complete set of formulas for multiple linear 
regression.

Can you help me?

Thanks


Date: 01/27/97 at 09:10:37
From: Doctor Mitteldorf
Subject: Re: multiple linear regression

Greetings from Philadelphia!

If you understand single-variable linear regression, then multiple
regression is just the same thing with matrices and vectors where you
had numbers before.  

Here are the formulas, first for single variable:

Say you have a collection of points (x,y), and you want the best line
through them.  The line will be: 

              y = ax + b

where a = (<xy>-<x><y>) / (<x^2>-<x>^2)
  and b = <y> - a<x>

The correlation coefficient r is given by:

      r = (<xy>-<x><y>) / sqrt{ (<x^2>-<x>^2) * (<y^2>-<y>^2) }

In the above, the notation <xy> means "average value of xy": in other 
words, for each point, multiply x for that point times y for that 
point, add up all the products, and divide by the number of points.  
Similarly, <x^2> is the mean value of x^2.  You'll recognize the 
denominator of the expression for a as the variance of x.  So you 
could rewrite formulas as:
 
    a = (<xy>-<x><y>) / var(x)
    r = a * sqrt{ var(x) / var(y) }

Now for the multivariate version of the formulas, you must think of
x as a vector, but y is still a scalar.  y is a function of multiple 
variables which together are called x.  I'll use capital letters for 
vectors and "." for the dot product of two vectors:

             A.X means A[1]*X[1] + A[2]*X[2] + ...

We're still looking for a linear relationship between x and y, and 
now it's of the form y = A.X + b.  Since X is a vector of n numbers, 
we look for n coefficients of proportionality, and make scalar a into 
vector A. 

In the formula for A, the numerator becomes:

               (<Xy>-<X><y>)

This is easy to interpret.  X is a vector, y is a scalar.  Every 
component of X is multiplied by the scalar y.

But the denominator takes a little more thought.  What do we mean by:

                        (<XX> - <X><X>)

This is a 2d rank tensor, which looks like a square matrix.  If X has
n components, then <XX> has n^2 components.  The (i,j) component of 
this object is made by averaging <X[i]X[j]> over all the points in 
your sample.  <X><X> is the matrix that you make just by multiplying 
out all possible combinations of the vectors X.  The (i,j) component 
of <X><X> is given by <X[i]><X[j]>; in other words, separately average 
the X[i] components for all points and the X[j] components for all 
points, then just multiply those two together.

<XX> and <X><X> are both matrices.  Subtract one from the other to 
get the "denominator" matrix corresponding to var(X).

Then you must "divide" this matrix into the numerator vector.  The 
way to do this is to invert the matrix, then multiply.  Symbolically, 
you could write the steps this way:

     Let vector V = (<Xy>-<X><y>) 
     Let matrix M = (<XX>-<X><X>)

     Then let vector A = Inv(M) * V
     Also, r^2 = Inv(M) * V

The inverse of the matrix M is another matrix.  The product of that
matrix with a vector is another vector.

Finally, b is just a scalar, and the formula for b is just as before, 
with A and X becoming vectors:

     b = <y> - A.<X>

I hope this helps.  Don't hesitate to write again if any part is still 
not clear.

-Doctor Mitteldorf,  The Math Forum
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