Regression Analysis of Four-Parameter Data

Date: 01/21/2000 at 00:47:31
From: David Choy
Subject: Regression Analysis of a Four-Parameter Logistic Equation

Dr. Math -

I am stuck. I am trying to determine if a biological relationship can 
be described by what has been referred to me as the Four Parameter 
Logistic Equation:

     y = [(A-D)/(1+{x/C}^B)]+D

The trouble here is that I don't have the mathematics savvy, with my 
biosciences background, to do the regression analysis to "fit" my 
data, and none of my biology texts makes reference to this model. How 
do I perform a regression analysis on my experimental data to 
determine A, B, C and D?

-David Choy

Date: 01/21/2000 at 05:27:04
From: Doctor Mitteldorf
Subject: Re: Regression Analysis of a Four-Parameter Logistic Equation

Dear David,

I'm a physicist who's moved into population genetics, so I'm glad to 
help.

If you understand single-variable linear regression, then multiple 
regression is just the same thing with matrices and vectors where you 
had numbers before. If you don't understand single-variable 
regression, you should find a statistics text and study the derivation 
of linear regression.

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 the 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 relation 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 second 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
  http://mathforum.org/dr.math/