How to compute the regression coefficient in Matlab with exp and ln?

I would like to compute the regression coefficients `a` and `b` for my data using this equation:

    y=exp(a * ln(1 - t / h) + b * ln(1 - t / t1))

and this data(example):

             t = [1,2,5,4,8,7,5,1,2,5,4,1,2,1,5]
             t1 = [1,2,4,4,5,3,7,5,6,8,7,1,2,1,5]
             h = [1,2,3,2,9,6,8,3,6,7,4,5,2,1,5]
             y = [1,2,1,4,4,6,5,8,5,7,3,1,4,1,5]

but I do not know how to include `exp` and `ln`. PLease help

[dpb]

Bad data...

>> log(1-t./h)
ans =
-Inf  -Inf   -0.4055 + 3.1416i  0 + 3.1416i  -2.1972          
-1.7918 + 3.1416i  -0.9808  -0.4055  -0.4055   -1.2528
-Inf             -0.2231   -Inf    -Inf      -Inf

If you can get a reasonable set of data for which (1-t/h) and (1-t/t1) are both >0 then you can try nlinfit but I suspect it'll be very difficult to separate out the two additive exponential terms.

Or, you can transform and solve for the log terms w/ OLS using

ln y = a ln(1-w) + b ln(1-x) where w=t/h and x=t/t1

with likely similar difficulties.

[san der]

I have a reasonable set of data for which (1-t/h) and (1-t/t1) are both >0. Thanks why shall I use the nonlinear fit? What you mean by this:it'll be very difficult to separate out the two additive exponential terms

[dpb]

a) Well, 'cuz it's a nonlinear equation, maybe? :)

b) The difficulty in estimating your a and b separately is that since they're an additive term in the exponential they combine as a single argument.

It would help if the data were collected for a design matrix that has those two terms evaluated independently altho I've not taken the time to try to work out a specific design for the case.

[san der]

in fact the original equation is like this {ln y = a ln(1-w) + b ln(1-x) where w=t/h and x=t/t1} so I need to find a and b based on initial values of y then I use a and b to find y. but then result will reflect ln(y) not y. what do you think?

Answer 1

[Keith Dalbey]

let t, h, t1, and y be column vectors then

if true
  % code
  G=[log(1-t./h) log(1-t./t1)];  ab=(G'*G)\(G'*log(y))
end

[Matt J]

ab=(G'*G)\(G'*log(y))

This should probably just be

 ab=G\log(y);

There's no need for the system of equations to be made square.

[Keith Dalbey]

perhaps we should tell him that I did least squares, and that you used the overloaded \ operator which does the equivalent of a pseudo inverse for non-square matrices. The answers should be identical for small matrices, for large ones your approach would be more accurate (less round off error), mine would likely be faster.

[san der]

Thanks but this will produce ab! I want a and b separately

[dpb]

Did you try it?