how to define link function in glmfit

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Grace
Grace 2012-2-14
Dear all,
I am employing glmfit to solve Generalized Linear Models, and the link function I choose is Weibull, which is also known as Gompertz,
y=1-exp(-exp(a+b*log10(x))) ;
The guiding information I find in help reads "cell array of the form {FL FD FI}, containing three function handles, created using @, that define the link (FL), the derivative of the link (FD), and the inverse link (FI)." And I try to write some code:
[b,dev,stats]=glmfit(X,Y,'normal','link',{@(y)(10^((log(-1og(1-y))-a)/b)),@(y)(10.^((log(-1og(1-y))-a)/b)/(b*(-1og(1-y)*(1-y)))),@(y)(1-exp(-exp(a+b*log10(y))))})
There are always errors. I am not familiar with function handles. And I want to know if y should be a matrix, how to define the link function in the right way?
Thanks a lot.

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回答（1 个）

Tom Lane
Tom Lane 2012-2-14
Did you really mean to specify the normal distribution? This type of link is more commonly used with the binomial distribution. In fact, it is the complementary log-log link using log data:
>> x = 10*rand(1000,1);
>> a = 1; b = -2;
>> p = 1-exp(-exp(a+b*log10(x)));
>> y = binornd(1,p);
>> glmfit(log10(x),y,'binomial','link','comploglog')
ans =
0.8951
-1.8827
This is possible with the normal distribution also, but there is a risk of problems as the fitted values approach 0 and 1. Here's an illustration on a well-conditioned problem:
>> x = 2+2*rand(1000,1);
>> p = 1-exp(-exp(a+b*log10(x)));
>> y = normrnd(p,.01);
>> glmfit(log10(x),y,'normal','link','comploglog')
ans =
0.9999
-1.9990

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