Solved – use glm algorithms to do a multinomial logistic regression

generalized linear modellogistic

I am using spotfire (S++) for statistical analysis in my project and I have to run multinomial logistic regression for a large data set. I know best algorithm would have been mlogit, but unfortunately that is not available in s++. However, I have an option of using glm algorithm for this regression. I want to clarify two things here:

1.Is my understanding correct that glm also can be used to run Multinomial Logistic Regression?

  1. If answer to previous question is yes, then what parameters should be used in glm algo?

Thanks,

Best Answer

Yes, with a Poisson GLM (log linear model) you can fit multinomial models. Hence multinomial logistic or log linear Poisson models are equivalent.

You need to see random counts $y_{ij}$ as Poisson random variables with means $μ_{ij}$ and specify the following the following log-linear model

$\log(μ_{ij}) = o + p_i + c_j + x_iβ_j$

To get a multinomial logit model the parameters are:

A parameter $p_i$ for each multinomial observation, for example individuals or group. This assures exact reproduction of the multinomial denominators and actually establishes the equivalence of Poisson and multinomial model. They are fixed in the multinomial likelihood, but random in the Poisson likelihood.

A parameter $c_j$ for each response category. This way the counts can be different for each response category and the margins can be non-uniform.

What you are really interested in are the interaction terms $x_iβ_j$ that represent the effects of $x_i$ on the log-odds of response $j$.

The log-odds can be simply calculated by $\log(μ_{ij}/μ_{ik}) = (c_j-c_k) +x_i(β_j-β_k)$. It is the log odds that observation i will fall in response category j relative to the response category $k$.

Then, the parameters in the multinomial logit model (denoted in latin letters) can be obtained as differences between the parameters in the corresponding log-linear model, i.e. $a_j = α_j-α_k$ and $b_j = β_j-β_k$.