# Why does Poisson regression not have a closed form solution?

I am trying to understand GLMs by trying to run them on my own on some Poisson data. My understanding is that if I have my Y values and X values, then using the log-link function, $$log[Y] = mX+b$$ for some $$m, b$$. Now the question is to find the coefficients which best explain my Y variable. When I read about it online, they will exponentiate this expression to optimize the likelihood over $$m, b$$. My question is, why can this not be done through simple linear regression on the $$log[Y]$$ values and then exponentiated after fitting a simple linear curve? Does this not result in a simple closed-form solution? Is there something wrong about this approach?

• GLMs consider $\log \mu$, where $\mu$ is the expected value of $Y$. That's different than considering $\mathbb{E}[\log Y]$, which has the expectation and the logarithm interchanged.
– Ben
Jun 15, 2022 at 15:44
• What would you do with the logarithm upon observing a count of zero, which always has a positive probability for any Poisson distribution? For various accounts of what Poisson regression actually is, please search this site. stats.stackexchange.com/questions/69820 answers the same question for logistic regression.
– whuber
Jun 15, 2022 at 16:08

You could use a linearized form of the equation and apply ordinary linear regression. But, this makes two errors:

• You ignore non-homogeneity of the error distribution. The variance in $$Y$$, and also the variance in the transformed variable $$\log(Y)$$, is not the same for different values of $$X$$. You need to perform the linear regression with weights $$w_i =\hat{Y}_i$$ to correct for this, where $$\hat{Y}_i$$ is the estimated value (more about that later).

• A non-linearity in the effect of errors on the response when a link function is applied to the response. (ie. the model of the error distribution is for $$Y$$ and not for $$\log(Y)$$ and that needs to be corrected).

This is a bit the same as the lognormal distribution (the exponetiation of a normal distribution with mean $$\mu$$ and deviation $$\sigma$$) not having a mean $$\exp(\mu)$$ but instead a mean $$\exp(\mu + 0.5 \sigma^2)$$.

That is an additional correction, and instead of using $$Y_i^\prime = \log(Y_i)$$ you use $$Y_i^\prime = \log(Y_i) + (Y_i-\hat{Y_i})/\hat{Y_i}$$.

So to find the solution $$\hat{Y}_i$$ with a linear regression, you need to use two corrections that depend on the solution $$\hat{Y}_i$$ itselve. That is why there is no closed form and an itterative procedure is used. We use a starting point, assume some initial value as solution and use that to compute a new value. Based on that new value we compute a new value, and so on.

• I understand. I was assuming log[Y] = mx + b, where errors in the log[Y] term are typical gaussian noise (as in linear regression). In fact it is log[E[Y]] = mx+b .
– bGe
Jun 16, 2022 at 3:27