# Basic R-Squared in Poisson Regression

I have read one cannot/should not calculate the basic R-Squared used in linear regression for a Poisson generalized linear regression model.

It is logical to me that one cannot determine the basic R-Squared for a logistic regression, since the R-Squared gives the proportion of explained variance of the dependent variable and the dependent variable is not metric in a logistic regression.

However, the dependent variable in a Poisson regression is clearly metric. It might not be fully continuous, but that should not be a requirement to calculate the variance. So I do not understand why one should not calculate the basic R-Squared for a Poisson regression model.

## 2 Answers

A Poisson regression is nonlinear. Yes, it’s called a generalized linear model, but it has a nonlinear link function. Poisson regression is not linear.

When a regression is nonlinear, the residuals and predictions and not orthogonal. This means that the total sum of squares does not decompose into the sum of squares of the regression and the sum of squares of the residuals. There is a third term.

Therefore, $$R^2=SSReg/SSTotal$$ does not represent the proportion of variance explained.

This has to do with the “other” I mention here: Nonlinear quantile regression SSReg analogue.

• Ok, thx a lot for the quick answer! – Benkyozamurai Jun 28 '20 at 14:52
• I think there are some generalizations of the r-squared for logistic regression expressed in terms of ratios of chi-square statistics from the log likelihood. I'm not confident, but my gut says that since the poisson is in the same family, those expressions might also be of some use. – Demetri Pananos Jun 28 '20 at 17:13
• en.wikipedia.org/wiki/Logistic_regression#Pseudo-R2s – Dave Jun 28 '20 at 17:22

Apparently for a Poisson regression model no R-square but a so-called deviance statistic has been proposed and is computed with many software packages.

Further to quote:

The deviance is a measure of how well the model fits the data - if the model fits well, the observed values $$Y_{i}$$ will be close to their predicted means $$\mu_{i}$$, causing both of the terms in D to be small, and so the deviance to be small.

The cited formula is:

For a fitted Poisson regression the deviance is equal to

$${D = 2 \sum^{n}_{i=1} \{ Y_{i} \log(Y_{i}/\mu_{i}) - (Y_{i}-\mu_{i}) \}}$$

where

$${\mu_{i} = \exp(\hat{\beta}_{0} + \hat{\beta}_{1}X_{1} + ... + \hat{\beta}_{p} X_{p})}$$

denotes the predicted mean for observation i based on the estimated model parameters.