# Negative-Binomial Method of moments with an offset

Given the method-of-moments approach to estimate the parameters of the NB-2 distribution $$\mu$$ and $$\phi$$: $$\mu = \bar{y}$$ $$\phi = \frac{\bar{y}^2}{s^2 -\bar{y}}$$

How can this be extended to include an offset/rate parameter (i.e., where $$\log(\mu) = X'\beta + \log(t)$$)?

• What is $t$? It looks like you might want to fit a negative binomial regression, Hilbe's eponymous textbook is the go-to resource. If I understand your problem correctly, it would be a question of log-transforming $t$ and then constraining the parameter to be $1$. – Stephan Kolassa May 3 at 9:00
• $t$ is the offset/exposure variable, for when the outcome y is the number of occurrences per unit $t$. More information in this post. The estimators in my post are equivalent to an intercept-only negative-binomial regression, but I'm trying to extend this to an intercept-and-offset model – AndrewrJ May 3 at 9:17
• Thanks. That link says pretty much the same thing as my comment: feed the exposure into your model (log transform it since you are using a log link), then fix the parameter to be $1$. Is that what you are looking for? – Stephan Kolassa May 3 at 9:19
• Not quite. I know that it can be fit in a regression model, but I'm trying to adapt the method of moments approach so I can estimate the distribution parameters without needing maximum-likelihood/optimisation – AndrewrJ May 3 at 9:31
• OK, thanks, that helps. When you write in your comment that you have an intercept-only model, that is $\log\mu=\beta_0+\log t$, it looks like you are trying to solve two equations (for the observed mean $\bar{y}$ and the observed variance $s^2$) using a single unknown $\beta_0$ plus an overdispersion parameter. So you can simply solve the equation for the mean to give you $\hat{\beta}_0$, then set the overdispersion parameter to whatever you observe. Would that work? – Stephan Kolassa May 3 at 9:39

Your model is $$\log\mu=\beta_0+\log t$$, since for an offset (which you log-transform since you are working with a log link) you constrain the corresponding parameter to be $$1$$. On the original scale (where we want to match moments), this means for the $$i$$-th observation

$$\mu_i = t_i\cdot\exp\beta_0.$$

Since you want to match moments, you can estimate $$\beta_0$$ so that $$\hat\mu$$ matches the mean of the observations, or equivalently by taking averages,

$$\bar y = \bar t\cdot\exp\beta_0,$$

so you set

$$\hat\beta_0 = \log\big(\frac{\bar y}{\bar t}\big)=\log\bar y-\log\bar t.$$

Now, for a negative binomial model, you have overdispersion, or

$$E(y_i-\mu_i)^2 =\mu_i+\frac{\mu_i^2}{\phi}$$

for some overdispersion parameter $$\phi>0$$, which is just a reformulation of your second formula, or

$$\phi = \frac{\mu_i^2}{E(y_i-\mu_i)^2-\mu_i}.$$

A possible moments estimator would then be

$$\hat\phi = \frac{\sum_{i=1}^n\hat\mu_i^2}{\sum_{i=1}^n(y_i-\hat\mu_i)^2-\hat\mu_i}.$$

There is likely some bias involved here, so I would recommend you think about "real" maximum likelihood estimation. Hilbe's textbook Negative Binomial Regression is very helpful.

• Excellent! Thanks so much for the quick help! – AndrewrJ May 3 at 11:51
• Sorry, quick follow-up. Where you wrote for overdispersion: $E(y_i - \mu_i)^2 = \mu_i(1+\frac{1}{\phi})$, shouldn't the variance be defined as: $\mu_i(1+\frac{\mu_i}{\phi})$? Or are you referring to a different quantity? – AndrewrJ May 4 at 6:45
• Yes, you are right, I'm sorry, my bad, I got confused. There are too many different parameterizations of the negbin... I edited my answer. – Stephan Kolassa May 4 at 15:29
• No worries at all, your help is very much appreciated – AndrewrJ May 4 at 23:27