How to find the elasticity of a negative binomial regression when the independent variables are numeric, categorical, or dummy variables?

Edit: For example,

m1 <- glm.nb(No_of_accidents ~ JunctionType + CollisionType)

This is a negbin regression if I don't get it wrong. Now after fitting the model with data, I get two coefficients and an intercept. But the coefficients cannot be explained like the linear regression model. I can't say that if the JunctionType is a roundabout, then the no of accidents will increase by 2 times the base condition (assuming that there exists a junction type called roundabout and the beta = 2 for roundabout after fitting the model. In such a case, how do I quantify the relation between JunctionType (X) and no. of accidents (Y)? I hope it clarifies my question.

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    $\begingroup$ What do you mean by the "elasticity of a negbin regression"? $\endgroup$ – Stephan Kolassa Oct 5 '20 at 4:57
  • $\begingroup$ I mean the effect of a 1% change in the X variable on the expected value of Y considering X as variables and Y the dependent variable if I fit the variables in a negbin regression. For example, m1 <- glm.nb(NoOfAccidents ~ JunctionType + CollisionType , data = Data) in this model m1, the coefficients values will give an indication of increase or decrease of Y with X. But how can explain the change in Y with X in quantitative terms? $\endgroup$ – sophie Oct 6 '20 at 16:41
  • $\begingroup$ Thank you, that clarifies matters. I have voted to reopen, and if it is reopened, I'll write an answer. $\endgroup$ – Stephan Kolassa Oct 7 '20 at 7:42
  • $\begingroup$ Hi @StephanKolassa, the question has been reopened. I am looking forward to hearing your answer. Thank you. $\endgroup$ – sophie Oct 9 '20 at 11:15

First off, a pointer to literature: Hilbe, Negative Binomial Regression is the standard textbook on negbin regression. Very much recommended.

Now, the standard way of setting up a negative binomial regression (there are less common others) is via a log link, i.e., the mean $\mu$ is parameterized via a design matrix $X$ and coefficients $\beta$ as $$ \log\mu = X\beta, \text{ or } \mu=\exp(X\beta). $$

I'll not distinguish between the true parameters $\beta$ and the estimates $\hat{\beta}$ in the following.

We can now distinguish different kinds of elasticity, depending on whether we are looking at a categorical/dummy predictor, or a numerical one.

Change in a dummy predictor

Assume we have $\mu_1=\exp(x_1\beta)$ and $\mu_2=\exp(x_2\beta)$, two fitted means, where the design matrix row vectors $x_1$ and $x_2$ differ in that one dummy flips from $0$ (in $x_1$) to $1$ (in $x_2$). That is, $x_2-x_1$ is a vector of $0$s and a single $1$ in the place at which the flipping predictor sits. Assume that this is the $j$-th entry. Then

$$\frac{\mu_2}{\mu_1} = \frac{\exp(x_2\beta)}{\exp(x_1\beta)} = \exp\big((x_2-x_1)\beta\big) = \exp(\beta_j). $$

That is, flipping the $j$-th dummy from $0$ to $1$ will multiply the fitted mean by $\exp(\beta_j)$.

Change in a categorical predictor

Categorical predictors with $k$ factor levels are (again: commonly) represented internally as $k-1$ dummy predictors, for which the calculation above holds. Thus, switching from the reference category to the $j$-th (non-reference) category will change the fitted mean multiplicatively by $\exp(\beta_j)$. Changing from the $k$-th (non-reference) category to the $j$-th (non-reference) category will change the mean multiplicatively by $\exp(\beta_j-\beta_k)$.

Additive change in a numerical predictor

We again have to deal with design matrix row vectors $x_1-x_2$, which will again be a vector of $0$s except for the entry in which the changing predictor sits. So we can just assume that we have a single predictor that first takes a value $x$ (for a fitted mean of $\mu_1=\exp(x\beta)$) and then changes to $x+\Delta x$ (for a fitted mean of $\mu_2 = \exp\big((x+\Delta x)\beta\big)$). We obtain

$$ \frac{\mu_2}{\mu_1} = \frac{\exp\big((x+\Delta x)\beta\big)}{\exp(x\beta)} = \exp(\Delta x\beta).$$

So changing a predictor by an additive $\Delta x$ will yield a multiplicative change of $\exp(\Delta x\beta)$ in the fitted mean, where $\beta$ is the coefficient corresponding to the changing predictor.

Multiplicative or percentage change in a numerical predictor

Assume a predictor's value changes multiplicatively, from $x$ to $cx$. As above, we get

$$ \frac{\mu_2}{\mu_1} = \frac{\exp(cx\beta)}{\exp(x\beta)} = \exp\big((c-1) x\beta\big).$$

So in this case, the multiplicative change depends on the initial value $x$ of the predictor, in contrast to the effect of an additive change in the predictor, which per above does not depend on this initial value $x$.

Infinitesimal change in a numerical predictor

The economic definition of elasticity is the change in the effect as the predictor changes by an infinitesimal amount:

$$ \frac{\partial\mu/\mu}{\partial x/x} = \frac{\partial\mu}{\partial x}\frac{x}{\mu}. $$

For $\mu=\exp(X\beta)$, we have

$$ \frac{\partial\mu}{\partial x_j} = \frac{\partial}{\partial x_j}\exp(X\beta) = \beta_j\exp(X\beta) = \beta_j\mu, $$


$$ \frac{\partial\mu}{\partial x_j}\frac{x_j}{\mu} = \beta_j\mu\frac{x_j}{\mu} = \beta_jx_j. $$

Thus, the elasticity of $\mu$ with respect to the $j$-th predictor is $\beta_jx_j$ - so it again depends on the value $x_j$ of the $j$-th predictor at which we calculate the elasticity.

  • $\begingroup$ Thank you so much for your answer. It clarifies all my confusion. $\endgroup$ – sophie Oct 11 '20 at 7:59

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