An Example

Consider the Poisson regression model (King, 1989: Chapter 5). The stochastic component is Poisson:

$$Y \sim \frac{e^{-\lambda}\lambda^y}{y!}$$ (20)

and the systematic component is exponential:

$$\lambda = \exp(X\beta)$$ (21)

The likelihood function is then

$$L(\beta\vert y) = \prod_{i=1}^n \frac{\exp(-e^{X\beta})\exp(X\beta)^y}{y!}$$ (22)

The fundamental variability in the Poisson model happens to equal the expected value, so that

$$V(Y\vert\lambda) = \sum_{y=1}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!} - \left(\sum_{y=1}^\infty y\frac{e^{-\lambda}\lambda^y}{y!} \right)^2$$ (23)

$$= (\lambda+\lambda^2)-\lambda^2$$

$$= \lambda$$

Where the summation is used in place of integration since the distribution is discrete. The vector $\lambda$ is $(N\times 1)$. By the usual no autocorrelation assumptions of the Poisson regression model, the covariances (based on fundamental variability) between different predicted values are zero. Thus, one can write the $(N\times N)$ fundamental variance matrix as $\lambda I$, where $I$ is an $(N\times N)$ identity matrix. To estimate this fundamental variability, one would use the $(N\times N)$ matrix $\exp(X^pb)I$.

To calculate the estimation variability, we need the first derivative matrix

$$g'(\beta) = \frac{\partial\exp(X\beta)}{\partial\beta}$$ (24)

$$= X\cdot e^{X\beta}$$

Note that this is a slight abuse of standard mathematical notation, used in order to make the transition to computation easier. $X$ is $(N\times k)$ and $e^{X\beta}$ is $(N\times 1)$. The notation ``$\cdot$'' in this equation refers to multiplying each column of $X$ by $e^{X\beta}$ element by element.²

Thus, the estimated variance matrix of the $k$-vector of predicted values $\hat Y^p$ is then as follows:

$$V(\hat Y^p) = (X^p\cdot e^{X^pb})\widehat{V(b)}(X^p\cdot e^{X^pb})'$$ (25)

One would add $\hat\lambda$ or, equivalently, $\exp(X^pb)I$ to the diagonal elements of this matrix in order to calculate the total variability.