How to use inverse information matrix and Delta method to find sample variation?

Question

Explain how the inverse of the information matrix and the Delta Rule be used to generate an approximate sampling variance for the estimated proportion exp(beta 0 + beta 1)/(1+exp( beta 0 + beta 1))?

I believe I am needing to find the estimated proportion of the exposed group/when x=1. I think I need to take the inverse logit with respect to beta 1 and use this in the Delta rule somehow.

Any help would be appreciated on how to do this.

Answer 1

[The picture does not show the complete setup, but it seems to be a logistic regression problem with $y \mid x \sim \text{Bernoulli}(\pi(x))$ where $\pi(x) = \exp[\beta_0 + \beta_1 x] / (1 + \exp[\beta_0 + \beta_1 x])$, and $x$ is a binary predictor. The model assumption is necessary to ensure that the regularity conditions of CLT for the MLE holds.]

First, we need the asymptotic distribution of the MLE $\hat \beta = (\hat \beta_0, \hat \beta_1)$. If $\beta$ is the true value of the regression parameter vector, then from the CLT of the MLE (Wikipedia) we have (the regularity conditions justifying the CLT holds under a logistic regression framework) $$ \hat \beta \stackrel{a}{\sim} N\left(\beta, I(\beta)^{-1}\right) $$

where $X \stackrel{a}{\sim} f$ means $X$ is approximately distributed as $f$, and $I(\beta)^{-1}$ is as given. (Derivation of this inverse Fisher information matrix is cumbersome. Fortunately, the final form of the matrix is given here, saving us a substantial amount of work.)

Let $g(\beta) = \exp[\beta_0 + \beta_1] / (1 + \exp[\beta_0 + \beta_1]) = 1 - 1/(1 + \exp[\beta_0 + \beta_1])$ (which is $\pi(1)$ in my notation). Then by the delta method (Wikipedia) we have ($g$ is a smooth function with infinitely many derivatives, so we don't have to worry about the regularity conditions) $$ g(\hat \beta) \stackrel{a}{\sim} N \left(g(\beta), \left( \frac{\partial g}{\partial \beta} \right) I(\beta)^{-1} \left(\frac{\partial g}{\partial \beta} \right)^T \right). $$

Here $$ \frac{\partial g}{\partial \beta} = \left( \frac{\exp[\beta_0 + \beta_1]}{(1+\exp[\beta_0 + \beta_1])^2}, \frac{\exp[\beta_0 + \beta_1]}{(1+\exp[\beta_0 + \beta_1])^2} \right) = \frac{\exp[\beta_0 + \beta_1]}{(1+\exp[\beta_0 + \beta_1])^2} (1, 1) $$ (Verify!) Use the above gradient with the given inverse Fisher information matrix $I(\beta)^{-1}$ to get the approximate variance of $g(\hat \beta)$, which is $\left( \frac{\partial g}{\partial \beta} \right) I(\beta)^{-1} \left(\frac{\partial g}{\partial \beta} \right)^T$. [You may find the identity $$ (1, 1) \begin{pmatrix} a_{11} & a_{12} \\ a_{12} & a_{22} \end{pmatrix} \begin{pmatrix} 1 \\ 1 \end{pmatrix} = a_{11} + 2 a_{12} + a_{22} $$
helpful.]