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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))?

For reference

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.

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  • $\begingroup$ This looks like a homework/self-study problem, in which case the self-study tag needs to be added. You also need to show your attempt(s) at solving this problem if you want a good response. $\endgroup$ – rishic Sep 29 at 6:07
  • $\begingroup$ Thanks , have edited question now. $\endgroup$ – KiwiStats Sep 29 at 6:35
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[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.]

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