# Logistic Regression: Asymptotic confidence interval for the lethal dose

For the logistic model:

$$\log \Big( \frac{\pi(x)}{1-\pi(x)}\Big) = b_0 +b_1x$$

I want to construct a asymptotic confidence interval for the ratio of the m.l.e's of $$b_0$$, $$b_1$$:

$$LD50 = -\frac{\hat b_0}{\hat b_1}$$

I want to use the delta method.

I know that $$\hat b -b \xrightarrow[\text{}]{\text{D}} \mathcal{N_p}(0,X^TW(b)X)$$

with $$W = \operatorname{diag}(\pi (x_1,b)(1-\pi(x_1,b)), \pi (x_2,b)(1-\pi(x_2,b)),... \pi (x_n,b)(1-\pi(x_n,b)))$$

Where do I start?

• You start with finding the gradient of $f(x,y):=x/y$. – d.k.o. Dec 14 '18 at 18:36

## 1 Answer

Define $$g(x,y) = - x/y$$, its gradient is $$\nabla g = ( - 1/y, x/y^2 ) ^ T$$, hence $$\sqrt{n}\left( g(b_0, b_1)-g(\hat b_0, \hat b_1) \right) \xrightarrow{D}N(0, \nabla g ^ T\Sigma_{b}\nabla g),$$ where $$\Sigma_b$$ is the covariance matrix of $$(\hat{b}_0, \hat{b}_1) ^ T$$, thus $$\lim_{n \to \infty} \mathbb{P}\left( \| \Sigma_b^{1/2}\nabla g \|Z_{a/2} \le \sqrt{n}\left( g(b_0, b_1)-g(\hat b_0, \hat b_1) \right) \le \| \Sigma_b^{1/2}\nabla g \|Z_{1 -a/2} \right) = 1 - a.$$ Try to finish the derivation of the CI...