# Wald confidence interval with delta method

Using the delta method, show that the Wald confidence interval for the logit of a binomial parameter $\pi$ is $$\log \left(\frac{\hat{\pi}}{1-\hat{\pi}} \right) \pm z_{\alpha/2} \sqrt{\frac{1}{n\hat{\pi}(1-\hat{\pi})}}$$ Explain how to use this interval to obtain one for $\pi$ itself.

Since $Y\sim \mathrm{Bin}(n,\pi)$ and $\hat{\pi}=\frac{y}{n}$ then $$\begin{cases} \operatorname{E}[y]=n\pi\\ \operatorname{Var}(y)=n\pi(1-\pi)\end{cases}\qquad \begin{cases} \operatorname{E}[\hat{\pi}]=\pi\\ \operatorname{Var}(\hat{\pi})=\frac{\pi(1-\pi)}{n} \end{cases}$$

Let $g(\hat{\pi})=\log(\frac{\hat{\pi}}{1-\hat{\pi}})$ then $g'(\hat{\pi})=\frac{1}{\pi(1-\pi)}$

The delta method says that variance of $g(\hat{\pi})$ is $$\operatorname{Var}(g(\hat{\pi}))=(g'(\hat{\pi}))^2\times \sigma^2(\hat{\pi})$$ $$=\frac{1}{\hat{\pi}^2(1-\hat{\pi})^2}\times\frac{\hat{\pi}(1-\hat{\pi})}{n}=\frac{1}{n\hat{\pi}(1-\hat{\pi})}$$

So

$$\sqrt{n}\left(\log\left(\frac{\hat{\pi}}{1-\hat{\pi}}\right)-\log\left(\frac{\pi}{1-\pi}\right)\right) \overset{D}\rightarrow N\left(0,\frac{1}{\hat{\pi}(1-\hat{\pi})}\right)$$

Finally a Wald confidence interval is $$\operatorname{E}[g(\hat{\pi})]\pm z_{\alpha/2} \sqrt{\operatorname{Var}(g(\hat{\pi}))}$$ $$\log \left(\frac{\hat{\pi}}{1-\hat{\pi}}\right) \pm z_{\alpha/2} \sqrt{\frac{1}{n\hat{\pi}(1-\hat{\pi})}}$$

Is this right?

Your final result is ok but there's a little mistake in the procedure. Escentially, $\mathrm{Var}(g(\hat{\pi}))$ is a function of the parameter $\pi$. Only $\widehat{\mathrm{Var}}(g(\hat{\pi}))$, the estimator of this variance, is a function of $\hat{\pi}$, the maximum likelihood estimator of $\pi$.
Finally, to show how to use this interval to obtain one for $\pi$ itself, notice that you have a confidence region, say $R1$ , for $\log(\frac{\pi}{1-\pi})$. Now, you can obtain a second confidence region $R2$ for $\pi$ so that every value of $\pi$ in this region generates a value of $\log(\frac{\pi}{1-\pi})$ that belongs to $R1$.