Alternative Confidence interval for Odds Ratio $\hat{p}\over{1-\hat{p}}$ from Logistic Regression?

I have a question about confidence interval calculations for the odds ratio $\hat{p}\over{1-\hat{p}}$ from a logistic regression model (perhaps obtained from the method of Generalized Estimating Equations, but that's of secondary importance). It's best asked after a bit of background and an example:

Let's assume we have a simple logistic regression model with a single binary independent variable ($X={0,1}$ ). For $X=1$, we can estimate the log odds by:

$\log({\hat{p}\over{1-\hat{p}}})=\hat{\beta}_0+\hat{\beta}_1$

To obtain a 95% confidence interval (CI) for the odds ratio, textbooks that I've consulted indicate that you calculate the 95% CI for the odds-ratio, ${\hat{p}\over{1-\hat{p}}}$, by:

Lower Bound= $e^{[\hat{\beta}_0+\hat{\beta}_1-1.96Var(\hat{\beta}_0+\hat{\beta}_1)]}$

Upper Bound= $e^{[\hat{\beta}_0+\hat{\beta}_1+1.96Var(\hat{\beta}_0+\hat{\beta}_1)]}$

What I am wondering is if it is appropriate to use the the Delta Method to obtain the variance of the odds ratio $\hat{p}\over{1-\hat{p}}$ $= \exp{(\hat{\beta}_0+\hat{\beta}_1)}$ instead? For example, could obtain a 95% confidence interval using the following method?:

(1) First, exponentiate to obtain the point estimate of the odds ratio: $\hat{p}\over{1-\hat{p}}$$=\exp(\hat{\beta}_0+\hat{\beta}_1)$=$f(\hat{\beta}_0,\hat{\beta}_1)$.

(2) Find the approximte large sample variance of this non-linear point estimate by the Delta methods by finding the gradient of $f$ as: $\nabla f =[\exp(\hat{\beta}_0+\hat{\beta}_1), \exp(\hat{\beta}_0+\hat{\beta}_1)]$

(3) Calculate $Var({\hat{p}\over{1-\hat{p}}})$ $\approx$ $\nabla f \hat{\Sigma} \nabla f^T$. where $\hat{\Sigma}$ is the empirical covariance matrix of $\hat{\beta}$.

(4) Find the approximate standard error, $SE$ by taking the square root of the approximate variance estimate $SE = \sqrt{\nabla f \Sigma \nabla f^T}$

(5) Finally, obtain the 95% CI by $\exp(\hat{\beta}_0+\hat{\beta}_1)\pm Z_{1-.05/2}SE$.

Is the delta method I have described appropriate to find a 95% confidence interval, and if it is, why is it never described in textbooks on logistic regression? If it isn't, why not, when it seems you can use it to linearize it for similar problems.

UPDATE: I updated the step 3 to make it clear that I'm using the estimated, empirical covariance matrix.

UPDATE 2

I still haven't received a very satisfactory answer to this question. It has even made me wonder why people don't use the Delta Method to compute confidence intervals around estimates of just $\hat{p}$ now, instead of talking the inverse of the logit link of the upper and lower confidence intervals. Maybe @FrankHarrel can shed some light as he seems especially knowledgeable about these types of calculations?

• if you use R , the function 'glm' with family is binary, then if your estimated model is m<-glm(....) then vcov(m) gives you the variance-covariance matrix of the coeffcients – user83346 Apr 5 '16 at 7:45
• I think you will find that Stata uses the delta method to report standard errors for odds ratios from logistic regression but estimates the confidence intervals on the log scale. This often confuses people as they do not agree. – mdewey Apr 5 '16 at 10:28
• @fcop and mdewey, can you comment on whether or not the steps I've listed above seem like a valid approach to modelling the confidence intervals? – StatsStudent Apr 5 '16 at 16:09
• @Analyst1: can you explain where you take the covariance between $\hat{\beta}_i$ into account and how you estimate it? – user83346 Apr 5 '16 at 17:43
• @fcop, the covariance between the $\hat{\beta}_i$ are taken into account in step 3 (and 4), with $\hat{\Sigma}$. $\hat{\Sigma}$ is the empirical covariance matrix of the $\hat{\beta}_i$'s. Is this what you are asking? – StatsStudent Apr 5 '16 at 17:58

• Thank you. What if I were comparing the difference of two proportions? estimated from a logistic regression model? Would that method seem more plausible with regard to (b)? In other words, could I carry out essentially the same process, only with a modified estimating function such as $\hat{p}_1\over{1-\hat{p}_1}$ - $\hat{p}_2\over{1-\hat{p}_2}$ and modified gradient? – StatsStudent Apr 5 '16 at 16:38
• I should have said, the difference of two odds ratios in my comment above or the difference of two proportions (and provided the estimating function $\hat{p}_1-\hat{p}_2$. Sorry for the confusion. – StatsStudent Apr 5 '16 at 17:00