Different ways to produce a confidence interval for odds ratio from logistic regression

I am studying how to construct a 95% confidence interval for odds ratio from the coefficients obtained in the logistic regression. So, considering the logistic regression model,


\log\left(\frac{p}{1 – p}\right) = \alpha + \beta x \newcommand{\var}{\rm Var}
\newcommand{\se}{\rm SE}

such that x = 0 for control group and x = 1 for case group.

I have already read that the simplest way is to construct a 95% CI for \beta then we applied the exponential function, that is,


\hat{\beta} \pm 1.96\times \se(\hat{\beta}) \rightarrow \exp\{\hat{\beta} \pm 1.96\times \se(\hat{\beta})\}

My questions are:

  1. What is the theoretical reason that justifies this procedure? I know \mbox{odds ratio} = \exp\{\beta\} and maximum likelihood estimators are invariant. However, I do not know the connection among these elements.

  2. Should the delta method produce the same 95% confidence interval as the previous procedure? Using the delta method,

    \exp\{\hat{\beta}\} \dot{\sim} N(\beta,\ \exp\{\beta\}^2 \var(\hat{\beta}))

    Then,

    \exp\{\hat{\beta}\} \pm 1.96\times \sqrt{\exp\{\beta\}^2 \var(\hat{\beta})}

    If not, which is the best procedure?

Answer

  1. The justification for the procedure is the asymptotic normality of the MLE for \beta and results from arguments involving the Central Limit Theorem.

  2. The Delta method comes from a linear (i.e first order Taylor) expansion of the function around the MLE. Subsequently we appeal to the asymptotic normality and unbiasedness of the MLE.

Asymptotically both give the same answer. But practically, you would favor the one which looks more closely normal. In this example, I would favor the first one because the latter is likely to be less symmetric.

Attribution
Source : Link , Question Author : Márcio Augusto Diniz , Answer Author : Amir

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