Confidence Intervals around Backtransformed Log-linear Regression

Suppose we are interested in the percentage effect of a binary $$D \in \{0,1\}$$ on an outcome $$Y \in \mathbb{R}$$, which might motivate a simple regression of the form: $$\log Y = \beta_0 + \beta_1 D + \epsilon$$ Since $$\beta_1$$ may be interpreted as the marginal effect of $$D$$ on $$\log Y$$ (which is not very helpful), we might wish to consider inference on $$e^{\beta_1}$$ which represents a multiplicative factor - i.e. if $$D=1$$, then we expect the outcome to change by a factor of $$e^{\beta_1}$$, or a percentage change of $$100 * (e^{\beta_1}-1)$$.

Now suppose we know that $$\sqrt{n}(\hat{\beta}_1 - \beta_1) \to^d \mathcal{N}(0, \sigma^2)$$ By the Delta Method, this implies $$\sqrt{n}(e^{\hat{\beta}_1} - e^{\beta_1}) \to^d \mathcal{N}(0, e^{2\beta_1}\sigma^2)$$ Thus confidence intervals for $$e^{\beta_1}$$ at a $$\alpha=0.05$$ significance level may be obtained by $$\Big[ \frac{ e^{\hat{\beta}_1} }{ 1+\frac{\hat{\sigma}}{\sqrt{n}}z_{1-\alpha/2} }, \frac{ e^{\hat{\beta}_1} }{ 1+\frac{\hat{\sigma}}{\sqrt{n}}z_{\alpha/2} } \Big]$$ where $$z$$ are the critical values from the quantile of a standard normal.

Would someone mind checking this calculation? It looks a bit strange as it is asymmetric, but perhaps that's just the nature of the log-exp transform.

From $$\sqrt{n}(e^{\hat{\beta}_1} - e^{\beta_1}) \to^d \mathcal{N}(0, e^{2\beta_1}\sigma^2)$$ we obtain, plugging in the estimator for $$\beta_1$$ in the denominator and using Slutzky's theorem, $$P\left(z_{\alpha/2}\leq \frac{\sqrt{n}(e^{\hat{\beta}_1} - e^{\beta_1})}{\sqrt{e^{2\hat\beta_1}\hat\sigma^2}}\leq z_{1-\alpha/2}\right)\to 1-\alpha$$ or $$P\left(\frac{z_{\alpha/2}e^{\hat\beta_1}\hat\sigma}{\sqrt{n}}-e^{\hat{\beta}_1}\leq - e^{\beta_1}\leq \frac{e^{\hat\beta_1}\hat\sigma z_{1-\alpha/2}}{\sqrt{n}}-e^{\hat{\beta}_1}\right)\to 1-\alpha$$ or $$P\left(e^{\hat{\beta}_1}-\frac{z_{\alpha/2}e^{\hat\beta_1}\hat\sigma}{\sqrt{n}}\geq e^{\beta_1}\geq e^{\hat{\beta}_1}-\frac{e^{\hat\beta_1}\hat\sigma z_{1-\alpha/2}}{\sqrt{n}}\right)\to 1-\alpha,$$ such that the confidence interval would become $$\left(e^{\hat{\beta}_1}\left[1-\frac{\hat\sigma z_{1-\alpha/2}}{\sqrt{n}}\right];e^{\hat{\beta}_1}\left[1-\frac{z_{\alpha/2}\hat\sigma}{\sqrt{n}}\right]\right)$$
• Ah I had accidentally wrote the denominator in the second step as $\sqrt{e^{2\beta_1 \hat{\sigma}^2}}$! Forgot the hats :)