Confidence Intervals around Backtransformed Log-linear Regression

Question

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 $100 * (e^{\beta_1}-1)$.

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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.

Answer 1

I indeed arrive at an expression that looks different.

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 $$ \left(e^{\hat{\beta}_1}\left[1-\frac{\hat\sigma z_{1-\alpha/2}}{\sqrt{n}}\right];e^{\hat{\beta}_1}\left[1-\frac{z_{\alpha/2}\hat\sigma}{\sqrt{n}}\right]\right) $$