Given $X_1 ... X_n \sim \textrm{Exp}(\lambda)$, I found the MLE : $$\hat{\lambda} = \frac{1}{\bar{X}}$$

Now I need to find confidence intervals for: $$\eta = \lambda \cdot \log(\lambda)$$

To do so, I need to find the standard error for $\eta$, but first, I'll need the standard error for $\lambda$, and I found that via Fisher Information (which is a result of the log-liklihood function' second derivative): $$\operatorname{se}(\lambda) = \sqrt{\frac{1}{I_n(\lambda)}} = \sqrt{\frac{\lambda^2}{n}}$$


$$\operatorname {se}(\eta) = f'(\lambda) \cdot \operatorname {se}(\lambda) = -\frac{1}{\bar{X}^2} \cdot \sqrt{\frac{\lambda^2}{n}}$$

So I finally got, for $\alpha = 0.05$:

$$\hat{\eta} \pm Z_{\frac{\alpha}{2}} \cdot -\frac{1}{\bar{X}^2} \cdot \sqrt{\frac{\lambda^2}{n}}$$

I don't know if what I got is right. Can anyone please check if I used the delta method correctly? cause I actually didn't use this: $$\eta = \lambda \cdot \log(\lambda)$$

And I feel it's a mistake not to use it.


1 Answer 1


Are you finding $\mathrm{SE}(\widehat{\lambda})$ and $\mathrm{SE}(\widehat{\lambda}\log{\widehat{\lambda}})$ instead? Let $g(\widehat{\lambda})=\widehat{\lambda}\log\widehat{\lambda}$. By delta method, the variance of $g(\widehat{\lambda})$ is \begin{eqnarray*} \mathbb{V}(g(\widehat{\lambda}))&\approx&(g'(\lambda))^2\mathbb{V}(\widehat{\lambda})\\ &=&(1+\log\lambda)^2\frac{\lambda^2}{n} \end{eqnarray*} Then, $\mathrm{SE}(\widehat{\lambda}\log\widehat{\lambda})=\frac{\lambda(1+\log\lambda)}{\sqrt{n}}$.


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