The Effect of Outliers

Question

The following question comes up in robust statistics. There are two formula indicated below that I do NOT know how to derive. However, in order to make the context clear, let's start with the easiest case. One can explore the effect of a single outlier on an estimator $T_n(\mathbf y)$ by replacing say $y_n$ by $z$ and then see what will happen to $T_n(z)$ as $z$ approaches the infinity. As an example, consider independent exponential observations $\mathbf y=(y_1, \cdots, y_n)$ with density function $f(y, \theta)=\theta e^{-\theta y}, y>0, \theta>0$. Then by the standard procedure, the maximum likelihood estimate under this model is $$\widehat\theta(z)=\frac{1}{n^{-1}(z+\sum_{i=1}^{n-1} y_i)};$$ and an approximate $100(1-\alpha)\%$ confidence interval for $\theta$ is $$\left[\widehat{\theta}(z)\mp \Phi^{-1}\left(1-\frac{\alpha}{2}\right)\frac{\widehat{\theta}(z)}{\sqrt{n}}\right].$$ My questions start henceforth. Instead of finding the MLE standard error for the esitmator $\widehat{\theta}(z)$, one can also find its $\textbf{non-parametric}$ standard errors $\frac{\left[\widehat{\theta}(z)\right]^2\widehat{\sigma}(z)}{\sqrt{n}}$. That is, one now has the following confidence interval for $\widehat\theta(z)$ $$\left[\widehat{\theta}(z)\mp \Phi^{-1}\left(1-\frac{\alpha}{2}\right)\frac{\left[\widehat{\theta}(z)\right]^2\widehat{\sigma}(z)}{\sqrt{n}}\right].$$

$\textbf{Question 1:}$ I have not had much exposure to non-parametric statistics before. Can anyone tell me how to derive the non-parametric standard error listed above, please? Thank you!

Now to have a more robust estimator, one can use median (med) instead of mean to get the following confidence interval for $\theta$ $$\left[\frac{\log 2}{\mathrm{med}(\mathbf y)}\mp\Phi^{-1}\left(1-\frac{\alpha}{2}\right)\frac{1}{\mathrm{med}{\mathbf(y)}\sqrt{n}}\right].$$

$\textbf{Question 2:}$ I have no idea how the above confidence interval is derived, especially the standard error. The estimate seems to be derived using method of moment. As far as I know, one of the drawbacks of MOM estimates is its lack of variance estimate. Could anyone tell me what is going on here, please? Thank you!

Answer 1

The wiki page for the exponential distribution states that if $Y\sim\mbox{Exp}(\lambda)$, then

$$F^{-1}_q(\lambda)=-\frac{\ln(1-q)}{\lambda}$$

so that $\mbox{med}(Y)=\frac{\ln(2)}{\lambda}$ which implies that a good (median) based estimator of $\lambda$ is

$$\hat{\lambda}=\frac{\ln(2)}{\mbox{med}(Y)}$$

FWIW, plugging in this median based expression for $\lambda$ in the second leg of your first equation yields:

$$\left(\frac{\ln(2)}{\mbox{med}(Y)}\pm\Phi^{-1}\left(1-\frac{\alpha}{2}\right)\frac{\ln(2)}{\mbox{med}(Y)\sqrt{n}}\right)$$

and since $2<\exp(1)$ these obvious C.I.'s are tighter than the ones you have. Are you sure the $\ln(2)$ in the second term of your C.I.'s didn't accidentally get left out?