Stanford University, Spring 2016, STATS 205

## Robustness of Estimators

• Robustness properties of the three estimators the mean, the median, and the Hodgesâ€“Lehmann estimate.
• Main measures of robustness:
• efficiency,
• influence,
• and breakdown
• Today, we focus on influence and breakdown

## Sensitivity Curve

• Function of the observations
• The parameter $$\theta$$ is a function of the cdf $$F$$
• The observations $$x_1,x_2,\dots,x_n$$ are drawn from $$F$$
• Denote $$\widehat{\theta}$$ as the estimator of $$\theta$$ in the sample
• Measure change in estimator $$\widehat{\theta}_n$$ when outlier $$x$$ is added to a sample $$x_n$$ $x_n = ( x_1,x_2,\dots,x_n )^T$ $x_{n+1} = ( x_1,x_2,\dots,x_n,x )^T$
• The sensitivity curve of an estimator is $S(x,\widehat{\theta}) = \frac{\widehat{\theta}_{n+1} - \widehat{\theta}_n}{1/(n+1)}$

## Sensitivity Curve Example

x_n = c(1.85,2.35,-3.85,-5.25,-0.15,2.15,0.15,-0.25,-0.55,2.65)
mean(x_n)
## [1] -0.09
median(x_n)
## [1] 0
hl.loc(x_n)
## [1] 0

## Sensitivity Curve Example

• Mean is unbounded
• Median and Hodgesâ€“Lehmann is bounded (estimators change slightly and changes soon become constant)

## Influence Function

• Sensitivity curve depends on sample $$x_1,x_2,\dots,x_n$$
• Influence function depends on the population distribution $$F$$
• It's a functional: a function that takes another function as its argument
• Denote functional as $$T(F)$$, e.g. median $$T(F) = F^{-1}(1/2)$$
• Then using Gateaux derivatives (directional derivatives for functionals) $L_F(x) = lim_{\epsilon\to 0} \frac{T((1-\epsilon)F + \epsilon \delta_x)-T(F)}{\epsilon}$ with the point mass $$\delta_x$$ corresponding to adding an outlier $$x$$

## Influence Function

• An estimator is robust if its influence function is bounded
• The influence function for our estimators are (up to constant of proportionality and center)
• Mean: $$x$$,
• Median: $$\operatorname{sign}(x)$$, and
• HL: $$F(x) âˆ’ 0.5$$
• Hence, mean is not robust, but median and HL are robust

## Breakdown Point of an Estimator

• Suppose we contaminate $$n-m$$ points in our sample $\boldsymbol{x}_n^* = (x_1,\dots,x_m,x_{m+1}^*,\dots,x_n^*)$
• Consider the contamination to be as very large (close to $$\infty$$)
• Breaking point: the smallest value $$n-m$$ so that $$\widehat{\theta}(\boldsymbol{x}_n^*)$$ is meaningless
• Finite sample breakdown point: The ratio $$\frac{n-m}{n}$$
• Asymptotic breakdown point: If it converges to a finite value as $$n \to \infty$$

## Breakdown Point of an Estimator

• Sample mean:
• Finite sample BP: $$\frac{1}{n}$$
• Asymptotic BP: $$0$$
• Sample median:
• Finite sample BP: $$\operatorname{floor}\left( \frac{n âˆ’ 1}{2} \right)$$
• Asymptotic BP: $$0.5$$
• Hodgesâ€“Lehmann estimator:
• Asymptotic BP: $$0.29$$

## Breakdown Point of an Estimator

The median of the Walsh averages, stays with the majority.

\begin{align} \# \text{good points} & > \# \text{bad points} \\ \frac{m(m+1)}{2} & > \frac{n(n+1) - m(m+1)}{2} \\ m(m+1) & > \frac{n(n+1)}{2} \end{align}

## Breakdown Point of an Estimator

• Obtaining finite sample BP is messy, asymptotic is easier: $m(m+1) > \frac{n(n+1)}{2} \Leftrightarrow \frac{m(m+1)}{n(n+1)} > \frac{1}{2}$
• let $$x = m/n$$ and $$x \approx (m+1)/(n+1)$$
• then $$x^2 > \frac{1}{2} \Leftrightarrow x > \frac{1}{\sqrt{2}}$$
• convert to BP $$\frac{1}{\#bad} = 1 - \frac{1}{\#good}$$: $1-\frac{1}{\sqrt{2}} \approx 0.29$
• More details are here

## Breakdown Point of an Estimator

• Instead of testing for normality to decide whether to use a $$t$$-test or the Wilcoxon test
• It is easier to use the notion of robustness of an estimator and their associated procedures
• There are too many ways that a distribution can deviate from normality
• How much robustness do we need? Is $$0.29$$ enough or $$0.5$$ really necessary?
• Sample median preferred over Hodgesâ€“Lehmann estimator?
• Based on breakdown point: yes