Question

In Exercise \(10.9 .5\), the influence function of the variance functional was derived directly. Assuming that the mean of \(X\) is 0 , note that the variance functional, \(V\left(F_{X}\right)\), also solves the equation $$ 0=\int_{-\infty}^{\infty}\left[t^{2}-V\left(F_{X}\right)\right] f_{X}(t) d t $$ (a) Determine the natural estimator of the variance by writing the defining equation at the empirical cdf \(F_{n}(t)\), for \(X_{1}-\bar{X}, \ldots, X_{n}-\bar{X}\) iid with \(\operatorname{cdf} F_{X}(t)\) and solving for \(V\left(F_{n}\right)\). (b) As in Exercise \(10.9 .6\), write the defining equation for the variance functional at the contaminated \(\operatorname{cdf} F_{x, \epsilon}(t)\). (c) Then derive the influence function by implicit differentiation of the defining equation in part (b).

Short answer

Natural estimator of the variance: \(V\left(F_{n}\right)=\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\). The influence function is given by: \[IF(x, F_X) = \frac{(x-\bar{x})^2 - V(F_X)}{2}.\]

Step-by-step solution

Set Up Defining Equation at Empirical cdf

Accordingly, replace \(F_X(t)\) and \(f_X(t)\) in the integral equation by \(F_n(t)\) and the sum of Dirac functions, respectively. Therefore, we have: \[0=\sum_{i=1}^{n}\left[(x_{i}-\bar{x})^{2}-V\left(F_{n}\right)\right] / n. \]Rearrange to solve for \(V\left(F_{n}\right)\), we get:\[V\left(F_{n}\right)=\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\].

Write Defining Equation for the Contaminated cdf

The definition of \(F_{X, \epsilon}\) as a contaminated cdf and the corresponding pdf can be used to modify the integral equation as follows:\[0=\int[t^2-V(F_{X,\epsilon})]f_{X,\epsilon}(t) dt\]where:\[F_{X,\epsilon}(t) =(1-\epsilon)F_X(t) + \epsilon F_X^D(t)\] and\[f_{X,\epsilon}(t) = (1-\epsilon)f_X(t) + \epsilon f_X^D(t)\].

Derive the Influence Function

To derive the Influence Function (IF) take the derivative of the equation in Step 2 with respect to \(\epsilon\), and then let the limit as \(\epsilon \rightarrow 0\). Solve for the IF, we yield:\[\frac{dF_{X,\epsilon}}{d\epsilon} = \lim_{\epsilon\rightarrow 0} \frac{F_{X,\epsilon} - F_{X}}{\epsilon}\]Then after some mathematical manipulations:\[IF(x, F_X) = \lim_{\epsilon\rightarrow 0} \frac{F_{X,\epsilon}- F_{X}}{\epsilon} = \frac{(x-\bar{x})^2 - V(F_X)}{2}.\]