Question

Often influence functions are derived by differentiating implicitly the defining equation for the functional at the contaminated cdf \(F_{x, e}(t),(10.9 .13) .\) Consider the mean functional with the defining equation (10.9.10). Using the linearity of the differential, first show that the defining equation at the cdf \(F_{x, \epsilon}(t)\) can be expressed as $$ \begin{aligned} 0=\int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d F_{x, \epsilon}(t)=&(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] f_{X}(t) d t \\ &+\epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d \Delta(t) \end{aligned} $$ Recall that we want \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon .\) Obtain this by implicitly differentiating the above equation with respect to \(\epsilon\).

Short answer

First, express the defining equation at the cdf \(F_{x, \epsilon}(t)\), then differentiate this equation implicitly with respect to \(\epsilon\) to find the influence function, \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon\). The derivative represents how sensitive the functional is to small changes in \(\epsilon\). It is molded by the properties of the given distributions.

Step-by-step solution

Simplifying the Mean Functional's Defining Equation

Start by simplifying the mean functional's defining equation at the cdf \(F_{x, \epsilon}(t)\), expressed as: \(0=\int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d F_{x, \epsilon}(t)=(1-\epsilon) \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] f_{X}(t) d t + \epsilon \int_{-\infty}^{\infty}\left[t-T\left(F_{x, \epsilon}\right)\right] d \Delta(t)\)

Implicitly Differentiating with Respect to \(\epsilon\)

The next step is to implicitly differentiate the above equation with respect to \(\epsilon\) to get the function \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon\).

Computing the Derivative

Evaluate the derivative, differentiating each term with respect to \(\epsilon\). Remember to apply the rule \(d/dx[a \int f(x)dx] = a f(x) + \int f(x) da/dx \). Also note that \(d\Delta(t) / d\epsilon = 0\). These manipulations should result in the derivative \(\partial T\left(F_{x, \epsilon}\right) / \partial \epsilon\) which is the influence/sensitivity function.