# Infinitesimal Robustness, influence function of $T$ at $F$.

This text is taken from Introduction to robust estimation and hypothesis testing. Wilcox R.

First I will write down the description that leads to definition of relative influence on $$T(F)$$ and then I will state my questions.

Consider a mixture of two distributions where an observation is randomly sampled from distribution $$F$$ with probability $$1 − \epsilon$$, otherwise sampling is from the distribution $$\Delta_x$$ . That is, with probability $$\epsilon$$, the observed value is $$x$$. The resulting distribution is $$F_{x,\epsilon} = (1 − \epsilon)F + \epsilon \Delta_x$$. But $$F$$ and $$\Delta_x$$ are distributions, so |$$F(y) − \Delta_x(y)| ≤ 1$$. Consequently, the Kolmogorov distance between $$F_{x,\epsilon}$$, and $$F$$ is at most $$\epsilon$$. Moreover, $$F(x)$$, and $$F$$ can be made arbitrarily close by choosing sufficiently small. The relative influence on $$T(F)$$ of having the value $$x$$ occur with probability $$\epsilon$$ is

$$\frac{T(F_{x,\epsilon})-T(F)}{\epsilon}$$

and the influence function of $$T$$ at $$F$$ is $$IF(x) = \lim \frac{T(F_{x,\epsilon}) − T (F)}{\epsilon}$$

So, I can understand everything until the point where the relative influence on $$T(F)$$ of having the value $$x$$ occur with probability $$\epsilon$$ is defined. It is derivative at point $$x$$, but why the probability $$\epsilon$$ is at the same chosen to be the distance of change in $$x$$? How is it connected?