Sample variance robustness I am trying to understand the robustness of the sample variance. 
I want to calculate its influence function and in order to do so a previous step is to obtain the functional for the contaminated distribution: $$T_{\sigma^2}(F_\epsilon) =T_{\sigma^2}((1-\epsilon)F+ \epsilon \delta_x)$$
In particular, there are two things I do not fully understand regarding the derivation of  $T_{\sigma^2}(F_\epsilon)$:


*

*The references I am using state that the sample variance functional can be expressed as: $$T_{\sigma^2}(F) =\int{(y-T_{\mu}(F))^2dF(y)}= \frac{1}{2} \int\int (y-z)^2dF(y)dF(z)$$
Could anyone explain me how the second equality is obtained?

*Using the previous expression, the functional $T_{\sigma^2}(F_\epsilon)$ can be obtained as: $$T_{\sigma^2}(F_\epsilon) =T_{\sigma^2}((1-\epsilon)F+ \epsilon \delta_x) = (1-\epsilon)^2 T_{\sigma^2}(F) + \epsilon(1-\epsilon) \int (y-x)^2dF(y)$$ In this case, I do not fully understand where the term $\epsilon(1-\epsilon) \int (y-x)^2dF(y)$ comes from.
Once these equalities have been obtained, I have no further problems to derive the influence function for the sample variance, which final result is $IF(x, T, F) = (x-\mu)^{2} - \sigma^2$.
EDIT: two small typos corrected
 A: *

*We can look at both sides of the equality.  First, the left-hand side:
$$\begin{align}
\int \left(y - T_\mu(F)\right)^2 \, dF(y)
&= \int \left(y - \int z \, dF(z) \right)^2 \, dF(y) \\
&= \int \left(y^2 - 2y\int z \, dF(z) + \left(\int z \, dF(z)\right)^2\right) \, dF(y) \\
&= \int y^2 \, dF(y) - \int 2 y\int z \, dF(z) \, dF(y) + \left(\int z \, dF(z)\right)^2 \\
&= \int y^2 \, dF(y) - 2\int  y \, dF(y)\int z \, dF(z) + \left(\int z \, dF(z)\right)^2 \\
&= \int y^2 \, dF(y) - \left(\int z \, dF(z)\right)^2. \\
\end{align}
$$


Now, the right-hand side:
$$\begin{align}
\frac12\int \int\left(y - z\right)^2 \, dF(y) \,dF(z)
&= \frac12 \int\int \left(y^2-2yz+z^2\right) \, dF(y) \, dF(z) \\
&= \frac12 \int\left(\int y^2 \, dF(y)\right)\, dF(z)  - \int\int yz\, dF(y) \, dF(z)
 + \frac12\int \left(\int z^2 \, dF(z) \right) \, dF(y)\\
&= \int y^2 \, dF(y) - \left(\int z \, dF(z)\right)^2. \\
\end{align}
$$
Here we made liberal use of the fact that $dF$ is a probability measure and that
$$T_\mu(F) = \int y \, dF(y).$$


*For this one, consider the definition of $F_\epsilon(x) = (1-\epsilon)F(x) + \epsilon\delta_x$.  Expanding
$$\frac12 \int\int\left(y-z\right)^2 \, dF_\epsilon(y) \, dF_\epsilon(z)$$
will give
$$\frac12 \int\int\left(y-z\right)^2 \, ((1-\epsilon)dF(y) + \epsilon\delta_x) \, ((1-\epsilon)dF(z) + \epsilon\delta_x).$$


When you expand this out using linearity, by the first identity you will get a term
$$(1-\epsilon)^2 \frac12 \int\int\left(y-z\right)^2 \,dF(y)  \,dF(z) = (1-\epsilon)^2T_{\sigma^2}(F),$$
then a cross product term that evaluates at only a single point of one of the variables of integration ($x$) due to the delta function
$$2 \times \frac12 \epsilon(1-\epsilon) \int(y-x)^2dF(y),$$
and finally a term that evaluates to 0 assuming the measure $dF$ assigns 0 probability to point masses.
