650 Nonparametric and Robust Statistics10.9.3.Consider the influence function for the Hodges–Lehmann estimator given
in expression (10.9.21). Show for it that property (10.9.22) is true. Next, evaluate
expression (10.9.23) and, hence, obtain the asymptotic distribution of the estimator
as given in expression (10.9.25). Does it agree with the result derived in Section
10.3?10.9.4.LetFx,
(t) be the point-mass contaminated cdf given in expression (10.9.13).
Show that
|Fx,
(t)−FX(t)|≤ ,
for allt.
10.9.5.SupposeXis a random variable with mean 0 and varianceσ^2. Recall that
the functionFx,
(t) is the cdf of the random variableU=I 1 − X+[1−I 1 − ]W,
whereX,I 1 − ,andW are independent random variables,X has cdfFX(t),W
has cdf Δx(t), andI 1 − has a binomial(1, 1 − ) distribution. Define the functional
Var(FX)=Var(X)=σ^2. Note that the functional at the contaminated cdfFx,
(t)
has the variance of the random variableU=I 1 − X+[1−I 1 − ]W.Toderivethe
influence function of the variance, perform the following steps:
(a)Show thatE(U)= x.(b)Show that Var(U)=(1− )σ^2 + x^2 − 2 x^2.(c)Obtain the partial derivative of the right side of this equation with respect to. This is the influence function.
Hint: BecauseI 1 − is a Bernoulli random variable,I 12 − =I 1 − .Why?
10.9.6.Often influence functions are derived by differentiating implicitly the defin-
ing equation for the functional at the contaminated cdfFx,
(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 cdfFx,
(t) can be expressed
as
0=∫∞−∞[t−T(Fx,
)]dFx,
(t)=(1− )∫∞−∞[t−T(Fx,
)]fX(t)dt+∫∞−∞[t−T(Fx,
)]dΔ(t). (10.9.52)Recall that we want∂T(Fx,
)/∂ . Obtain this by implicitly differentiating the above
equation with respect to.
10.9.7.In Exercise 10.9.5, the influence function of the variance functional was de-
rived directly. Assuming that the mean ofXis 0, note that the variance functional,
V(FX), also solves the equation
0=∫∞−∞[t^2 −V(FX)]fX(t)dt.