3.4 Let X1, . . . ,Xn be an independent sample from a population with density f(x − ) and let T(X1. . . . ,Xn) be a translation equivariant estimator of , then Tn has a continuous distribution function.

Hint: T(x1, . . . , xn) = t if and only if x1 = t−T(0, x2−x1, . . . , xn−x1).

Hence, given X2 − X1, . . . ,Xn − X1 = (y2, . . . , yn) and t 2 R, there is exactly one point x for which T(x) = t. Hence, P{T(X) = t|X2 − X1 =

y2, . . . ,Xn − X1 = yn} = 0 for every (y2, . . . , yn) and t, thus P{T(X) =

t} = 0 8t.