Suppose a random sample $x_1,\dots, x_n$ (i.i.d.) from a random variable $X$ defined over $(\Omega,\mathcal{F},P)$ whose probability density function is $f(x_1,\dots, x_n;\theta)$ and $T(x_1,\dots, x_n)$ is a complete statistic for this family. Let $f_A(x_1,\dots, x_n;\theta)$ be the probability density function associated with $P$ when we truncate it over $A$, a subset of the sample space $\Omega$, given by $$f_A(x_1,\dots, x_n;\theta) = \frac{f(x_1,\dots, x_n;\theta)I_{A}(x_1,\dots, x_n)}{P_\theta (A)}$$ such that $I_{A}(x_1,\dots, x_n)$ is the indicator function that is 1 when every point belongs to $A$ and 0 if at least one isn't and $P_\theta(A)$ is the probability of set $A$ that acts as a normalizing constant. Show that $T(x_1,\dots, x_n)$ is also complete for the truncated family.

I was studying mathematical statistics and found this problem. The problem set also had a exercise about showing that if $T$ is sufficient for a distribution then it is sufficient for the truncated one that is easily solved by using Fisher-Neyman factorization. I tried to show it through the definition of completeness, $\mathbb{E}_\theta(g(t))0 \xrightarrow{} P(g(t)=0)=1$, but this takes me nowhere as we know nothing about the pdf of $T$.


1 Answer 1


Consulting the Theory of Point Estimation (Lehmann and Casella, 1999) from which this exercise is taken

enter image description here

the second question contains the term in addition, which means that $T$ is sufficient.

If $T$ is incomplete and sufficient for the family of truncated distributions, there exists a non-zero function $g$ such that the expectation of $g(T)$ is zero for all $f_A(\cdot;\theta)$'s. This implies $$\mathbb E_\theta[g(T(X_1,\ldots,X_n)\mathbb I_A(X_1,\ldots,X_n)]=0\quad\forall\theta$$ for the untruncated distributions.

Hint 1: Apply the law of the total expectation to the above to derive a function of $T$ with expectation zero

Hint 2: Show that $g$ is not constant.

  • $\begingroup$ Thanks for the answer! The first part is very clear to me and also the logic of hint 2, however i don't get it how should i condition using the law of the total expectations. Would it be on $X_1,\dots,X_n \in A$? $\endgroup$ Nov 12, 2020 at 11:53
  • 1
    $\begingroup$ You have to solve Hint 1. The term inside the expectation need be turned into a function of $T$ only. $\endgroup$
    – Xi'an
    Nov 12, 2020 at 12:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.