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[From Theory of Point Estimation (Lehmann and Casella, 1999, Exercise 6.37)]

Let $P=\{P_\theta:\theta \in \Theta\}$ be a family of probability distributions and assume that $P_\theta$ has pdf $p_\theta$. Let $A$ be a fixed Borel subset of the sample space. For each distribution $P_\theta$, we consider the truncated distribution $P_\theta^\star$ on $A$, i.e., $$ P_\theta^\star (B) = \frac{P_\theta(A\cap B)}{P_\theta(A)} $$ or $$ p^\star_\theta(x)=\frac{p_\theta(x)}{\int_Ap_\theta(y)dy}, x\in A $$ and $0$ otherwise.

Denote $P^\star=\{P^\star_\theta:\theta \in \Theta\}$. Show that

(a) If $T$ is sufficient for $P$, then it is sufficient for $P^\star$

(b) if, in addition, $T$ is complete for $P$ it is also complete for $P^\star$.

I know the same question was asked and given a hint here, but I can't really understand it. Here's how I do (a):
$T$ is sufficient for $P$ iff it is sufficient for $\theta$. Then we have $$ p(x|\theta) = g(T(x)|\theta)h(x)$$ Let $l(x)=h(x)$ if $x\in A$ and $0$ otherwise, then $$p^\star(x|\theta) = g(T(x)|\theta) l(x)/c$$ for some constant c, so $p^\star$ is sufficient for $\theta$ and thus for $P$.
Is this proof valid?
Also, it would be really nice if someone could show me how to prove part (b). I am thinking about extending the nonzero $g$ if $T$ is not complete for $P^\star$, but I don't know if that is possible.

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    $\begingroup$ Hint: I would not validate this answer because $c$ is not a constant. $\endgroup$
    – Xi'an
    Commented Dec 7, 2022 at 16:05
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    $\begingroup$ Hint (b): start from the definition of completeness applied to $P^*$, by assuming $g\circ T$ having zero expectation under $P^*$ and deduce what happens to $(g\circ T)\mathbb I_A$ under $P$. $\endgroup$
    – Xi'an
    Commented Dec 7, 2022 at 16:12

1 Answer 1

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You have the right general idea, but your "constant" is not actually constant, so you need to be more careful. Since $T$ is sufficient in the initial problem, let $p_\theta(x) = g_\theta(T(x)) h(x)$ be the Fisher-Neyman decomposition for that density. Since $B$ is assumed to be known$^\dagger$ you have:

$$\begin{align} p_\theta^\star(x) &= \frac{p_\theta(x)}{\int_B p_\theta(x) \ dx} \cdot \mathbb{I}(x \in B) \\[6pt] &= \frac{g_\theta(T(x)) h(x)}{\int_B g_\theta(T(x)) h(x) \ dx} \cdot \mathbb{I}(x \in B) \\[12pt] &= g_\theta^\star(T(x)) h^\star(x), \\[6pt] \end{align}$$

where the functions in the new decomposition are:

$$g_\theta^\star(x) \equiv \frac{g_\theta(T(x))}{\int_B g_\theta(T(x)) h(x) \ dx} \quad \quad \quad \quad \quad h^\star(x) \equiv h(x) \cdot \mathbb{I}(x \in B).$$


$^\dagger$ Note that if $B$ is an unknown parameter then $T$ is not sufficient.

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