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What follows is a task from an old university course on estimation theory. I would like to know whether my solution is correct.

Let $(\mathbb{H}, \mathcal{H})$ be a measurable space, and let $\mathcal{W} = \{P_\gamma | \gamma \in \Gamma\}$ be a family of probability measures, with $\Gamma$ being an index set. $h: \Gamma \rightarrow \mathbb{R}$ is a function that should be estimated. Let $(T_n)$ be a sequence of estimators for $h$. We now define weak consistency and asymptotic unbiasedness:

$(T_n)$ is defined to be weakly consistent if for all $\varepsilon > 0, \gamma \in \Gamma$: \begin{equation*} \lim_{n \rightarrow \infty} P_\gamma(\{x \in \mathbb{H} | |T_n(x) - h(\gamma)| \geq \varepsilon \}) = 0 \end{equation*} $(T_n)$ is defined to be asymptotically unbiased if for all $\gamma \in \Gamma$: \begin{equation*} \lim_{n \rightarrow \infty} E_\gamma(T_n) = h(\gamma) \end{equation*} We now consider an asymptotically unbiased sequence of estimators $S_n: \mathbb{H} \rightarrow \mathbb{R}$ where for all $\gamma \in \Gamma$: \begin{equation*} \lim_{n \rightarrow \infty} V_\gamma(S_n) = 0 \end{equation*} It should be shown that $(S_n)$ is also weakly consistent for $h$.

My proof: Let us first define the term $F_{\gamma,n}(t) := P_\gamma(x |(S_n - h(\gamma))^2 \geq t)$. It can be shown using the layer-cake theorem that \begin{equation*} E_\gamma((S_n - h(\gamma))^2) = \int_0^\infty F_{\gamma,n}(t) d \lambda(t) \end{equation*} We now consider the term \begin{equation*} \lim_{n \rightarrow \infty} E_\gamma((S_n - h(\gamma))^2) = \lim_{n \rightarrow \infty} \int_0^\infty F_{\gamma,n}(t) d \lambda(t) \end{equation*} We can now show that the left-hand term approximates 0, since: \begin{equation*} \lim_{n \rightarrow \infty} E_\gamma((S_n - h(\gamma))^2) = \lim_{n \rightarrow \infty} E_\gamma((S_n - E_\gamma(S_n) + E_\gamma(S_n) - h(\gamma))^2) \end{equation*} \begin{equation*} = \lim_{n \rightarrow \infty} E_\gamma((S_n - E_\gamma(S_n))^2) + \lim_{n \rightarrow \infty} 2 \cdot E_\gamma((S_n - E_\gamma(S_n))(E_\gamma(S_n) - h(\gamma))) + \lim_{n \rightarrow \infty} E_\gamma((E_\gamma(S_n) - h(\gamma))^2) \end{equation*} In the sum at the end, the last two terms go to 0 because $(S_n)$ is asymptotically unbiased, and therefore $E_\gamma(S_n) \rightarrow h(\gamma)$. The first term is the variance of $S_n$ which also goes to 0 for $n \rightarrow \infty$ by definition of $(S_n)$. Therefore, we have
\begin{equation*} \lim_{n \rightarrow \infty} \int_0^\infty F_{\gamma,n}(t) d \lambda(t) = 0 \end{equation*} It follows that $F_{\gamma,n}(t) = P_\gamma(x |(S_n - h(\gamma))^2 \geq t)$ must become arbitrary small for $n \rightarrow \infty$, and therefore that $(S_n)$ is weakly consistent.

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  • $\begingroup$ Your proof is quite convoluted and hard to follow, I posted something simpler below. $\endgroup$ – Gabriel Romon Aug 26 at 7:42
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Let $\epsilon >0$ and $\gamma \in \Gamma$ be fixed.
For $n\geq 1$, note the inclusions $$\begin{aligned} (|S_n-h(\gamma)|\geq \epsilon) &\subset \left(|S_n-E_\gamma(S_n)|+|E_\gamma(S_n)-h(\gamma)|\geq \epsilon\right)\\ &\subset \left(|S_n-E_\gamma(S_n)|\geq \frac{\epsilon}2\right) \cup \left(|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2 \right) \end{aligned}$$ The event $\left(|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2 \right)$ is deterministic. It is equal to $\mathbb H$ if $|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2$, and $\emptyset$ otherwise. Thus $$\begin{aligned}P(|S_n-h(\gamma)|\geq \epsilon) &\leq P\left(|S_n-E_\gamma(S_n)|\geq \frac{\epsilon}2\right) + P\left(|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2 \right) \\ &= P\left([S_n-E_\gamma(S_n)]^2\geq \frac{\epsilon^2}4\right) + 1_{|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2} \\ &\leq \frac{V_\gamma(S_n)}{\frac{\epsilon^2}4} + 1_{|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2} \end{aligned}$$

Since $S_n$ is asymptotically unbiased, $1_{|E_\gamma(S_n)-h(\gamma)| \geq \frac{\epsilon}2}\xrightarrow[n\to \infty]{} 0$ and the conclusion follows.

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