# Weak consistency from asymptotic unbiasedness

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.

• Your proof is quite convoluted and hard to follow, I posted something simpler below. – Gabriel Romon Aug 26 at 7:42

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.