# Asymptotic unbiasedness + asymptotic zero variance = consistency?

Here, Ben shows that an unbiased estimator $$\hat\theta$$ of a parameter $$\theta$$ that has an asymptotic variance of zero converges in probability to $$\theta$$. That is, $$\hat\theta$$ is a consistent estimator of $$\theta$$.

It makes sense that we should be able to relax the condition to asymptotic unbiasedness: $$\underset{n\rightarrow\infty}{\lim}\mathbb E\left[\hat\theta_n - \theta\right] = 0$$. That seems to stay within the spirit of the estimator closing-in on the true parameter value...

...but math has surprised me before.

Can we relax the condition to asymptotic unbiasedness? What is the proof?

Your question may be restated as

Let $$\langle\hat{\theta}_n\rangle_{n\in\mathbb N}$$ be a sequence of random variables such that \begin{align*} & \theta_n := E[\hat{\theta}_n] \to \theta & \text{ as } n \to \infty. \tag{1}\label{1} \\ & \operatorname{Var}(\hat{\theta}_n) \to 0 & \text{ as } n \to \infty. \tag{2}\label{2} \end{align*} Does $$\eqref{1}$$ and $$\eqref{2}$$ imply $$\hat{\theta}_n \to_p \theta$$?

The answer is in the affirmative and the proof is similar to the original case (i.e., using Chebyshev's inequality, but with slightly more work to split the deviance $$|\hat{\theta}_n - \theta|$$).

Given $$\varepsilon > 0$$, by $$\eqref{1}$$, there exists $$N$$ sufficiently large such that $$|\theta_n - \theta| < \varepsilon/2$$ for all $$n > N$$, it then follows that for all $$n > N$$, \begin{align*} & P(|\hat{\theta}_n - \theta| > \varepsilon) \\ =& P(|\hat{\theta}_n - \theta_n + \theta_n - \theta| > \varepsilon) \\ \leq & P(|\hat{\theta}_n - \theta_n| > \varepsilon / 2) + P(|\theta_n - \theta| > \varepsilon / 2) \\ =& P(|\hat{\theta}_n - \theta_n| > \varepsilon / 2) \\ \leq & 4\operatorname{Var}(\hat{\theta}_n) / \varepsilon^2. \end{align*} The right-hand side of the inequality converges to $$0$$ as $$n \to \infty$$ by $$\eqref{2}$$, which implies $$\hat{\theta}_n \to_p \theta$$. This completes the proof.

An equivalent way of seeing it is by decomposing $$\hat{\theta}_n - \theta$$ as $$(\hat{\theta}_n - \theta_n) + (\theta_n - \theta)$$ then applying the Slutsky's theorem (or merely the simple asymptotic fact that if $$X_n \to_p X$$ and $$Y_n \to_p Y$$ then $$X_n + Y_n \to_p X + Y$$) -- the first part $$\hat{\theta}_n - \theta_n \to_p 0$$ is exactly the original case you linked, while the second part $$\theta_n - \theta \to 0$$ follows by the asymptotic unbiasedness condition $$\eqref{1}$$.

At my first reading, I mistakenly interpreted OP's problem as "Does asymptotic unbiasedness alone imply consistency?", which is certainly not true. One counterexample is: let $$\theta = 0$$, $$\hat{\theta}_n$$ is a Rademacher random variable, i.e., $$P(\hat{\theta}_n = 1) = P(\hat{\theta}_n = -1) = \frac{1}{2}$$ for all $$n$$. It is easy to verify that $$E[\hat{\theta}_n] \equiv \theta$$ while $$\hat{\theta}_n \not\to_p \theta$$. Note that in this case $$\operatorname{Var}(\hat{\theta}_n) \equiv 1$$.

• I thought the question was whether variance going to zero plus asymptotic unbiasedness was enough (as a slight relaxation of variance going to zero plus unbiasedness) [which is true] Commented May 1 at 1:14
• I have stated a more general result, Zhanxiong. Commented May 1 at 1:14
• @ThomasLumley reading the body, I thought whether asymptotic unbiasedness was enough. Commented May 1 at 1:16
• @User1865345 OK, I restored it (btw, do you know how to make smaller fonts? I hope by doing so the appendix wouldn't distract OP's true question). Commented May 1 at 1:49
• @mhdadk Appreciate it. Commented May 1 at 12:09

The following general result would be relevant here:

Given a measure space $$(\Omega,\boldsymbol{\mathfrak A},\mu),$$ consider a sequence of measurable functions $$\langle f_n\rangle_{n\in \mathbb N}$$ where $$f_n\in L^p(\Omega,\boldsymbol{\mathfrak A},\mu),~p\in[1,\infty).$$ If the sequence converges in the $$L^p$$ norm to $$f,$$ then $$f_n\overset{\mu}{\to}f.$$

To see that, note for an arbitrary $$\varepsilon> 0,$$ $$\Vert f_n-f\Vert_p^p\geq \int_{\{\Omega:|f_n-f|\geq \varepsilon\}}|f_n-f|^p~\mathrm d\mu\geq \varepsilon^p\mu\{\Omega:|f_n-f|\geq \varepsilon\},$$ but by assumption $$\Vert f_n-f\Vert_p\to 0,$$ so $$f_n$$ converges in measure to $$f$$ on $$\Omega.$$

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## Reference:

$$\rm[I]$$ Real Analysis: Theory of Measure and Integration, J. Yeh, World Scientific, $$2014,$$ Th. $$16.25.$$