# Consistent or inconsistent estimator

If $$\hat{\theta}_n$$ is an estimator for the parameter $$\theta$$, then the two sufficient conditions to ensure consistency of $$\hat{\theta}_n$$ are:

Bias($$\hat{\theta}_n)\to 0$$ and Var$$(\hat{\theta}_n)\to 0$$,

then we will have $$\lim_{n\to\infty}Pr(|\hat{\theta}_n-\theta|>\varepsilon)=0, \forall\varepsilon>0$$.

Suppose $$X_1,\ldots X_n$$ be iid samples drawn from the Binomial($$2,p$$), where $$p$$ is the unknown parameter in $$[0,1]$$. So, $$E(X_1)=2p$$, $$Var(X_1)=2p(1-p)$$. Let $$\hat{p}=\frac{X_1+X_2}{4}$$ be the estimator for $$p$$. I computed that $$Bias(\hat{p})=0$$ and $$Var(\hat{p})=\frac{p(1-p)}{4}$$.

In this case, I cannot make use of the theorem above as $$Var(\hat{p})$$ does not converge to $$0$$. By the Chebyshev's inequality, for all $$\varepsilon>0$$, we have $$Pr(|\hat{p}-p|>\varepsilon)\leq \frac{p(1-p)}{4\varepsilon^2}.$$ I am somehow stuck at this stage and cannot make any conclusion whether $$\hat{p}$$ is consistent.

The next question is that what if $$\hat{p}_1 = X_1+X_2$$ is another estimator, in this case it is biased and I cannot tell how should I use Chebyshev's inequality to prove/disprove consistency.

• Consistency is an asymptotic property, which means, it only makes sense when your estimator is a function of sample size $n$ which tends to infinity. Your proposed estimator, however, is independent of $n$. Hence, it is not appropriate to discuss if it is consistent. Feb 22 at 16:22
• Your $\hat{p}$ is a little sad because it has all those $n-2$ observations to use, but you've only picked the first 2! Feb 22 at 16:28
• @Zhanxiong Those are excellent points -- but they do not preclude applying the definition of consistency, because this estimator is (technically) a function of the entire sample. It just isn't affected by the last $n-2$ observations!
– whuber
Feb 22 at 16:47
• @whuber I see your point. Perhaps it is better to say "it is trivial to discuss if $\hat{p}$ is consistent". I have added an answer to clarify it. Feb 22 at 17:47

Technically speaking, $$\hat{p}$$ is not consistent of $$p$$ (when the parameter space is $$[0, 1]$$), because $$P[|\hat{p} - p| > \epsilon]$$ is a fixed positive value that does not depend on $$n$$ (hence it cannot converge to $$0$$). In detail, for fixed $$p \in (0, 1)$$ and $$\epsilon = \frac{1}{2}(1 - p) > 0$$, since $$X_1 + X_2 \sim B(4, p)$$, it follows that \begin{align} & P[|\hat{p} - p| > \epsilon] = P[|X_1 + X_2 - 4p| > 4\epsilon] \\ =& \sum_{k: |k - 4p| > 4\epsilon} \binom{4}{k}p^k(1 - p)^{4 - k} \geq p^4. \tag{1} \end{align}
The last inequality holds because $$|4 - 4p| - 4\epsilon = 2(1 - p) > 0$$ implies that $$4$$ belongs to the set $$\{k: |k - 4p| > 4\epsilon\}$$. As the consistency requires $$\hat{p}$$ converges to $$p$$ in probability for every $$p$$ in $$[0, 1]$$, $$(1)$$ shows that $$\hat{p}$$ is inconsistent of $$p$$.
As many comments under the post indicated, the proposed estimator $$\hat{p}$$ is essentially independent of $$n$$ in that it only used the first two observations (despite mathematically it can still be viewed as a function of the whole sample $$\{X_1, \ldots, X_n\}$$). In other words, the precision of this estimator does not improve at all as the sample size increases, hence you shall not expect it would be consistent.
• +1. But doesn't consistency require convergence for every value of the parameter? Thus, it's irrelevant that things work out when $p\in\{0,1\}$ unless you wish to limit the parameter space to these values.