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.

  • 2
    $\begingroup$ 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. $\endgroup$
    – Zhanxiong
    Commented Feb 22, 2023 at 16:22
  • 2
    $\begingroup$ 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! $\endgroup$
    – AdamO
    Commented Feb 22, 2023 at 16:28
  • $\begingroup$ @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! $\endgroup$
    – whuber
    Commented Feb 22, 2023 at 16:47
  • $\begingroup$ @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. $\endgroup$
    – Zhanxiong
    Commented Feb 22, 2023 at 17:47

1 Answer 1


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.

  • $\begingroup$ +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. $\endgroup$
    – whuber
    Commented Feb 22, 2023 at 18:22
  • $\begingroup$ @whuber You're right. I will update the answer. $\endgroup$
    – Zhanxiong
    Commented Feb 22, 2023 at 18:27

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