Let $X_1,X_2,X_3,..,X_n,X_{n+1}$ be random samples from $N(\mu,1)$. Let us define $\bar {X}_n = \frac{\sum X_i}{n}$ and $T = \frac{1}{2}(\bar {X}_n + X_{n+1})$. It is required to test whether $T$ is consistent and unbiased for $\mu$.

We can clearly see that T is unbiased for $\mu$. For testing the consistency, I have compute the varianve of the $T$ as

$V(T) = \frac{1}{4}(\frac{1}{n} + 1)$ which as $n$ is large does not converge to 0. Henec, from this definition, $T$ is not consistent.

But if we see the invariance property of consistent estimators, it says any continuous function of consistent estimator is consistent. Can't we view the $T$ as the function of consistent estimator $\bar {X}_n$ for $\mu$?


1 Answer 1


$X_{n+1}$ converges not to a constant, but to a distribution with variance 1.

Therefore, Slutsky's theorem (with the two "estimators" $\dfrac{1}{2}\overline{X}_n$ and $\dfrac{1}{2}X_{n+1}$ being two random elements) does not apply. For that, you would need to have $\dfrac{1}{2}X_{n+1}$ converging to a constant.

Intuitively, your estimator $T$ gives constant (and therefore "not vanishing") weight to $X_{n+1}$.

  • $\begingroup$ Ohh. I see. So, it is $X_{n+1}$ which is causing it to become inconsistent. If it was a constant, it would have been consistent. Can it be any constant? Let us suppose we have $n$ in place of $X_{n+1}$. $\endgroup$
    – userNoOne
    Commented Feb 13, 2021 at 9:52
  • 1
    $\begingroup$ In your context, it would need to converge to $\mu$ in order for $T$ to be consistent for $\mu$. $\endgroup$ Commented Feb 14, 2021 at 7:57

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