# Doubt in the Invariance Property of Consistent Estimators

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 $$\mu$$?

$$X_{n+1}$$ 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 $$X_{n+1}$$.
• 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 $X_{n+1}$. Feb 13, 2021 at 9:52
• In your context, it would need to converge to $\mu$ in order for $T$ to be consistent for $\mu$. Feb 14, 2021 at 7:57