Problem statement: "To show that an estimator can be consistent without being unbiased or even asymptotically unbiased, consider the following estimation procedure: To estimate the unknown mean, $\mu$ of a population with the finite variance $\sigma^2$, we first take a random sample of size $n$. Then we randomly draw one of $n$ slips of paper numbered from $1$ through $n$, and if the number we draw is $2, 3,\ldots$ or $n$, we use as our estimator the mean of the random sample, $\bar X$; otherwise, we use the estimate $n^2$. Show that the estimation procedure is consistent and neither unbiased nor asymptotically unbiased."
I have figured out the following:
$E[X^*|2,3,\ldots,n]=E[\bar X]=\mu$
$E[X^*|1]=E[n^2]=n^2$
$E[X^*]=E[\bar X]\cdot\frac{n-1}{n}+E[n^2]\cdot\frac{1}{n}=\mu\frac{n-1}{n}+n^2\frac{1}{n}$ by the law of total expectation. Thus, $X^*$ is neither unbiased or asymptotically unbiased since the expected value is not equal to $\mu$ and the limit as $n$ approaches infinity is not equal to $0$.
I am not sure how to show that the procedure is consistent. Any suggestions on a starting place?
Let $Y$ be the estimator of the paramenter $\mu$ , $A_\epsilon$ be the unfortunate event of having $|Y - \mu| \ge \epsilon$ for some $\epsilon >0$.
First a little of intuition: we know that the sample mean $Y_1=\bar X$ is a good (consistent) estimator (weak law of large numbers). Hence, we know that, if we used that estimator, $\displaystyle P(A_\epsilon) \to 0$ as ${n\to\infty}$. The problem here is that sometimes we don't use that nice estimator, but a ridiculously bad one ($Y_2=n^2$); fortunately, the bad estimator is used rarely, and more rarely as $n$ grows. Is this enough to (asympotically) dismiss it, and assert that we have a good estimator? Let's see.
Let $Z$ be the indicator value of the event that we draw the number 1 ($Z=1$ if we had to used the bad estimator, $Z=0$ otherwise).
$$ \begin{align} P(A_\epsilon) &= \sum P(A_\epsilon , Z) \\ & = P(A_\epsilon \mid Z=1) P(Z=1)+P(A_\epsilon \mid Z=0) P(Z=0)\\ & = P(A_\epsilon \mid Z=1) \frac{1}{n}+P(A_\epsilon \mid Z=0) \frac{n}{n-1} \end{align}$$
We want the left side to tend to zero. Now, we know the term $P(A_\epsilon \mid Z=0)$ tends to zero (because of what we said above), hence the problematic term is $P(A_\epsilon \mid Z=1)$ which can be quite high... but not higher than $1$ (no?)
Hence we can bound
$$P(A_\epsilon) \le \frac{1}{n} + P(A_\epsilon \mid Z=0)$$
Because both terms on the right side tend to zero, so does the left side.