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I'm working on a homework problem.

Let $X_1,\dots,X_n$ be i.i.d Bernoulli random variables with $\mathbb{P}(X_i=1)=p $ and $\mathbb{P}(X_i=0)=1-p$, where $p\in(0,1)$. However, we can only observe $Y_i=\textbf{1}_{(X_i=1,X_{i+1}=1)},i=1,2,\dots,n-1$, where $\textbf{1}$ denotes the indicator function.

Question 1 is to find a moment estimator of $p$ based on $Y_1,\dots,Y_{n-1}$.

Question 2 is to find the (nondegenerate) asymptotic distribution of the estimator in Q1.

I'm not quite sure whether my understanding of Q1 is right. Here's my thought.

For Q1, we can easily compute $\mathbb{E}(Y_i)=p^2, Var(Y_i)=p^2(1-p^2), Cov(Y_i,Y_{i+1})=p^3-p^4$. By Chebyshev's Inequality, we can show $\frac{\sum\limits_{i=1}^{n-1}Y_i}{n-1}\stackrel{P}{\rightarrow} p^2$. Therefore, a moment estimator can be $\hat{p}=(\frac{\sum\limits_{i=1}^{n-1}Y_i}{n-1})^{\frac{1}{2}}$.

However, for Q2, the variance of $\hat{p}^2$ goes to zero as $n$ goes to infinity. By Lindeberg-Feller CLT, we can only get $\frac{\hat{p}^2-p^2}{\sqrt{Var(\hat{p}^2)}}\stackrel{d}{\rightarrow}\mathcal{N}(0,1)$, but cannot find the asymptotic distribution of $\hat{p}^2$.

Any help will be appreciated.

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  • $\begingroup$ Indicator...sorry for the typos. I've already fixed them. $\endgroup$ – Kangping Yan May 6 '18 at 4:38
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You should inderstand the second question as: find the linear transformation of $\hat p$ s.t. it converges in distribution to some non-degenerate distribution. It is impossible to understand this question directly: since $\hat p\stackrel{P}{\rightarrow}p$, then $\hat p\stackrel{d}{\rightarrow}p$ and the limiting distribution is degenerate.

In view of asymptotic normality of $\hat p^2$, we can find some $\sigma(n)$ s.t. $$ \frac{\hat p - p}{\sigma(n)} \stackrel{d}{\rightarrow} \mathcal N(0,1) $$ To do it, one can either use delta-method or apply Slutsky's theorem as the following: $$ \frac{\hat{p}-p}{\sigma(n)} =\frac{\hat p^2-p^2}{\sqrt{Var(\hat{p}^2)}}\cdot\frac{\sqrt{Var(\hat{p}^2)}}{\sigma(n)}\cdot\underbrace{\frac{1}{\hat p+p}}_{\stackrel{P}{\rightarrow} \frac1{2p}} \stackrel{d}{\rightarrow}\mathcal{N}(0,1) $$ if $$ \sigma(n) = \frac{\sqrt{Var(\hat{p}^2)}}{2p}=\sqrt{\frac{Var(\hat{p}^2)}{4p^2}} $$ The answer is $$\tag{1}\label{1} \frac{\hat{p}-p}{\sqrt{\frac{Var(\hat{p}^2)}{4p^2}}}\stackrel{d}{\rightarrow}\mathcal{N}(0,1) $$ I have seen many times that this answer is written in the form $$ \hat{p} \stackrel{d}{\rightarrow}\mathcal{N}\left(p,\tfrac{Var(\hat{p}^2)}{4p^2}\right) $$ and then $\mathcal{N}\left(p,\tfrac{Var(\hat{p}^2)}{4p^2}\right)$ is declared to be the limiting distribution of $\hat{p}$.

This is a bit strange, because $n$ presents in the right hand side. Such a statement should be understood in the sense of (\ref{1}).

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  • $\begingroup$ So you think I correctly understand question 1? And other moment estimators' asymptotic distributions are also in such a "strange" form? $\endgroup$ – Kangping Yan May 6 '18 at 11:04
  • $\begingroup$ Yes, you correctly find MME for $p$. The variance of $Y_i$ and covariances of neibours are also correct. Not entirely sure that the Lindeberg - Feller CLT is applicable, but completely agreed that CLT is valid for this weak dependent (so called 'one-dependent') r.v.'s. This meaning of words "asymptotic distribution of smth" is standard. Look here for example math.stackexchange.com/q/2425696/413376 $\endgroup$ – NCh May 6 '18 at 12:54

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