Question

Exercise 6.14 Let y be distributed Bernoulli P(y = 1) unknown 0<p<1 p and P(y = 0) = 1-p f or Some

Answer 1

Part 1

$\mathbb{E}(Y)=\sum y\times\mathbb{P}(Y=y)=0*(1-p)&plus;1*p=p$

Part 2

Suppose we have iid observations $Y_1,Y_2,...,Y_n$ from the distribution of $Y$.

Define $\bar{Y}_n=\frac1n\sum Y_i$ .

Then, $\mathbb{E}(\bar{Y}_n)=\frac1n \sum\mathbb{E}(Y_i)=\frac1n\sum p=p$ .

So, the natural moment estimator $\hat{p}=\bar{Y}_n$

Part 3

Let us first derive the variance of $Y$.

$Var(Y)=\mathbb{E}(Y^2)-(\mathbb{E}(Y))^2=(0^2*(1-p)&plus;1^2*p)-(p)^2=p-p^2=p(1-p)$

Now, $Var(\hat{p})=Var(\bar{Y}_n)=\frac1nVar(Y)=\frac{p(1-p)}n$

Part 4

Let $T=\sqrt{n}(\hat{p}-p)$ .

Then, $T=\sqrt{n}(\bar{Y}_n-p)=\sqrt{n}(\frac1n\sum Y_i-p)=\sqrt{n}\times\frac1n\sum(Y_i-p)$ .

Now, $(Y_i-p)$s are iid random variables, with $\mathbb{E}(Y_i-p)=\mathbb{E}(Y_i)-p=p-p=0$ and $Var(Y_i-p)=Var(Y_i)=p(1-p)$ .

Hence, by Lindeberg Levy Central Limit Theorem, asymptotic distribution of T is $\mathbb{N}(0,p(1-p))$ .