Question

Q2 Suppose X1, X2, ..., Xn are i.i.d. Bernoulli random variables with probability of success p. It is known ΣΧ; is an unbiased estimator for p. that = n

Answer 1

Answer:

X1, X2......Xn ud Bernoulli(p) E(X;) =p V(X;) = p(1-P)

We know that < x;~ Binomial(n, p) i=1

(2)

$We\ know\ that\ T=\sum X_{i}\sim B(n,p)$

E(T) + Ex = np

Var(T) + Var (< x; x) = np(1-2)

$E(T^{2})=np-np^{2}+n^{2}p^{2}\ \ \ \ \ \ \ \ [\because Var(T)=E(T^{2})-(E(T))^{2}]$

$E(T^{2}-T)=E(T^{2})-E(T)=np-np^{2}+n^{2}p^{2}-np$

  $=np^{2}(n-1)$

$\therefore E\left [ \frac{T^{2}-T}{n(n-1)} \right ]=p^{2}$

$ie\ \ \frac{T^{2}-T}{n(n-1)}\ is\ an\ ube\ of\ p^{2}$

or

$\frac{\sum X_{i}[\sum X_{i}-1]}{n(n-1)}\ ube\ of\ p^{2}$

(3)

$\tilde{p}=\frac{\sum X_{i}+2}{n+4}$

$E(\hat{p})=\frac{1}{n+4}E\left [ \sum X_{i}+2 \right ]=\frac{1}{n+4}\left [ E\left ( \sum X_{i} \right )+2 \right ]$

  $=\frac{1}{n+4}(np+2)$

  $=\frac{np+2}{n+4}$

  $E(\hat{p})\neq p$

$\therefore \tilde{p}\ is\ a\ biased\ estimator\ of\ p$

(4)

$Now\ Var(\tilde{p})=\frac{1}{(n+4)^{2}}Var\left ( \sum X_{i}+2 \right )$

  $=\frac{Var\left ( \sum X_{i} \right )}{(n+4)^{2}}=\frac{np(1-p)}{(n+4)^{2}}$

$Var(\hat{p})=Var\left ( \frac{\sum X_{i}}{n} \right )=\frac{1}{n^{2}}Var\left ( \sum X_{i} \right )=\frac{np(1-p)}{n^{2}}$

  $=\frac{p(1-p)}{n}$

$Var(\tilde{p})<Var(\hat{p})\ when$

$\frac{np(1-p)}{(n+4)^{2}}<\frac{p(1-p)}{n}$