Question

Problem:2 Let Xi, X2,... , Xn be a random sample from Bernoulli(p) and consider es- timators iand p2- i. Compute the mean square error (MSEp) for both estimators p? and p2 Note that you must show the details of the calculation to receive full credit. ii. Use R to plot MSE, for both estimators using sample sizes n 20 and n- 300. Comment on the plots. iii. Use R to simulate 10,000 different Bernoulli samples of n 300 with success probability p- .45. For each trial, compute both estimators pi and p2. Estimate the true (MSEp) by computing MSE,(pl) 10000 an 10000 10000 2 - 45)2 where p and p are the estimates for the rth trial. Are the empirical MSEp values consistent with your plots from Part ii?

Answer 1

$E(\hat{p}_1)=p,~E(\hat{p}_2)=\frac{np+\sqrt{n/4}}{n+\sqrt{n}}\\ Var(X_i)=p(1-p),~Var(\hat{p}_1)=\frac{p(1-p)}{n},~Var(\hat{p}_2)=\frac{np(1-p)}{(n+\sqrt{n})^2}\\ MSE_p=\frac{p(1-p)}{n}~for~\hat{p}_1\\ MSE_p=Var(\hat{p}_2)+Bias^2(p)~for~\hat{p}_1\\$

$=\frac{np(1-p)}{(n+\sqrt{n})^2}+\left (\frac{np+\sqrt{n/4}}{n+\sqrt{n}}-p \right )^2$

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

(b) Now we take p=(0.4,0.6,0.9), n=(20, 300)

Then the values of $\hat{p}_1$ is

n=20 n=300

0.5 0.42
0.5 0.62
0.9 0.91

Then the values of $\hat{p}_2$ is

n=20 n=300

0.34 0.40

0.50 0.57
0.83 0.91

n 300 n=20 04 05 0.6 0.7 08 0.9

It is observed that value of $\hat{p}_2$ for n=300 is consistently dominated the value of $\hat{p}_2$ for n=20 whereas it is not observed for $\hat{p}_1$ .