Question

1. Let $X_1, dots, X_n$ be an i.i.d. sample from a Bernoulli distribution with parameter $p$. That is, egin{equation} X_i = egin{cases} 1, & ext{with probability } p, \ 0, & ext{with probability } 1 - p. end{cases} end{equation} Let $ar{X}_n = frac{1}{n} sum_{i=1}^n X_i$ be the sample mean. a) (8 pts) Consider the statistic $T_n = ar{X}_n (1 - ar{X}_n)$. Calculate $E(T_n)$. b) (7 pts) Based on your answer to part a, formulate an unbiased estimator of $V(X_i)$. c) (8 pts) Let $U_n$ be an estimator of $p$, where egin{equation} U_n = frac{1 + sum_{i=1}^n X_i}{n + 2}. end{equation}Calculate $E(U_n)$ and the bias of $U_n$. d) (7 pts) Calculate the variance of $U_n$. e) (5 pts) Suppose that $p = frac{1}{2}$. Which estimator has the lowest MSE: $ar{X}_n$ or $U_n$?

          1. Let $X_1, dots, X_n$ be an i.i.d. sample from a Bernoulli distribution with parameter $p$. That is,
egin{equation*} X_i = egin{cases} 1, & 	ext{with probability } p, \ 0, & 	ext{with probability } 1 - p. end{cases} end{equation*}
Let $ar{X}_n = frac{1}{n} sum_{i=1}^n X_i$ be the sample mean.
a) (8 pts) Consider the statistic $T_n = ar{X}_n (1 - ar{X}_n)$. Calculate $E(T_n)$.
b) (7 pts) Based on your answer to part a, formulate an unbiased estimator of $V(X_i)$.
c) (8 pts) Let $U_n$ be an estimator of $p$, where
egin{equation*} U_n = frac{1 + sum_{i=1}^n X_i}{n + 2}. end{equation*}Calculate $E(U_n)$ and the bias of $U_n$.
d) (7 pts) Calculate the variance of $U_n$.
e) (5 pts) Suppose that $p = frac{1}{2}$. Which estimator has the lowest MSE: $ar{X}_n$ or $U_n$?

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$1. Let X1, dots, Xn be an i.i.d. sample from a Bernoulli distribution with parameter p. That is, eginequation* Xi = egincases 1, extwith probability p, 0, extwith probability 1 - p. endcases endequation* Let arXn = frac1n sumi=1^n Xi be the sample mean. a) (8 pts) Consider the statistic Tn = arXn (1 - arXn). Calculate E(Tn). b) (7 pts) Based on your answer to part a, formulate an unbiased estimator of V(Xi). c) (8 pts) Let Un be an estimator of p, where eginequation* Un = frac1 + sumi=1^n Xin + 2. endequation*Calculate E(Un) and the bias of Un. d) (7 pts) Calculate the variance of Un. e) (5 pts) Suppose that p = frac12. Which estimator has the lowest MSE: arXn or Un?$

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Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Sri K

01/28/2023

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Let X1, - Xn be an i.i.d. sample from a Bernoulli distribution with parameter p_ That is with probability P, Xi = 3 0. with probability 1 = p. Show more…

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Let X1, ..., Xn be an i.i.d. sample from a Bernoulli distribution with parameter p. That is, Xi = 1 with probability p, and Xi = 0 with probability 1 - p. Let Xn = (1/n) * ̑Xi be the sample mean. a) (8 pts) Consider the statistic Tn = Xn(1 - Xn). Calculate E(Tn). b) (7 pts) Based on your answer to part a, formulate an unbiased estimator of V(Xi). c) (8 pts) Let Un be an estimator of p, where Un = (1 + ̑Xi) / (n + 2). Calculate E(Un) and the bias of Un. d) (7 pts) Calculate the variance of Un. e) (5 pts) Suppose that p = 1/2. Which estimator has the lowest MSE: Xn or Un?

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0:00 Hello students.

00:02 Today we will discuss about this question.

00:04 In this question we are given that let x1 up to xn be an identically independent distribution sample for a bernoulli distribution with parameter p that is xi that is equals to 1 with probability p and 0 with probability 1 minus p and let x n bar that is equal to 1 divide by n summation of i is equals to 1 to n x i that be a sample mean sample mean so here in the part a we need to consider that statistic that is tn is equal to x n bar 1 minus x n bar and and then we need to calculate e of tn that is equals to question mark.

01:09 So here first of all, we can say that e of tn that can be given is e of xn bar minus xn bar whole square.

01:23 That is equal to e of xn bar minus e of xn bar whole square.

01:30 So that is equals to we can write e of xn bar minus v of xn bar plus e square of xn bar.

01:41 We can write like this.

01:43 Now here we can write e of xn bar that is equals to e of 1 divide by n.

01:50 Summation i is equals to 1 to 1 to n xi that is equals to 1 divide by n.

01:56 Summation i is equal to 1 to n e of x i now e of x i that is equals to 1 multiplied by p plus 0 multiplied by 1 minus p is so therefore e of x n bar is equal to 1 divide by n summation i is equals to 1 to n p so that is equals to np divided by n that is equal to p now e of x i square sorry e of x i square that is equals to one square multiplied by p plus zero square multiplied by one minus p that is equals to p so therefore here we can now we can write variance of x i is equals to e of x i so that is equal to p minus p square is equals to p 1 minus p.

02:56 Now here variance of x n bar is equals to variance of 1 divide by n, summation of i is equals to 1 to 1 2 n x i.

03:07 So that is equals to 1 divide by n square, summation i is equals to 1 to n v of x i...

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