Suppose that, for some unknown parameter ?

Suppose that, for some unknown parameter ? > 0, X1, ..., Xn i.i.d.~ U(0, ?), that is, X1, ..., Xn are a random sample from a (continuous) uniform distribution on the interval (0, ?). You are given that E(Xi) = ?/2 and var(Xi) = ?^2/12, for i = 1, ..., n. 1. The estimator T1 of ? Consider the estimator T1 = (2/n) ?_{i=1}^n Xi. (a) Find E(T1). (b) Find var(T1). (c) Find MSE(T1). 2. The estimator T2 of ? Consider the estimator T2 = ((n+1)/n) X_(n), where X_(n) = max(X1, ..., Xn) is the largest of X1, ..., Xn. You are given that E(X_(n)) = (n/(n+1)) ? and var(X_(n)) = (n/((n+1)^2 (n+2))) ?^2. (a) Find E(T2). (b) Find var(T2). (c) Find MSE(T2).

Transcript

00:01 Okay, so in this question we have some unknown parameter theta, and we're told that we have the expectation value of xi is theta over 2, and the variance of xi is theta squared over 12 for i equals 1 up to n, and all of the xi are independent.

00:35 So now first of all, let's consider t1, which is 2 over n times the sum from i equals 1 to n of xi.

00:44 First of all, let's find the expected value of t1.

00:50 This is going to be, because the expectation value is a linear operator, we can put it inside the sum.

01:00 And since the expected value of xi is theta over 2, this becomes theta over n times the sum from i equals 1 to n of 1.

01:09 And this is just n, so the expected value of t1 is theta.

01:18 Now we want the variance of t1.

01:21 Variance of t1 is the variance of the sum from i equals 1 to n, 2xi over n.

01:29 Since all of these variables are independent, we can bring the variance inside the sum, and then since the variances of each of these xi is the same, we can pull the 2 over n out by squaring it, and then the variance of the xi is theta squared over 12.

01:52 So this gives us, and because each of these are the same, we get theta squared over 3n squared times n.

02:03 So the variance of t1 is theta squared over 3n.

02:12 So the mean squared error of t1 is the variance of t1 plus the bias of t1 squared, which is the expected value of t1 minus theta squared.

02:27 But the expected value of t1 is theta because it's an unbiased estimator.

02:32 So the mean squared error of t1 is just its variance, which is theta squared over 3m.

02:46 Okay, so now let's have a look at an estimator t2, which is n plus 1 over n times the nth order statistic, which is the maximum of the set of observations.

03:03 And we're told that the expected value of that nth order statistic is n over n plus 1 times theta, and its variance is n over n plus 1 squared n plus 2 theta squared.

03:23 So first of all, let's find the expected value of t2...

Question

Please give Ace some feedback