View Question
 Question
 Subject: statistics Category: Miscellaneous Asked by: brkyrhrt99_00-ga List Price: \$6.00 Posted: 21 Oct 2002 12:25 PDT Expires: 20 Nov 2002 11:25 PST Question ID: 86073
 ```x is uniformly distributed (0, theta) there are 2 ways to estimate theta: 1) Theta hat=2 xbar 2) theta hat=amax(x1,....,xsubn) find: 1) a such that amax(x1,....,xsubn)is an unbiased estimate 2) standard deviation of theta hat 3) in what cases which estimate is better``` Request for Question Clarification by rbnn-ga on 21 Oct 2002 16:59 PDT ```Hi, Not sure what "amax(x1,...,xsubn)" means. I understand you are sampling n times from the distribution to get x1...xn, but what is "amax"? Similarly for one, what is "a" in the "a such that"?``` Clarification of Question by brkyrhrt99_00-ga on 21 Oct 2002 19:16 PDT ```I am not quite sure what it means myself. At the lecture the teacher just said amax and that was it.``` Clarification of Question by brkyrhrt99_00-ga on 21 Oct 2002 21:38 PDT `I believe what he means is medean. But then again I might be wrong`
 ```Thank you for another most interesting problem in probability/statistics. It sounds like the professor is choosing unusually interesting problems actually. Anyway, let's begin with some terminology (and of course, please remember to please ask for clarification if anything is unclear). I will interpret the word "amax" to mean "a * max" here, for some positive constant real number a. We are given a random variable X uniformly distributed from 0 to theta. We are given a positive integer constant n. We are given two other random variables: Y= 2*(x_1+x_2+...+x_n)/n and Z=a*max(x_1,...,x_n) where each x_i is independently distributed according to X. In words, for example, to compute Z we think of randomly drawing n numbers from 0 to theta, taking their max, then multiplying the result by a. We want to find the mean and standard deviation of the resulting numbers. We also define the random variables: W=max(x_1,...,x_n) --------------------------------- Part 1) We want to find an a such that Z is and unbiased estimate of theta. An unbiased estimate means that E(Z)=theta . In other words, if we do the experiment I mentioned above a bunch of times, we should get theta. To do this, we will begin by computing the probability density function of W . We will do that by first computing the probability DISTRIBUTION of W, then differentiating the distribution to get the density. (Recall the fact of probability that the derivative of distribution is density). Now the probability distribution of W is F(z) = P(W<=z) for some real number z. It is easy to compute this manually: F(z) =P(W<=z) =P(max(x_1,...,x_n))<=z 0 if z<=0 (z/theta)^n if 0<=z<=theta 1 if z>=theta Hence, the probability density for W, f, is the derivative: f(z)= F'(z)= n/theta^n * z^{n-1} if 0<=z<=theta 0 otherwise Now that we have the probability density, we can compute the expectation: E(W) =Integral from z= 0 to theta of z f(z) =Integral n/theta^n z^n = n/(n+1) * theta Hence, E(Z)=E(aW)=a E(W)= a*n/(n+1) * theta The estimate E(Z) for theta is unbiased if this value equals theta: E(Z)=theta if and only if a=(n+1)/n . Hence, the value for a that produces an unbiased estimate for theta is a= (n+1)/n By the way, I wrote a simple Java progam to test it. If you know Java or have it installed, here it is: public class MaxMean{ public static void main(String[] args){ int n=Integer.parseInt(args[0]); double a= (double)(n+1)/(double)n; double theta=Double.parseDouble(args[1]); int iterations=1000000; if (args.length>2)iterations=Integer.parseInt(args[2]); System.out.println("Starting trials: n: "+n+"\ntheta: "+theta+"\niterations: "+iterations); double sum=0; //sum of all the values so far for (int i=0;i2)iterations=Integer.parseInt(args[2]); System.out.println("Starting trials: n: "+n+"\ntheta: "+theta+"\niterations: "+iterations); double sum=0; //sum of all the values so far double sum2=0; //sum of squares for (int i=0;i
 brkyrhrt99_00-ga rated this answer: `excellent`