What Defines a Maximum Likelihood Estimator in Ordered Statistics?

[cse63146]

Homework Statement

Let Y1<Y2<...<Yn be the order statistics of a random sample from a distribution with pdf $$f(x; \theta) = 1$$, $$\theta - 0.5 < x < \theta + 0.5$$. Show that every statistic u(X1,X2,...,Xn) such that $$Y_n - 0.5<u(X_1,X_2,...,X_n)<Y_1 + 0.5$$ is a mle of theta. In particular $$(4Y_1 + 2Y_n + 1)/6, (Y_1 + Y_n)/2, (2Y_1 + 4Y_n - 1)/6$$ are three such statistics.

Homework Equations

The Attempt at a Solution

I don't even know how to start this problem. Can someone please point me in the right direction?

 

[mighty2000]

Thank you for reaching out with your question. It seems like you are trying to understand the properties of order statistics and how they relate to maximum likelihood estimation (MLE). Let's break down the problem step by step to help you get started.

First, let's define some notation. The notation Y1<Y2<...<Yn represents the order statistics of a random sample from a distribution with pdf f(x; \theta). This means that Y1 is the smallest value in the sample, Y2 is the second smallest value, and so on, with Yn being the largest value. The notation u(X1,X2,...,Xn) represents any statistic that can be calculated using the sample data.

Next, the given pdf f(x; \theta) = 1, \theta - 0.5 < x < \theta + 0.5 indicates that the distribution is uniform over the interval [\theta - 0.5, \theta + 0.5]. In other words, any value within this interval is equally likely to be observed.

Now, let's think about what it means for a statistic u(X1,X2,...,Xn) to be a MLE of \theta. This means that the statistic u is the most likely value of \theta given the observed data. In other words, it maximizes the likelihood function L(\theta; x1,x2,...,xn) = f(x1;\theta)f(x2;\theta)...f(xn;\theta) for a given set of data x1,x2,...,xn.

To show that a statistic u is a MLE of \theta, we need to show that it satisfies the following two conditions:
1. u(X1,X2,...,Xn) is a function of the order statistics Y1,Y2,...,Yn.
2. u(X1,X2,...,Xn) maximizes the likelihood function L(\theta; x1,x2,...,xn).

Now, let's apply these conditions to the given statistics (4Y1 + 2Yn + 1)/6, (Y1 + Yn)/2, and (2Y1 + 4Yn - 1)/6.

For the first statistic, (4Y1 + 2Yn + 1)/6, we can rewrite it as u(Y1,Y2,...,Yn) = (4Y