95% confidence interval for \theta_{2}-\theta_{1} from uniform (\theta_{1}, \theta_{2})

bittercynicn7f

bittercynicn7f

Answered question

2022-03-03

95% confidence interval for θ2θ1 from uniform (θ1,θ2)
where X1,,Xn are the random sample from U(θ1,θ2), the uniform distribution with two parameters θ1,θ2.
I have the maximum likelihood estimators θ^1=X(1), the minimum order statistic, and θ^2=X(n), the maximum order statistic. But I don't know how to compute the confidence interval of θ2θ1.

Answer & Explanation

Jenson Mcculloch

Jenson Mcculloch

Beginner2022-03-04Added 9 answers

Let δ =θ2θ1 and Zi =θ1+Xi. We have ZiUnif(0,δ), and X(1)=θ1+Z(1) and X(n)=θ1+Z(n). Then,
R =X(n)X(1)=Z(n)Z(1)
is a good estimate of δ.
We note that Z(n)Z(1)δ is a pivot, i.e. its distribution does not depend on δ and it is easy to find out. This can be used to construct the CI.
Hint: Note that Ui =Ziδ is uniform (0,1). You are basically computing the distribution of U(n)U(1). You can either use the known joint distribution of (U(1),U(n)) or do it directly yourself:
(U(n)U(1)t) =(U(n)t+U(1))  

  =01[U(n)t+u, U(1)(u,u+du)]
The integrand is the probability that the minimum is about u and the rest of them are (u,t+u). There are n possible choices for which one of the variables is the minimum, and all these events have the same probability du(min{t+u,1}u)n1. Hence,
p(t) =P(U(n)U(1)t)=n01(min{t+u,1}u)n1du.
which can be simplified and explictly computed =n(1t)tn1+tn for t[0,1].
Choose t1 and t2 such that p(t2)p(t1)=0.95. Then [Rt2,Rt1] is a 95% CI for δ. You can try to minimize the length, by minimizing |1t21t1| subject to the given constraint.

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