Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from the uniform distribution with pdf \(f\left(x ; \theta_{1}, \theta_{2}\right)=1 /\left(2 \theta_{2}\right), \theta_{1}-\theta_{2}0\) and the pdf is equal to zero elsewhere. (a) Show that \(Y_{1}=\min \left(X_{i}\right)\) and \(Y_{n}=\max \left(X_{i}\right)\), the joint sufficient statistics for \(\theta_{1}\) and \(\theta_{2}\), are complete. (b) Find the MVUEs of \(\theta_{1}\) and \(\theta_{2}\).

Short answer

The minimum and maximum of the sample, \(Y_{1}=\min (X_{i})\) and \(Y_{n}=\max (X_{i})\), are joint sufficient statistics and they are also complete. The MVUEs for the parameters \(\theta_{1}\) and \(\theta_{2}\) of the uniform distribution are given by \(\hat{\theta}_1 = \frac{Y_1 + Y_n}{2}\) and \(\hat{\theta}_2 = \frac{Y_n - Y_1}{2}\) respectively.

Step-by-step solution

- Testing for Completeness

First, show that \(Y_{1}=\min (X_{i})\) and \(Y_{n}=\max (X_{i})\) are complete. Towards the end, we need to show that the expectation of any unbiased estimator of the parameters based on \(Y_{1}\) and \(Y_{n}\) is zero. In this distribution, consider any unbiased estimator g(Y_1,Y_n). Its expectation will be:\[E[g(Y_1,Y_n)]=0\]If this holds true for all \(Y_1\) and \(Y_n\), then \(Y_1\) and \(Y_n\) can be considered complete for this distribution.

- Sufficient Statistics

To show that \(Y_{1}\) and \(Y_{n}\) are sufficient statistics, use the factorization theorem which states that a statistics T(X) is sufficient for θ iff the joint pdf can be factorized as:\[f(x_1,x_2,...,x_n;θ) = g(T(x_1,x_2,...,x_n);θ) * h(x_1,x_2,...,x_n)\]Here \(T(x_1,x_2,...,x_n)\) is the sufficient statistics, in this case it would be \(T(X) = (Y_1, Y_n)\). Therefore, the joint pdf can be expressed in terms of \(Y_1\) and \(Y_n\) (minimum and maximum values of the sample).

- MVUEs of \(\theta_{1}\) and \(\theta_{2}\)

To find the minimum variance unbiased estimators (MVUEs) of \(\theta_{1}\) and \(\theta_{2}\), we know from Lehmann-Scheffe theorem that if a statistic is complete and sufficient, then it is the unique best unbiased estimator of its expected value. Hence, we can evaluate the expected value of \(Y_{1}\) and \(Y_{n}\) which will give the MVUEs for \(\theta_{1}\) and \(\theta_{2}\) respectively. \[E[Y_1] = \theta_1 - \theta_2\]\[E[Y_n] = \theta_1 + \theta_2\]Which shows us that the MVUEs for \(\theta_{1}\) and \(\theta_{2}\) can be expressed in terms of \(Y_{1}\) and \(Y_{n}\):\[\hat{\theta}_1 = \frac{Y_1 + Y_n}{2}\]And \[\hat{\theta}_2 = \frac{Y_n - Y_1}{2}\]