Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\) be independent random samples from \(N\left(\theta_{1}, \theta_{3}\right)\) and \(N\left(\theta_{2}, \theta_{4}\right)\) distributions, respectively. (a) If \(\Omega \subset R^{3}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{2}, \theta_{3}\right):-\infty<\theta_{i}<\infty, i=1,2 ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}, \theta_{2}\), and \(\theta_{3}\). (b) If \(\Omega \subset R^{2}\) is defined by $$ \Omega=\left\\{\left(\theta_{1}, \theta_{3}\right):-\infty<\theta_{1}=\theta_{2}<\infty ; 0<\theta_{3}=\theta_{4}<\infty\right\\} $$ find the mles of \(\theta_{1}\) and \(\theta_{3}\).

Short answer

For the first condition, \(\hat{\theta_{1}} = \frac{1}{n} \sum_{i=1}^{n} X_{i}\), \(\hat{\theta_{2}} = \frac{1}{m} \sum_{i=1}^{m} Y_{i}\), and \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{2}})^{2}\). For the second condition, \(\hat{\theta_{1}} = \frac{1}{n+m} \sum_{i=1}^{n} X_{i} + \sum_{j=1}^{m} Y_{j}\) and \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{1}})^{2}\)

Step-by-step solution

Compute mles for the first condition

For \(X_{1}, X_{2}, \ldots, X_{n}\sim N(\theta_{1}, \theta_{3})\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\sim N(\theta_{2}, \theta_{3})\), we calculate the sample means as the mle estimates for \(\theta_{1}\) and \(\theta_{2}\) as these are independent normal samples. Thus, \(\hat{\theta_{1}} = \frac{1}{n} \sum_{i=1}^{n} X_{i}\) and \(\hat{\theta_{2}} = \frac{1}{m} \sum_{i=1}^{m} Y_{i}\). Compute also for \(\theta_{3}\) as \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{2}})^{2}\)

Compute mles for the second condition

For \(X_{1}, X_{2}, \ldots, X_{n}\sim N(\theta_{1}, \theta_{3})\) and \(Y_{1}, Y_{2}, \ldots, Y_{m}\sim N(\theta_{1}, \theta_{3})\), the mle estimate for \(\theta_{1}\) is the combined sample mean. Thus, \(\hat{\theta_{1}} = \frac{1}{n+m} \sum_{i=1}^{n} X_{i} + \sum_{j=1}^{m} Y_{j}\). Compute for \(\theta_{3}\) as \(\hat{\theta_{3}} = \frac{1}{n+m} \sum_{i=1}^{n} (X_{i} - \hat{\theta_{1}})^{2} + \sum_{j=1}^{m} (Y_{j} - \hat{\theta_{1}})^{2}\)