91Ó°ÊÓ

The distribution of X1 and X2 are uniform in the interval \(\left( {0,\theta } \right)\,and\,\left( {2\theta ,3\theta } \right)\).

The joint p.d.f of X1 and X2 is \(\frac{1}{{4{\theta ^2}}}\) provided that each observation lies in either the interval\(\left( {0,\theta } \right)\) or the interval \(\left( {2\theta ,3\theta } \right)\). Thus M.L.E of \(\theta \) will be the smallest value of \(\theta \) for which these restrictions are satisfied.

If we take it \(3\hat \theta = 9,\,or\,\hat \theta = 3\), then \(\hat \theta \) it will be as small as possible, and the restrictions will be satisfied because both observed values will lie in the interval\(\left( {2\hat \theta ,3\hat \theta } \right)\).

It is not possible that both X1 and X2 lie in the interval \(\left( {2\theta,3\theta } \right)\)because it is necessary r that to be true.

Here, however, \({X_2}/{X_1} = 9/4\).

Therefore, if we take \(\hat \theta = 4, then\,\hat \theta \) will be as small as possible, and restrictions will be satisfied because X1 will lie in the interval\(\left( {0,\hat \theta } \right)\) and X2 will lie in \(\left( {2\hat \theta,3\hat \theta } \right)\).

X1 and X2 may lie in the interval\(\left( {2\theta,3\theta } \right)\)for a reason in part (b).

It is not possible that X1 lies in\(\left( {0,\theta } \right)\) and X2 lies in\(\left( {2\theta ,3\theta } \right)\)because that must be true\({X_2}/{X_1} \ge 2\). Here, however\({X_2}/{X_1} = 9/5\).

Hence, it must be true that both X1 and X2 lie in the interval \(\left( {0,\theta } \right)\). Under this condition, the smallest possible value of \(\theta \) is \(\hat \theta = 9\).