91Ó°ÊÓ

Referring to example 8.5.11, there is a uniform distribution on the interval \(\left( {\theta - \frac{1}{2},\theta + \frac{1}{2}} \right)\;,\;\left( { - \infty < \theta < \infty } \right)\), which \(\theta \) is unknown. The two observations such that \({X_1} = 4.7\;and\;{X_2} = 5.3\) are taken randomly from the uniform distribution.

Let us consider,

\(\begin{align}{Y_1} &= \min \left( {{X_1},{X_2}} \right)\\ &= \min \left( {4.7,5.3} \right)\\ &= 4.7\end{align}\)

And

\(\begin{align}{Y_2} &= \max \left( {{X_1},{X_2}} \right)\\ &= \max \left( {4.7,5.3} \right)\\ &= 5.3\end{align}\)

Thus, the confidence interval between larger and smaller values is \(\left( {4.7,5.3} \right)\).

The consistent with the observed data values \(\theta \) are between \(\theta - \frac{1}{2}\;and\;\theta + \frac{1}{2}\)

Therefore,

\(\begin{align}\left( {\theta - \frac{1}{2}} \right) &= 5.3 - \frac{1}{2}\\ &= 4.8\end{align}\)

And

\(\begin{align}\left( {\theta + \frac{1}{2}} \right) &= 4.7 + \frac{1}{2}\\ &= 5.2\end{align}\)

Thus, the interval of possible \(\theta \) values consistent with the observed data is \(\left( {4.8,5.2} \right)\).

In part b, there is the interval for possible \(\theta \) values , and in part a, there is the 50% confidence interval \(\left( {4.7,5.3} \right)\).

So, by both intervals, it is observed that the 50% confidence interval is larger than the set of possible \(\theta \) values.

Consider, \(Z = {Y_2} - {Y_1}\).

Now referring to part a,\({Y_1} = 4.7\;and\;{Y_2} = 5.3\)

So, the value of the random variable is,

\(\begin{align}Z &= 5.3 - 4.7\\ &= 0.6\end{align}\)

By the theorem, the conditional probability that\({{\bf{\bar X}}_{\bf{2}}}\)is close to\({\bf{\theta }}\)given Z=z is

\({\bf{Pr}}\left( {\left| {{{{\bf{\bar X}}}_{\bf{2}}} - {\bf{\theta }}} \right|{\bf{ < c}}\left| {{\bf{Z = z}}} \right.} \right){\bf{ = }}\left\{ \begin{align}{l}\frac{{{\bf{2c}}}}{{{\bf{1}} - {\bf{z}}}}\;{\bf{if}}\;{\bf{c}} \le \frac{{{\bf{1}} - {\bf{z}}}}{{\bf{2}}}\\{\bf{1}}\;{\bf{if}}\;{\bf{c > }}\frac{{{\bf{1}} - {\bf{z}}}}{{\bf{2}}}\end{align} \right.\)

Since,

\(\begin{align}0.1 \le \frac{{1 - 0.6}}{2}\\ \Rightarrow 0.1 \le 0.2\end{align}\)

Therefore, the probability that\(\left( {\left| {{{\bar X}_2} - \theta } \right| < 0.1\left| {z = 0.6} \right.} \right)\)is,

\(\begin{align}\Pr \left( {\left| {{{\bar X}_2} - \theta } \right| < 0.1\left| {Z = 0.6} \right.} \right) &= \frac{{2c}}{{1 - z}}\\ &= \frac{{2 \times 0.1}}{{1 - 0.6}}\\ &= 0.5\end{align}\)

Thus, the required conditional probability is 0.5.