91Ó°ÊÓ

\(V\) is a random variable and \(r\left( {v,x} \right)\) is strictly decreasing in v for each x .Then it is required to prove that the following are the endpoints of an exact coefficient \(\gamma \) confidence interval for \(g\left( \theta \right)\) :

\(\begin{align}A &= r\left( {{G^{ - 1}}\left( {{\gamma _1}} \right),X} \right)\\B &= r\left( {{G^{ - 1}}\left( {{\gamma _2}} \right),X} \right)\end{align}\)

Let \(r\left( {v,x} \right)\) is strictly decreasing in v for each x then if

\(V\left( {X,\theta } \right) < c\) ,it follows that \(g\left( \theta \right) > r\left( {c,X} \right)\) and vice-versa.

It follows that \(V\left( {X,\theta } \right) < c\)

From the definition of A and B by setting \(c = {G^{ - 1}}\left( {{\gamma _i}} \right),i = 1,2\)

Then the following must be true

\(\begin{align}P\left( {g\left( \theta \right) > B} \right) &= {\gamma _1}\,\,\,\,\,\,{\rm{and}}\\P\left( {g\left( \theta \right) > A} \right) &= {\gamma _2}\end{align}\)

Now using the fact that \(r\left( {v,x} \right)\)is strictly decreasing in v for each x it follows that

\(\begin{align}P\left( {A = g\left( \theta \right)} \right) = P\left( {V\left( {X,\theta } \right) = {G^{ - 1}}\left( {{\gamma _2}} \right)} \right) = 0\,\,\,\,\,{\rm{and}}\\{\rm{P}}\left( {B = g\left( \theta \right)} \right) = P\left( {V\left( {X,\theta } \right) = {G^{ - 1}}\left( {{\gamma _1}} \right)} \right) = 0\end{align}\)

Which finally gives the following the result:

\(P\left( {A < g\left( \theta \right) < B} \right) = {\gamma _2} - {\gamma _1} = \gamma \)

Hence the proof.