Question

Question: Suppose that in a problem of simple linear regression, a confidence interval with confidence coefficient \({\bf{1 - }}{{\bf{\alpha }}_{\bf{0}}}\)\(\left( {{\bf{0 < }}{{\bf{\alpha }}_{\bf{0}}}{\bf{ < 1}}} \right)\) is constructed for the height of the regression line at a given value of \(x\). Show that the length of this confidence interval is shortest when \({\bf{x = \bar x}}\).

Short answer

It is proved that the length of the confidence interval is shortest when \(x = \bar x\).

Step-by-step solution

Determine the formula for confidence interval for predicted value

The\({\bf{(1 - \alpha )100\% }}\)confidence interval for the mean of\({\bf{Y}}\)when\({\bf{x}}\)is of the form:

\(\left( {{{{\bf{\hat \beta }}}_{\bf{0}}}{\bf{ + }}{{{\bf{\hat \beta }}}_{\bf{1}}}{\bf{x}}} \right){\bf{ \pm \sigma '}}\left( {\sqrt {\frac{{\bf{1}}}{{\bf{n}}}{\bf{ + }}\frac{{{{{\bf{(x - \bar x)}}}^{\bf{2}}}}}{{{\bf{s}}_{\bf{x}}^{\bf{2}}}}} } \right){\bf{T}}_{{\bf{n - 2}}}^{{\bf{ - 1}}}\left( {{\bf{1 - }}\frac{{\bf{\alpha }}}{{\bf{2}}}} \right)\)

where,

and \(T_{n - 2}^{ - 1}\left( {1 - \frac{\alpha }{2}} \right)\) is a value such that the area under the density curve of a Student’s \(t\)-distribution left of that value, equals \(1 - \frac{\alpha }{2}\) with following degrees of freedom,

\(\begin{array}{c}n - 2 = 10 - 2\\ = 8\end{array}\)

State the length of the confidence interval

The length of this confidence interval is the difference between its upper and lower limit.

\({l_0}(x) = 2\sigma '\sqrt {\frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}} T_{n - 2}^{ - 1}\left( {1 - \frac{\alpha }{2}} \right)\)

To determine the value of \(x\)which minimizes function \({l_0}\).

\({l_1}(x) = \sqrt {\frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}} \)

Since the square root function is strictly increasing, then its equivalent to minimize the function under the square root. We get,

\({l_2}(x) = \frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}\)

Hence, the length of the confidence interval is \({l_0}(x) = 2\sigma '\sqrt {\frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}} T_{n - 2}^{ - 1}\left( {1 - \frac{\alpha }{2}} \right)\).

Find the shortest length in the confidence interval

Cancel out the common terms from \({l_2}(x) = \frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}\).

Finally, we conclude that the same value of \(x\)which minimizes \({l_2}\)also minimizes

\(l(x) = {(x - \bar x)^2}\)

The derivative of \(l\)with respect to \(x\)is,

\(l'(x) = 2(x - \bar x)\)

Stationary points are the null points of the above derivative,

\(\begin{array}{c}l'(x) = 0\\2(x - \bar x) = 0\\x = \bar x.\end{array}\)

The second derivative is,

\(l''(x) = 2 > 0\)

This implies that \(l\)is strictly convex.

Thus, at the point \(x = \bar x\) it reaches its global minimum.

Hence we proved.