Question

Consider again the conditions of Exercise 12, For what value of the temperature \({\bf{x}}\) can the durability of a specimen of the alloy be predicted with the smallest M.S.E.?

Short answer

The smallest M.S.E of the prediction is obtained when the new observation equals the sample mean \(\bar x = 2.25\)

Step-by-step solution

Definition of Mean Squared Error.

The mean squared error of the prediction\(\hat y(x)\)using linear regression is,

\({\mathop{\rm MSE}\nolimits} (\hat y(x)) = {\sigma ^2}\left( {1 + \frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}} \right)\)

Where,

\(\begin{array}{l}{\sigma ^2}:{\rm{variance}}\\n:{\rm{sample}}\;{\rm{size}},\\\bar x:{\rm{mean}},\\s_x^2:s{\rm{ample}}\;{\rm{variance}},\\x:{\rm{observations}}\end{array}\)

It is required to minimize the value of MSE to obtain the smallest value of x which minimizes the MSE

Define the function which is required to be minimized,\(f:\mathbb{R} \to \langle 0, + \infty \rangle ,\quad f(x) = {\sigma ^2}\left( {1 + \frac{1}{n} + \frac{{{{(x - \bar x)}^2}}}{{s_x^2}}} \right)\)

Determine the value of the durability of the specimen.

The first derivative of\(f\)is,

\(f'(x) = {\sigma ^2}\left( {\frac{{2(x - \bar x)}}{{s_x^2}}} \right)\)

Notice that neither\(\bar x\)nor\({s_x}\)depend on\(x\)- they are computed using some fixed values\({x_1}, \ldots ,{x_n}\).

Equate the first derivative to 0,

\(\begin{array}{c}f'(x) = 0\\{\sigma ^2}\left( {\frac{{2(x - \bar x)}}{{s_x^2}}} \right) = 0\\x - \bar x = 0\\x = \bar x\end{array}\)

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

\(\begin{array}{c}f''(x) = {\sigma ^2}\left( {\frac{2}{{s_x^2}}} \right)\\ > 0\end{array}\)

Thus, the value \({x^*} = \bar x\) is the global minima.

From exercise 12, the x-mean is 2.25.

Therefore, the smallest M.S.E of the prediction is obtained when the new observation equals the sample mean \(\bar x = 2.25\)