91Ó°ÊÓ

A person either can pay a cost c for the opportunity of observing the value of X before predicting the value of y or simply predict the value of Y.

From exercise 9, it was found that when Y is predicted from X,

The overall M.S.E is \(\frac{1}{{12}} - \frac{{\log 3}}{{144}} + c\)

Using X without Y is predicted; the M.S.E. is\({\mathop{\rm var}} \left( Y \right)\)

It is found that

Calculate\(E\left( Y \right)\)

\(\begin{align}E\left( Y \right) &= \int\limits_0^1 {\int\limits_0^1 {y\left( {x + y} \right)} } dxdy\\ &= \int {\left( {{y^2} + \frac{y}{2}} \right)dy} \end{align}\)

\(\begin{align}E\left( Y \right) &= \int {{y^2}dy + \frac{1}{2}\int {ydy} } \\ &= \frac{{{y^3}}}{3} + \frac{{{y^2}}}{2}\end{align}\)

\(E\left( Y \right) = \frac{7}{{12}}\)

Calculating\(E\left( {{Y^2}} \right)\)

\(\begin{align}E\left( {{Y^2}} \right) = \int\limits_0^1 {\int\limits_0^1 {{y^2}\left( {x + y} \right)dxdy} } \\\int {\left( {{y^3} + \frac{{{y^2}}}{2}} \right)dy} \end{align}\)

\(\begin{align}E\left( {{Y^2}} \right) &= \int {{y^3}dy + \frac{1}{2}\int {{y^2}dy} } \\ &= \frac{{{y^4}}}{4} + \frac{{{y^3}}}{6}\end{align}\)

\(E\left( {{y^2}} \right) = \frac{5}{{12}}\)

Hence

\(Var\left( x \right) = E\left( {{Y^2}} \right) - E{\left( Y \right)^2}\)

\(\begin{align}Var\left( Y \right) &= \frac{5}{{12}} - {\left( {\frac{7}{{12}}} \right)^2}\\ &= \frac{5}{{12}} - \frac{{49}}{{144}}\\ &= \frac{{11}}{{144}}\end{align}\)

The total loss when X is used for predicting Y will be less than \(Var\left( Y \right)\) if and only if \(c < \frac{{\log 3 - 1}}{{144}}\)

Hence, the maximum value of c that she should be willing to pay is\(\frac{{\log 3 - 1}}{{144}}\)