91Ó°ÊÓ

Referring to exercise 7,

The probability density function of X and Y is,

\(\) \(f\left( {x,y} \right) = \left\{ {\begin{align}{}{x + y}&{for\,\,0 \le x \le 1\,\,and\,\,0 \le y \le 1,}\\0&{otherwise.}\end{align}} \right.\)

The marginal p.d.f. of X is,

\(\begin{align}f\left( x \right) &= \int_0^1 {f\left( {x,y} \right)dy} \\ &= \int_0^1 {\left( {x + y} \right)dy} \\ &= \left( {x + \frac{{{y^2}}}{2}} \right)_0^1\\ &= x + \frac{1}{2}\,\,\,;\,\,0 \le x \le 1.\end{align}\)

The conditional p.d.f. of Y, given that,\(X = x\)is,

\(\begin{align}g\left( {y\left| x \right.} \right) &= \frac{{f\left( {x,y} \right)}}{{f\left( x \right)}}\\ &= \frac{{\left( {x + y} \right)}}{{x + \frac{1}{2}}}\\ &= \frac{{2\left( {x + y} \right)}}{{\left( {2x + 1} \right)}}\end{align}\)

\(\begin{align}E\left( {Y\left| X \right.} \right) &= \int_0^1 {\frac{{2y\left( {x + y} \right)}}{{\left( {2x + 1} \right)}}dy} \\ &= \frac{1}{{\left( {2x + 1} \right)}}\int_0^1 {\left( {2xy + 2{y^2}} \right)dy} \\ &= \frac{1}{{\left( {2x + 1} \right)}}\left( {\frac{{2x{y^2}}}{2} + \frac{{2{y^3}}}{3}} \right)_0^1\end{align}\)

\(\begin{align} &= \frac{1}{{\left( {2x + 1} \right)}}\left( {\frac{{2x}}{2} + \frac{2}{3}} \right)\\ &= \frac{1}{{\left( {2x + 1} \right)}}\left( {x + \frac{2}{3}} \right)\\ &= \frac{{\left( {3x + 2} \right)}}{{3\left( {2x + 1} \right)}}\end{align}\)

\(E\left( {Y\left| X \right.} \right) = \frac{{\left( {3x + 2} \right)}}{{3\left( {2x + 1} \right)}}\)

So,

\(\begin{align}E\left( {{Y^2}\left| X \right.} \right) &= \int_0^1 {\frac{{2{y^2}\left( {x + y} \right)}}{{\left( {2x + 1} \right)}}dy} \\ &= \frac{1}{{\left( {2x + 1} \right)}}\int_0^1 {\left( {2x{y^2} + 2{y^3}} \right)dy} \\ &= \frac{1}{{\left( {2x + 1} \right)}}\left( {\frac{{2x{y^3}}}{3} + \frac{{2{y^4}}}{4}} \right)_0^1\\ &= \frac{1}{{\left( {2x + 1} \right)}}\left( {\frac{{2x}}{3} + \frac{1}{2}} \right)\end{align}\)

\(\begin{align} &= \frac{1}{{\left( {2x + 1} \right)}}\left( {\frac{{4x + 3}}{6}} \right)\\ &= \frac{{\left( {4x + 3} \right)}}{{6\left( {2x + 1} \right)}}\end{align}\)

\(E\left( {{Y^2}\left| X \right.} \right) = \frac{{\left( {4x + 3} \right)}}{{6\left( {2x + 1} \right)}}\)

So,

\(\begin{align}Var\left( {Y\left| X \right.} \right) &= E\left( {{Y^2}\left| X \right.} \right) - {\left( {E\left( {Y\left| X \right.} \right)} \right)^2}\\ &= \frac{{\left( {4x + 3} \right)}}{{6\left( {2x + 1} \right)}} - {\left( {\frac{{\left( {3x + 2} \right)}}{{3\left( {2x + 1} \right)}}} \right)^2}\\ &= \frac{1}{{36}}\left( {3 - \frac{1}{{{{\left( {2x + 1} \right)}^2}}}} \right)\end{align}\)

\(Var\left( {Y\left| X \right.} \right) = \frac{1}{{36}}\left( {3 - \frac{1}{{{{\left( {2x + 1} \right)}^2}}}} \right)\)

The overall M.S.E is,

\(\begin{align}E\left( {Var\left( {Y\left| X \right.} \right)} \right) &= \int_0^1 {\frac{1}{{36}}\left( {3 - \frac{1}{{{{\left( {2x + 1} \right)}^2}}}} \right)f\left( x \right)dx} \\ &= \int_0^1 {\frac{1}{{36}}\left( {3 - \frac{1}{{{{\left( {2x + 1} \right)}^2}}}} \right)\left( {x + \frac{1}{2}} \right)dx} \\ &= \int_0^1 {\frac{1}{{36}}\left( {3 - \frac{1}{{{{\left( {2x + 1} \right)}^2}}}} \right)\left( {\frac{{2x + 1}}{2}} \right)dx} \end{align}\)

\(\begin{align} &= \frac{3}{{72}}\int_0^1 {\left( {2x + 1} \right)dx} - \frac{1}{{72}}\int_0^1 {\frac{1}{{\left( {2x + 1} \right)}}dx} \\ &= \frac{1}{{24}}\int_1^3 {\left( {z \times \frac{{dz}}{2}} \right)} - \frac{1}{{72}}\int_0^3 {\left( {\frac{{dz}}{{2z}}} \right)} \\ &= \frac{1}{{48}}\left( {\frac{{{z^2}}}{2}} \right)_1^3 - \frac{1}{{144}}\left( {\log z} \right)_1^3\\ &= \frac{1}{{96}}\left( {9 - 1} \right) - \frac{1}{{144}}\left( {\log 3 - 0} \right)\\ &= \frac{1}{{12}} - \frac{{\log 3}}{{144}}\\ &= 0.0757\end{align}\)

\(E\left( {Var\left( {Y\left| X \right.} \right)} \right) = 0.0757\)

Therefore, the minimum value of overall M.S.E is 0.0757