91Ó°ÊÓ

Referring to the exercise 7,

The probability density function of X and Y is,

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

a.

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,

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

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

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

For,\(X = \frac{1}{2}\)

\(\begin{align}E\left( {Y\left| {X = \frac{1}{2}} \right.} \right) &= \frac{{\left( {\frac{{3 \times 1}}{2} + 2} \right)}}{{3\left( {\frac{{2 \times 1}}{2} + 1} \right)}}\\ &= \frac{{\left( {\frac{3}{2} + 2} \right)}}{{3\left( {1 + 1} \right)}}\\ &= \frac{{\frac{7}{2}}}{6}\\ &= \frac{7}{{12}}\end{align}\)

\(\begin{align}E\left( {Y\left| {X = \frac{1}{2}} \right.} \right) &= \frac{7}{{12}}\\ &= 0.583333\end{align}\)

Therefore, the predicted value of Y is 0.58333

b.

M.S.Eis the\(Var\left( {Y\left| {X = \frac{1}{2}} \right.} \right)\)

So,

\(\begin{align}Var\left( {Y\left| X \right.} \right) &= E\left( {{Y^2}\left| X \right.} \right) - {\left( {E\left( {Y\left| X \right.} \right)} \right)^2}\\ &= \frac{{\left( {4x + 3} \right)}}{{6\left( {2x + 1} \right)}} - {\left( {\frac{{\left( {3x + 2} \right)}}{{3\left( {2x + 1} \right)}}} \right)^2}\\ &= \frac{{\left( {4x + 3} \right)}}{{6\left( {2x + 1} \right)}} - \frac{{{{\left( {3x + 2} \right)}^2}}}{{9{{\left( {2x + 1} \right)}^2}}}\\ &= \frac{{3\left( {4x + 3} \right)\left( {2x + 1} \right) - {{\left( {3x + 2} \right)}^2}}}{{18{{\left( {2x + 1} \right)}^2}}}\\ &= \frac{{3\left( {8{x^2} + 4x + 6x + 3} \right) - \left( {9{x^2} + 12x + 4} \right)}}{{18{{\left( {2x + 1} \right)}^2}}}\\ &= \frac{{24{x^2} + 30x + 9 - 9{x^2} - 12x - 4}}{{18{{\left( {2x + 1} \right)}^2}}}\\ &= \frac{{15{x^2} + 18x + 5}}{{18{{\left( {2x + 1} \right)}^2}}}\end{align}\)

\(Var\left( {Y\left| X \right.} \right) = \frac{{15{x^2} + 18x + 5}}{{18{{\left( {2x + 1} \right)}^2}}}\)

Then,

\(X = \frac{1}{2}\)

\(\begin{align}Var\left( {Y\left| {X = \frac{1}{2}} \right.} \right) &= \frac{{15{x^2} + 18x + 5}}{{18{{\left( {2x + 1} \right)}^2}}}\\ &= \frac{{15\left( {\frac{1}{4}} \right) + \frac{{18}}{2} + 5}}{{18{{\left( {\frac{{2 \times 1}}{2} + 1} \right)}^2}}}\\ &= \frac{{\frac{{15 + 36 + 20}}{4}}}{{18 \times 4}}\\ &= \frac{{\frac{{71}}{4}}}{{72}}\\ &= \frac{{71}}{{288}}\\ &= 0.246528\end{align}\)

Therefore, the M.S.E is 0.246528