91Ó°ÊÓ

A score of a person in mathematics and music aptitude test is a number in the interval

The joint pdf of x and y is

\(f\left( {x,y} \right) = \left\{ \begin{align}{l}\frac{2}{5}\left( {2x + 3y} \right)\;\;\;\;\;\;\;for\,0 \le x \le 1\,and0 \le x \le 1\\0\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{align} \right.\)

The prediction is the mean of Y:

\(E\left( Y \right) = \int\limits_0^1 {\int\limits_0^1 {\frac{2}{5}\left( {2x + 3y} \right)dxdy} } \)

\(\begin{align}E\left( Y \right) &= \int {\frac{1}{5}y2\left( {2x + 3y} \right)dy} \\ &= \int {\left( {\frac{{6{y^2}}}{5} + \frac{{2y}}{5}} \right)} dy\end{align}\)

\(\begin{align}E\left( Y \right) &= \frac{6}{5}\int {{y^2}dy} + \frac{2}{5}\int {ydy} \\ &= \int {\frac{{2{y^3}}}{5} + \frac{{{y^2}}}{5}} + c\end{align}\)

\(E\left( Y \right) = \frac{3}{5}\)

Hence Predicted value of students' scores on the music test has the smallest M.S.E is \(\frac{3}{5}\).

The prediction is the median of X

The marginal pdf of x is:

\({f_1}\left( x \right) = \int\limits_0^1 {\frac{2}{5}\left( {2x + 3y} \right)dy} \)

\(\begin{align} &= \int\limits_0^1 {\frac{2}{5}\left( {2x + 3y} \right)dy} \\ &= \frac{2}{5}\int\limits_0^1 {\left( {2x + 3y} \right)dy} \\ &= \frac{6}{5}\int\limits_0^1 {ydy + \frac{{4x}}{5}\int\limits_0^1 {1dy} } \end{align}\)

\( = \frac{1}{5}\left( {4x + 3} \right)\)

We must have,

\(\int\limits_0^m {\frac{1}{5}\left( {4x + 3} \right)dx = \frac{1}{2}} \)

\(\begin{align}\int\limits_0^m {\frac{1}{5}\left( {4x + 3} \right)dx = \frac{1}{2}} \\\frac{1}{5}\left( {4\frac{{{x^2}}}{2} + 3x} \right)_0^m &= \frac{1}{2}\\2{m^2} + 3m = \frac{5}{2}\\\end{align}\)

Therefore,

\(4{m^2} + 6m - 5 = 0\)

\(m = \frac{{\sqrt {29} - 3}}{4}\)

Therefore, predicted value of students score on the mathematics test has the

smallest M.A.E. is\(\frac{{\sqrt {29} - 3}}{4}\)