Answer to Question #242535 in Statistics and Probability for PRS

Question

Obtain linear regression equation of Y on X and X on Y for the bivariate function expressed as follows

f(x,y)={ 2/3(x+2y) 0<x<1 0<y<1

{ 0 otherwise

Answer 1

1.partial densities

f(x)=

$\int _{-\infty}^{\infty}f(x,y)dy=\\ 1)0,\space x\notin\space[0,1]$

$2)\frac{2}{3}\cdot \int_0^1(x+2y)\cdot dy=(xy+y^2)|_0^1\cdot \frac{2}{3}=\\ \frac{2}{3}\cdot(x+1),x\in [0,1]$

f(y)=

$\int _{-\infty}^{\infty}f(x,y)dx=\\ 1)0,\space y\notin\space[0,1]$

$2)\frac{2}{3}\cdot \int_0^1(x+2y)\cdot dx=(x^2/2+2yx)|_0^1\cdot \frac{2}{3}=\\ \frac{1+4y}{3},y\in [0,1]$

$2)\frac{2}{3}\cdot \int_0^1(x+2y)\cdot dx=(x^2/2+2yx)|_0^1\cdot \frac{2}{3}=\\ \frac{1+4y}{3},y\in [0,1]$

2 Partial means

$mx=\frac{2}{3}\int_0^1(x+1)\cdot x\cdot dx=\\ \frac{2}{3}\cdot (x^2/3+x^2/2)|_0^1=5/9$

$my=\frac{1}{3}\int_0^1(4y+1)\cdot y\cdot dy=\\ \frac{1}{3}\cdot (4y^3/3+y^2/2)|_0^1=11/18$

3 Partial dispersions

$\sigma_x^2=\frac{2}{3}\int_0^1(x+1)\cdot x^2\cdot dx-(5/9)^2=\\ \frac{2}{3}\cdot (x^4/4+x^3/3)|_0^1-\frac{25}{81}=\frac{14}{36}-\frac{25}{81}=\\ \frac{126-100}{324}=\frac{13}{162}$

$\sigma_y^2=\frac{1}{3}\int_0^1(4y+1)\cdot y^2\cdot dx-(11/18)^2=\\ \frac{1}{3}\cdot (y^4+y^3/3)|_0^1-\frac{121}{324}=\frac{4}{9}-\frac{121}{324}=\\ \frac{144-121}{324}=\frac{23}{324}$

4 Correlation moment

$K=\frac{2}{3}\int_0^1\int_0^1(x-5/9)(y-11/18)\cdot(x+2y)dxdy=\\ \frac{2}{3}\cdot \int_0^1(y-\frac{11}{18})\int_0^1(x^2-\frac{5}{9}\cdot x+2xy-10y/9)dxdy=\\ \frac{2}{3}\cdot \int_0^1(y-\frac{11}{18})(1/3-5/18+y-10y/9)dy=\\ \frac{1}{27}\cdot \int_0^1(y-\frac{11}{18})(1-2y)/18\cdot dy=\\$

$=\frac{-2/3+11/18+1/2-11/18}{27}=-\frac{1}{162}$

Regression y on x

$x=\frac{K_{xy}}{\sigma_y^2}(x-m_y)+m_x=$

$=\frac{-1/162}{13/162}(x-5/9)+11/18=$

$-\frac{1}{13}\cdot x+153/234$

Regression x on y

$x=\frac{K_{xy}}{\sigma_y^2}(y-m_y)+m_x=$

$=\frac{-1/162}{23/324}(y-11/18)+5/9=\\ =-\frac{2}{23}\cdot y+14/23$