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

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Answer in Statistics and Probability for PRS #242535

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

{ 0 otherwise

1.partial densities

f(x)=

“\int _{-\infty}^{\infty}f(x,y)dy=\\n1)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}=\\n\frac{2}{3}\cdot(x+1),x\in [0,1]”

f(y)=

“\int _{-\infty}^{\infty}f(x,y)dx=\\n1)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}=\\n\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}=\\n\frac{1+4y}{3},y\in [0,1]”

2 Partial means

“mx=\frac{2}{3}\int_0^1(x+1)\cdot x\cdot dx=\\n\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=\\n\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=\\n\frac{2}{3}\cdot (x^4/4+x^3/3)|_0^1-\frac{25}{81}=\frac{14}{36}-\frac{25}{81}=\\n\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=\\n\frac{1}{3}\cdot (y^4+y^3/3)|_0^1-\frac{121}{324}=\frac{4}{9}-\frac{121}{324}=\\n\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=\\n\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=\\n\frac{2}{3}\cdot \int_0^1(y-\frac{11}{18})(1/3-5/18+y-10y/9)dy=\\n\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=\\n=-\frac{2}{23}\cdot y+14/23”

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