The joint probability density function of π and π is given by π(π₯, π¦) = { π₯ + π¦ 3 0 β€ π₯ β€ 1, 0 β€ π¦ β€ 2 0 πππ ππ€βπππ Find the two lines of regression.
lines of regression:
"y=a_1x+b_1"
"a_1=\\sigma_{XY}\/\\sigma^2_X,b_1=\\mu_Y-a_1\\mu_X"
"x=a_2y+b_2"
"a_2=\\sigma_{XY}\/\\sigma^2_Y,b_2=\\mu_X-a_2\\mu_Y"
covariance:
"\\sigma_{XY}=\\iint (x-\\mu_X)(y-\\mu_Y)f(x,y)dydx"
"f_X(x)=\\int^2_0 f(x,y)dy=\\frac{1}{3}\\int^2_0 (x+y)dy=\\frac{1}{3}(2x+1)"
"f_Y(y)=\\int^1_0 f(x,y)dx=\\frac{1}{3}\\int^1_0 (x+y)dx=\\frac{1}{3}(1\/2+y)"
"\\mu_X=\\int^1_0 xf_X(x)dx=\\frac{1}{3}\\int^1_0 x(2x+1)dx=\\frac{1}{3}(2\/3+1\/2)=7\/18"
"\\mu_Y=\\int^2_0 yf_Y(y)dy=\\frac{1}{3}\\int^2_0 y(1\/2+y)dy=\\frac{1}{3}(1+8\/3)=11\/9"
"\\sigma_X=\\sqrt{E(X^2)-\\mu_X^2}"
"E(X^2)=\\int^1_0 x^2f_X(x)dx=\\frac{1}{3}\\int^1_0 x^2(2x+1)dx=\\frac{1}{3}(1\/2+1\/3)=5\/18"
"\\sigma_X=\\sqrt{5\/18-(7\/18)^2}=\\sqrt{41}\/18=0.356"
"\\sigma_Y=\\sqrt{E(Y^2)-\\mu_Y^2}"
"E(Y^2)=\\int^2_0 x^2f_Y(y)dy=\\frac{1}{3}\\int^2_0 y^2(1\/2+y)dy=\\frac{1}{3}(4\/3+4)=16\/9"
"\\sigma_Y=\\sqrt{16\/9-(11\/9)^2}=\\sqrt{23}\/9=0.533"
"\\sigma_{XY}=\\frac{1}{3}\\intop^1_0 \\intop^2_0 (x-7\/18)(y-11\/9)(x+y)dydx="
"=\\frac{1}{3}\\intop^1_0(x-7\/18)(xy^2\/2-11xy\/9+y^3\/3-11y^2\/18)|^2_0dx="
"=\\frac{1}{3}\\intop^1_0(x-7\/18)(2x-22x\/9+8\/3-44\/18)dx="
"=\\frac{1}{3}\\intop^1_0(x-0.39)(-0.44x+0.22)dx=\\frac{1}{3}(-0.15+0.11+0.09-0.09)=-0.013"
"a_1=-0.013\/0.356^2=-0.1"
"b_1=11\/9+0.1\\cdot7\/18=1.26"
"y=-0.1x+1.26"
"a_2=-0.013\/0.533^2=-0.05"
"b_1=7\/18+0.05\\cdot11\/9=0.45"
"x=-0.05y+0.45"
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