Answer to Question #275557 in Statistics and Probability for vikas

Question

The joint probability density function of 𝑋 and 𝑌 is given by

𝑓(𝑥, 𝑦) = {

𝑥 + 𝑦

3

0 ≤ 𝑥 ≤ 1, 0 ≤ 𝑦 ≤ 2

0 𝑒𝑙𝑠𝑒𝑤ℎ𝑒𝑟𝑒

Find the two lines of regression.

Answer 1

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:

$\sigma_{XY}=\iint (x-\mu_X)(y-\mu_Y)f(x,y)dydx$

$f_X(x)=\int^2_0 f(x,y)dy=\frac{1}{3}\int^2_0 (x+y)dy=\frac{1}{3}(2x+1)$

$f_Y(y)=\int^1_0 f(x,y)dx=\frac{1}{3}\int^1_0 (x+y)dx=\frac{1}{3}(1/2+y)$

$\mu_X=\int^1_0 xf_X(x)dx=\frac{1}{3}\int^1_0 x(2x+1)dx=\frac{1}{3}(2/3+1/2)=7/18$

$\mu_Y=\int^2_0 yf_Y(y)dy=\frac{1}{3}\int^2_0 y(1/2+y)dy=\frac{1}{3}(1+8/3)=11/9$

$\sigma_X=\sqrt{E(X^2)-\mu_X^2}$

$E(X^2)=\int^1_0 x^2f_X(x)dx=\frac{1}{3}\int^1_0 x^2(2x+1)dx=\frac{1}{3}(1/2+1/3)=5/18$

$\sigma_X=\sqrt{5/18-(7/18)^2}=\sqrt{41}/18=0.356$

$\sigma_Y=\sqrt{E(Y^2)-\mu_Y^2}$

$E(Y^2)=\int^2_0 x^2f_Y(y)dy=\frac{1}{3}\int^2_0 y^2(1/2+y)dy=\frac{1}{3}(4/3+4)=16/9$

$\sigma_Y=\sqrt{16/9-(11/9)^2}=\sqrt{23}/9=0.533$

$\sigma_{XY}=\frac{1}{3}\intop^1_0 \intop^2_0 (x-7/18)(y-11/9)(x+y)dydx=$

$=\frac{1}{3}\intop^1_0(x-7/18)(xy^2/2-11xy/9+y^3/3-11y^2/18)|^2_0dx=$

$=\frac{1}{3}\intop^1_0(x-7/18)(2x-22x/9+8/3-44/18)dx=$

$=\frac{1}{3}\intop^1_0(x-0.39)(-0.44x+0.22)dx=\frac{1}{3}(-0.15+0.11+0.09-0.09)=-0.013$

$a_1=-0.013/0.356^2=-0.1$

$b_1=11/9+0.1\cdot7/18=1.26$

$y=-0.1x+1.26$

$a_2=-0.013/0.533^2=-0.05$

$b_1=7/18+0.05\cdot11/9=0.45$

$x=-0.05y+0.45$