Question

Given joint density function of x, y:

$f(x,y)=\frac{3}{2}(x^2+y^2)$,  $0\leq x\leq 1, 0\leq y\leq 1$

Find the coefficients of the the best linear predictor y=a+bx+e

Answer 1

The estimates for the coefficients of linear predictor

$Y=a+bX+\epsilon$

are given by

$\hat{b}=\frac{Cov(X,Y)}{Var(X)}$

and

$\hat{a}=E(Y)-\hat{b}E(X)$

First we get the following

Marginal pdf of X

$\begin{align*} f(x)&=\int_0^1f(x,y)dy\\ &=\int_0^1\frac{3}{2}(x^2+y^2)dy\\ &=\frac{3}{2}\left(x^2\times \left.y\right|_0^1+\left.\frac{y^3}{3}\right|_0^1 \right )\\ &=\frac{3}{2}\left(x^2+\frac{1}{3} \right )\\ &=\frac{1}{2}(3x^2+1),\quad 0\le x\le 1 \end{align*}$

Marginal pdf of Y

$\begin{align*} f(y)&=\int_0^1f(x,y)dx\\&=\int_0^1\frac{3}{2}(x^2+y^2)dx\\ &=\frac{3}{2}\left(\left.\frac{x^3}{3}\right|_0^1+y^2\times \left.x\right|_0^1 \right )\\ &=\frac{3}{2}\left(y^2+\frac{1}{3} \right )\\ &=\frac{1}{2}(3y^2+1),\quad 0\le y\le 1 \end{align*}$

The expected value of X is

$\begin{align*} E(X)&=\int_0^1xf(x)dx\\ &=\int_0^1x\frac{1}{2}(3x^2+1)dx\\ &=\frac{1}{2}\left(\left.3\frac{x^4}{4}\right|_0^1+\left. \frac{x^2}{2}\right|_0^1 \right )\\ &=\frac{1}{2}\left(\frac{3}{4}+\frac{1}{2} \right )\\ &=\frac{5}{8} \end{align*}$

The expected value of $\begin{align*}X^2 \end{align*}$ is

$\begin{align*} E(X^2)&=\int_0^1x^2f(x)dx\\ &=\int_0^1x^2\frac{1}{2}(3x^2+1)dx\\ &=\frac{1}{2}\left(\left.3\frac{x^5}{5}\right|_0^1+\left. \frac{x^3}{3}\right|_0^1 \right )\\ &=\frac{1}{2}\left(\frac{3}{5}+\frac{1}{3} \right )\\ &=\frac{14}{30} \end{align*}$

the variance of X is

$\begin{align*} Var(X)=E(X^2)-[E(X)]^2&=\frac{14}{30}-\left(\frac{5}{8} \right)^2= 0.0760 \end{align*}$

The expected value of Y is

$\begin{align*} E(Y)&=\int_0^1yf(y)dy\\ &=\int_0^1y\frac{1}{2}(3y^2+1)dy\\ &=\frac{1}{2}\left(\left.3\frac{y^4}{4}\right|_0^1+\left. \frac{y^2}{2}\right|_0^1 \right )\\ &=\frac{1}{2}\left(\frac{3}{4}+\frac{1}{2} \right )\\ &=\frac{5}{8} \end{align*}$

The expectation of X,Y is

$\begin{align*} E(X,Y)&=\int_0^1\int_0^1xyf(x,y)dxdy\\ &=\int_0^1\int_0^1xy\frac{3}{2}(x^2+y^2)dxdy\\ &=\frac{3}{2}\int_0^1y\left(\left.\frac{x^4}{4}\right|_0^1+y^2\times \left.\frac{x^2}{2}\right|_0^1 \right )dy\\ &=\int_0^1\frac{3}{4}\left(y^3+\frac{y}{2}\right)dy\\ &=\frac{3}{4}\left(\left.\frac{y^4}{4}\right|_0^1+\left.\frac{y^2}{2\times 2}\right|_0^1\right)\\ &=\frac{3}{8} \end{align*}$

Covariance of X,Y is

$\begin{align*} Cov(X,Y)&=E(X,Y)-E(X)E(Y)&=\frac{3}{8}-\frac{5}{8}\times \frac{5}{8}=-0.0156 \end{align*}$

Now we have everything to get the estimates of coefficients

The slope estimate is

$\hat{b}=\frac{Cov(X,Y)}{Var(X)}=\frac{-0.0156}{0.0760}=-0.2055$

The estimate of intercept is

$\hat{a}=E(Y)-\hat{b}E(X)=\frac{5}{8}-(-0.2055)\times \frac{5}{8}=0.7534$

The best linear predictor is

$\hat{y}=0.7534-0.2055x$