12.1 Covariance and the correlation coefficient  (Page 2/2)

Uncorrelated but not independent

Suppose the joint density for $\{X,Y\}$ is constant on the unit circle about the origin. By the rectangle test, the pair cannot be independent. By symmetry, the $\rho =1$ line is $u=t$ and the $\rho =-1$ line is $u=-t$ . By symmetry, also, the variance about each of these lines is the same. Thus $\rho =0$ , which is true iff $\mathrm{Cov}[X,Y]=0$ . This fact can be verified by calculation, if desired.

Uniform marginal distributions

Consider the three distributions in [link] . In case (a), the distribution is uniform over the square centered at the origin with vertices at (1,1), (-1,1), (-1,-1),(1,-1). In case (b), the distribution is uniform over two squares, in the first and third quadrants with vertices (0,0), (1,0), (1,1), (0,1) and (0,0),

(-1,0), (-1,-1), (0,-1). In case (c) the two squares are in the second and fourth quadrants. The marginalsare uniform on (-1,1) in each case, so that in each case

This means the $\rho =1$ line is $u=t$ and the $\rho =-1$ line is $u=-t$ .

1. By symmetry, $E\left[XY\right]=0$ (in fact the pair is independent) and $\rho =0$ .
2. For every pair of possible values, the two signs must be the same, so $E\left[XY\right]>0$ which implies $\rho >0$ . The actual value may be calculated to give $\rho =3/4$ . Since $1-\rho <1+\rho$ , the variance about the $\rho =1$ line is less than that about the $\rho =-1$ line. This is evident from the figure.
3. $E\left[XY\right]<0$ and $\rho <0$ . Since $1+\rho <1-\rho$ , the variance about the $\rho =-1$ line is less than that about the $\rho =1$ line. Again, examination of the figure confirms this.

A pair of simple random variables

With the aid of m-functions and MATLAB we can easily caluclate the covariance and the correlation coefficient. We use the joint distribution for Example 9 in "Variance." In that example calculations show

so that $\rho =-0.04699$ .

An absolutely continuous pair

The pair $\{X,Y\}$ has joint density function ${\displaystyle f_{XY}(t,u)=\frac{6}{5}(t+2u)}$ on the triangular region bounded by $t=0,u=t$ , and $u=1$ . By the usual integration techniques, we have

From this we obtain $E\left[X\right]=2/5,\mathrm{Var}\left[X\right]=3/50,E\left[Y\right]=3/4$ , and $\mathrm{Var}\left[Y\right]=3/80$ . To complete the picture we need

Then

APPROXIMATION

$Y=g\left(X\right)$ But $\rho =0$

Suppose $X\sim$ uniform (-1,1), so that $f_X\left(t\right)=1/2,-1<t<1$ and $E\left[X\right]=0$ . Let $Y=g\left(X\right)=cosX$ . Then

Thus $\rho =0$ . Note that g could be any even function defined on (-1,1). In this case the integrand $tg\left(t\right)$ is odd, so that the value of the integral is zero.