2.5. The Correlation Coefficient 131Example 2.5.5.To illustrate how the correlation coefficient measures the intensity
of the concentration of the probability forXandYabout a line, let these random
variables have a distribution that is uniform over the area depicted in Figure 2.5.1.
That is, the joint pdf ofXandYis
f(x, y)={ 1
4 ah −a+bx < y < a+bx, −h<x<h
0elsewhere.We assume here thatb≥0, but the argument can be modified forb≤0. It is easy
to show that the pdf ofXis uniform, namelyf 1 (x)={∫
a+bx
−a+bx1
4 ahdy=1
2 h −h<x<h
0elsewhere.The conditional mean and variance are
E(Y|x)=bx and var(Y|x)=a^2
3.From the general expressions for those characteristics we know that
b=ρσ 2
σ 1anda^2
3=σ 22 (1−ρ^2 ).Additionally, we know thatσ^21 =h^2 /3. If we solve these three equations, we obtain
an expression for the correlation coefficient, namely
ρ=bh
√
a^2 +b^2 h^2.Referring to Figure 2.5.1, we note- Asagets small (large), the straight-line effect is more (less) intense andρis
closer to 1 (0). - Ashgets large (small), the straight-line effect is more (less) intense andρis
closer to 1 (0). - Asbgets large (small), the straight-line effect is more (less) intense andρis
closer to 1 (0).
Recall that in Section 2.1 we introduced the mgf for the random vector (X, Y).
As for random variables, the joint mgf also gives explicit formulas for certain mo-
ments. In the case of random variables of the continuous type,
∂k+mM(t 1 ,t 2 )
∂tk 1 ∂tm 2=∫∞−∞∫∞−∞xkymet^1 x+t^2 yf(x, y)dxdy,so that
∂k+mM(t 1 ,t 2 )
∂tk 1 ∂tm 2∣
∣
∣
∣
t 1 =t 2 =0=∫∞−∞∫∞−∞xkymf(x, y)dxdy=E(XkYm).