PROBABILITY
A biased die gives probabilities^12 p,p,p,p,p, 2 pof throwing1, 2, 3, 4, 5, 6respectively.
If the random variableXis the number shown on the die and the random variableYis
defined asX^2 , calculate the covariance and correlation ofXandY.We have already calculated in subsections 30.2.1 and 30.5.4 that
p=2
13
,E[X]=
53
13
,E
[
X^2
]
=
253
13
,V[X]=
480
169
.
Using (30.135), we obtain
Cov[X, Y]=Cov[X, X^2 ]=E[X^3 ]−E[X]E[X^2 ].NowE[X^3 ]isgivenby
E[X^3 ]=1^3 ×^12 p+(2^3 +3^3 +4^3 +5^3 )p+6^3 × 2 p=1313
2
p= 101,and the covariance ofXandYis given by
Cov[X, Y] = 101−53
13
×
253
13
=
3660
169
.
The correlation is defined by Corr[X, Y]=Cov[X, Y]/σXσY. The standard deviation of
Ymay be calculated from the definition of the variance. LettingμY=E[X^2 ]=^25313 gives
σ^2 Y=p
2(
12 −μY) 2
+p(
22 −μY) 2
+p(
32 −μY) 2
+p(
42 −μY) 2
+p(
52 −μY) 2
+2p(
62 −μY) 2
=
187 356
169
p=28 824
169
.
We deduce that
Corr[X, Y]=3660
169
√
169
28 824
√
169
480
≈ 0. 984.
Thus the random variablesXandYdisplay a strong degree of positive correlation, as we
would expect.
We note that the covariance ofXandYoccurs in various expressions. Forexample, ifXandYarenotindependent then
V[X+Y]=E[
(X+Y)^2]
−(E[X+Y])^2=E[
X^2]
+2E[XY]+E[
Y^2]
−{(E[X])^2 +2E[X]E[Y]+(E[Y])^2 }=V[X]+V[Y]+2(E[XY]−E[X]E[Y])=V[X]+V[Y]+2 Cov[X, Y].