r = Corr1X, Y2 = Cov1X, Y2
sXsY (4.34)
The correlation is the covariance divided by the standard deviations of the two ran-dom variables. This results in a standardized measure of relationship that varies from -1 to +1. The following interpretations are important:
1. A correlation of 0 indicates that there is no linear relationship between the two random variables. If the two random variables are independent, the correlation is equal to 0.
2. A positive correlation indicates that if one random variable is high (low), then the other random variable has a higher probability of being high (low), and we say that the variables are positively dependent. Perfect positive linear dependency is indi-cated by a correlation of +1.0.
3. A negative correlation indicates that if one random variable is high (low), then the other random variable has a higher probability of being low (high), and we say that the variables are negatively dependent. Perfect negative linear dependency is indi-cated by a correlation of -1.0.
The correlation is more useful for describing relationships than the covariance. With a cor-relation of +1 the two random variables have a perfect positive linear relationship, and, there-fore, a specific value of one variable, X, predicts the other variable, Y, exactly. A correlation of
-1 indicates a perfect negative linear relationship between two variables, with one variable, X, predicting the negative of the other variable, Y. A correlation of 0 indicates no linear relation-ship between the two variables. Intermediate values indicate that variables tend to be related, with stronger relationships occurring as the absolute value of the correlation approaches 1.
We also know that correlation is a term that has moved into common usage. In many cases correlation is used to indicate that a relationship exists. However, variables that have nonlinear relationships will not have a correlation coefficient close to 1.0. This distinction is important for us in order to avoid confusion between correlated random variables and those with nonlinear relationships.
4.7 Jointly Distributed Discrete Random Variables 183
Example 4.17 Joint Distribution of Stock Prices (Compute Covariance and Correlation)
Find the covariance and correlation for the stocks A and B from Example 4.16 with the joint probability distribution in Table 4.7.
Solution The computation of covariance is tedious for even a problem such as this, which is simplified so that all of the joint probabilities, P1x, y2, are 0.0625 for all pairs of values x and y. By definition, you need to find the following:
Cov1X, Y2 = a
x a
y
xyP1x, y2 - mXmY
= 0310210.06252 + 10.05210.06252 + 10.10210.06252 + 10.15210.062524 + 0.05310210.06252 + 10.05210.06252 + 10.10210.06252 + 10.15210.062524 + 0.10310210.06252 + 10.05210.06252 + 10.10210.06252 + 10.15210.062524 + 0.15310210.06252 + 10.05210.06252 + 10.10210.06252 + 10.15210.062524 - 10.0752 10.0752
= 0.005625 - 0.005625 = 0 Thus,
r = Corr1X, Y2 = Cov1X, Y2 sXsY = 0
Microsoft Excel can be used for these computations by carefully following the example in Figure 4.5.
Figure 4.5 Covariance Calculation Using Microsoft Excel Joint Probability Distribution of X and Y
Y Return %
X Return % 0 0.05 0.1 0.15 P(x) E(X)
0 0.0625 0.0625 0.0625 0.25
0.05 0.0625 0.0625 0.0625 0.25
0.1 0.0625 0.0625 0.0625 0.25
0.15 0.0625 0.0625 0.0625 0.25
0.25 0.25 0.25 0.075
0.075 E(Y)
Calculation of Covariance xyP(x,y) xyP(x,y)
xyP(x,y) xyP(x,y) xyP(x,y)
xyP(x,y) xyP(x,y) xyP(x,y)
0 0 0
0 0.000156 0.000313 0 0.000313 0.000625 0 0.000469 0.000938
Sum xyP(x,y) 0 0.000938 0.001875
0.0625 0.0625 0.0625 0.0625 0.25
0 0.000469 0.000938 0.001406
0.002813 0.005625
Covariance
Sum xyP(x,y) 2 E(X)E(Y) 5 0.005625 2 0.005625 0
184 Chapter 4 Discrete Probability Distributions
The reason a covariance of 0 does not necessarily imply statistical independence is that covariance is designed to measure linear association, and it is possible that this quan-tity may not detect other types of dependency, as we see in the following illustration.
Suppose that the random variable X has probability distribution P1 -12 = 1>4 P102 = 1>2 P112 = 1>4 Let the rand om variable Y be defined as follows:
Y = X2
Thus, knowledge of the value taken by X implies knowledge of the value taken by Y, and, therefore, these two random variables are certainly not independent. Whenever X = 0, then Y = 0, and if X is either -1 or 1, then Y = 1. The joint probability distribution of X and Y is
P1 -1, 12 = 1>4 P10, 02 = 1>2 P11, 12 = 1>4
with the probability of any other combination of values being equal to 0. It is then straight-forward to verify that
E3X4 = 0 E3Y4 = 1>2 E3XY4 = 0
The covariance between X and Y is 0. Thus we see that random variables that are not inde-pendent can have a covariance equal to 0.
To conclude the discussion of joint distributions, consider the mean and variance of a random variable that can be written as the sum or difference of other random variables.
These results are summarized below and can be derived using Equations 4.30, 4.31, and 4.32.
Covariance and Statistical Independence
If two random variables are statistically independent, the covariance between them is 0. However, the converse is not necessarily true.
Summary Results for Linear Sums and Differences of Random Variables
Let X and Y be a pair of random variables with means mX and mY and vari-ances s2X and sY2. The following properties hold:
1. The expected value of their sum is the sum of their expected values:
E3X + Y4 = mX + mY (4.35)
2. The expected value of their difference is the difference between their expected values:
E3X - Y4 = mX - mY (4.36)
3. If the covariance between X and Y is 0, the variance of their sum is the sum of their variances:
Var1X + Y2 = s2X + s2Y (4.37)
But if the covariance is not 0, then
Var1X + Y2 = s2X + s2Y + 2 Cov1X, Y2
4. If the covariance between X and Y is 0, the variance of their difference is the sum of their variances:
Var1X - Y2 = s2X + s2Y (4.38)
4.7 Jointly Distributed Discrete Random Variables 185 But if the covariance is not 0, then
Var1X - Y2 = s2X + s2Y - 2 Cov1X, Y2
Let X1, X2, c, XK be K random variables with means m1, m2, c , mK and variances s21, s22,c, sK2. The following properties hold:
5. The expected value of their sum is as follows:
E3X1 + X2 + g + XK4 = m1 + m2+ g + mK (4.39) 6. If the covariance between every pair of these random variables is 0, the
variance of their sum is as follows:
Var1X1 + X2 + g + XK2 = s21 + s22 + g + s2K (4.40) 7. If the covariance between every pair of these random variables is not 0,
the variance of their sum is as follows:
Var1X1 + X2 + g + XK2 = aK
i=1
s2i + 2 aK-1
i=1a
K
j7iCov1Xi, Yj2 (4.41)
Example 4.18 Simple Investment Portfolio (Means and Variances, Functions of Random Variables)
An investor has $1,000 to invest and two investment opportunities, each requiring a minimum of $500. The profit per $100 from the first can be represented by a random variable X, having the following probability distributions:
P1X = -52 = 0.4 and P1X = 202 = 0.6
The profit per $100 from the second is given by the random variable Y, whose probabil-ity distributions are as follows:
P1Y = 02 = 0.6 and P1Y = 252 = 0.4
Random variables X and Y are independent. The investor has the following possible strategies:
a. $1,000 in the first investment b. $1,000 in the second investment c. $500 in each investment
Find the mean and variance of the profit from each strategy.
Solution Random variable X has mean mX = E3X4 = a
x xP1x2 = 1 -5210.42 + 120210.62 = +10 and variance
s2
X= E31X - mx224 = a
x
1x - mx22P1x2 = 1 -5 - 102210.42 + 120 - 102210.62 = 150
Random variable Y has mean mY = E3Y4 = a
y
yP1y2 = 10210.62 + 125210.42 = +10
and variance
s2Y= E3(Y - mY224 = a
y 1y - mY22P1y2 = 10 - 102210.62 + 125 - 102210.42 = 150
186 Chapter 4 Discrete Probability Distributions