Page 256 - DCOM203_DMGT204_QUANTITATIVE_TECHNIQUES_I

Unit 12: Probability and Expected Value

Expression for Covariance Notes

E
Y
As a particular case, assume that X i ,Y j X i X Y j Y , where ( )E X X and ( ) Y
m n
Thus, E X X Y Y X i X Y j Y p ij
i 1 j 1
The above expression, which is the mean of the product of deviations of values from their
respective means, is known as the Covariance of X and Y denoted as Cov(X, Y) or XY . Thus, we

can write Cov X ,Y E X X Y Y

An alternative expression of Cov(X, Y)

Y
Cov ( , ) E X E ( ) Y E ( )
X
Y
X
Y
Y
X
E . X Y E ( ) E ( ). Y E ( )
Y
Y
E
E
E . X Y X . ( ) E ( . ) E ( ). ( )
Y
X
X
Note that E[{Y – E(Y)}] = 0, the sum of deviations of values from their arithmetic mean.
Mean and Variance of a Linear Combination
Let Z X ,Y aX bY be a linear combination of the two random variables X and Y, then
using the theorem of addition of expectation, we can write
Z
E ( ) E (aX bY ) aE ( ) bE ( ) a b
Y
X
Z X Y
Further, the variance of Z is given by
2 2 2 2
Z
Z E Z E ( ) E aX bY a X b Y E a X X b Y Y
2
2
a E X X 2 b E Y Y 2 2abE X X Y Y
a 2 2 X b 2 2 2ab XY
Y
Example: A random variable X has the following probability distribution:
X : 2 1 0 1 2
1 1 1
Probability : p p
6 4 6
1. Find the value of p.
2
2. Calculate E(X + 2) and E(2X + 3X + 5).
Solution:
Since the total probability under a probability distribution is equal to unity, the value of p
1 1 1
should be such that p p 1 .
6 4 6

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