A Probability Course for the Actuaries A Preparation for Exam P/1

16 EXPECTED VALUE OF A FUNCTION OF A DISCRETE RANDOM VARIABLE135
Solution.
Since each possible outcome has the probability 1 , we get
6
6∑
E[g(X)] = (2i 2 + 1) · 1
6
i=1
( )
= 1 6∑
6 + 2 i 2
6
i=1
= 1 (
)
6(6 + 1)(2 · 6 + 1)
6 + 2
6
6
= 94
3
As a consequence of the above theorem we have the following result.
Corollary 16.1
If X is a discrete random variable, then for any constants a and b we have
Proof.
Let D denote the range of X. Then
A similar argument establishes
E(aX + b) = aE(X) + b.
E(aX + b) = ∑ x∈D(ax + b)p(x)
=a ∑ xp(x) + b ∑ p(x)
x∈D
x∈D
=aE(X) + b
E(aX 2 + bX + c) = aE(X 2 ) + bE(X) + c.
Example 16.3
Let X be a random variable with E(X) = 6 and E(X 2 ) = 45, and let
Y = 20 − 2X. Find the mean and variance of Y.
Solution.
By the properties of expectation,
E(Y ) =E(20 − 2X) = 20 − 2E(X) = 20 − 12 = 8
E(Y 2 ) =E(400 − 80X + 4X 2 ) = 400 − 80E(X) + 4E(X 2 ) = 100
Var(Y ) =E(Y 2 ) − (E(Y )) 2 = 100 − 64 = 36