Page 106 - Probability and Statistical Inference

2. Expectations of Functions of Random Variables 83

which was exactly the same integral evaluated in (2.3.13). In other words,
from (2.3.14) we can immediately claim that M (t) = exp(1/2t ). Thus, one
2
Y
can rewrite (2.3.15) as follows:

Now, log(M (t)) = tµ + 1/2t σ so that dM (t)/dt = (µ + tσ ) M (t). Hence,
2
2
2
X
X
X
dM (t)/dt, when evaluated at t = 0, reduces to η = µ because one has
1
X
M (0) = 1.
X
Also, using the chain rule of differentiation we have d M (t)/dt = σ M (t) +
2
2
2
X
X
(µ + tσ ) M (t) so that d M (t)/dt , when evaluated at t = 0, reduces to η = σ 2
2 2
2
2
2
X
X
+ µ . In view of the Theorem 2.3.1, we can say that µ is the mean of X and V(X)
2
2
2
2
= E(X ) – µ = σ + µ – µ = σ . These answers were verified earlier in (2.2.30).
2
2
2
One can also easily evaluate higher order derivatives of the mgf. In view of
k
the Theorem 2.3.1 one can claim that d M (t)/dt with k = 3 and 4, when
k
X
4
3
evaluated at t = 0, will reduce to η or E(X ) and η or E(X ) respectively. Then,
α
4
3
in order to obtain the third and fourth central moments of X, one should pro-
ceed as follows:
by the Theorem 2.2.1. Similarly,
Look at the related Exercises 2.3.1-2.3.2.
Example 2.3.3 Suppose that Z is the standard normal variable. How should
one directly derive the expression for E(|Z|)? Here, the mgf of Z from (2.3.14) is
not going to be of much help. Let us apply the Definition 2.2.3 and proceed as
follows. As before, we substitute u = –z when z < 0, and express E(|Z|) as
In the last step of (2.3.19), one immediately realizes that the two integrals are
really the same that is, and
hence one has