Page 97 - Probability and Statistical Inference

74 2. Expectations of Functions of Random Variables

In order to find the mean and variance of this distribution, we make the one-
to-one substitution u = –z along the lines of (1.7.15) when z < 0, and write

But, the two integrals I , J are equal. Hence, E(Z) = 0. Next, in order to evalu-
1
1
ate E(Z ) we first write
2

Then, we proceed as before with the one-to-one substitution u = –z when z <
0, and rewrite

since the two integrals in the previous step are equal.
(2.2.23)
Now, in order to evaluate the last integral from (2.2.23), we further make the
one-to-one substitution υ = 1/2u when u > 0, and proceed as follows:
2

which reduces to 1/2 since Refer to (1.6.20). Now, combining
(2.2.23)-(2.2.24) we see clearly that E(Z ) = 1. In other words, the mean and
2
variance of the standard normal variable Z are given by

The result in (2.2.22), namely that E(Z) = 0, may not surprise anyone
because the pdf ϕ(z) is symmetric about z = 0 and E(Z) is finite. Refer to the
Exercise 2.2.13 in this context.