Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

1.9. Some Special Expectations 73

Example 1.9.5.It is known that the series

1

12

+

1

22

+

1

32

+···

converges toπ^2 /6. Then

p(x)=

{ 6

π^2 x^2 x=1,^2 ,^3 ,...

0elsewhere

is the pmf of a discrete type of random variableX. The mgf of this distribution, if

it exists, is given by

M(t)=E(etX)=

∑

x

etxp(x)

=

∑∞

x=1

6 etx

π^2 x^2

.

The ratio test of calculus^7 may be used to show that this series diverges ift>0.
Thus there does not exist a positive numberhsuch thatM(t)existsfor−h<t<h.
Accordingly, the distribution has the pmfp(x) of this example and does not have
an mgf.

Example 1.9.6.LetXbe a continuous random variable with pdf

f(x)=

1

π

1

x^2 +1

, −∞<x<∞. (1.9.2)

This is of course the Cauchy pdf which was introduced in Exercise 1.7.24. Lett> 0

be given. Ifx>0, then by the mean value theorem, for some 0<ξ 0 <tx,

etx− 1

tx

=eξ^0 ≥ 1.

Hence,etx≥1+tx≥tx. This leads to the second inequality in the following

derivation:

∫∞

−∞

etx

1

π

1

x^2 +1

dx ≥

∫∞

0

etx

1

π

1

x^2 +1

dx

≥

∫∞

0

1

π

tx

x^2 +1

dx=∞.

Becausetwas arbitrary, the integral does not exist in an open interval of 0. Hence,
the mgf of the Cauchy distribution does not exist.

Example 1.9.7.LetXhave the mgfM(t)=et

(^2) / 2
,−∞<t<∞. As discussed in
Chapter 3, this is the mgf of a standard normal distribution. We can differentiate
M(t) any number of times to find the moments ofX. However, it is instructive to
(^7) For example, see Chapter 2 ofMathematical Comments.