Page 116 - Elements of Distribution Theory

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052184472Xc04 CUNY148/Severini May 24, 2005 2:39

102 Moments and Cumulants

Example 4.8 (Inverse gamma distribution). Consider the inverse gamma distribution con-
sidered in Example 4.7. The Laplace transform of this distribution is given by
√
L(t) = 2tK 2 (2 t)
where K 2 denotes a modified Bessel function. The following properties of the modified
Bessel functions are useful; see, for example, Temme (1996) for further details.
The derivative of K ν satisfies
ν

K (t) =−K ν−1 (t) − K ν (t).
ν
t
The function K ν for ν = 0, 1, 2 has the following behavior near 0:
 2
 2/t if ν = 2
K ν (t) ∼ 1/t if ν = 1 as t → 0.

− log(t)if ν = 0
Using these results it is easy to show that

√ √
L (t) =−2 tK 1 (2 t)

and that L (0) =−1. Hence, E(X) = 1. Similarly,

√
L (t) = 2K 0 (2 t) ∼ log(1/t)as t → 0

2
so that L (0) does not exist. It follows that E(X )is not finite.

The Laplace transform of a sum of independent random variables is easily determined
from the Laplace transforms of the individual random variables. This result is stated in the
following theorem; the proof is left as an exercise.

Theorem 4.7. Let X and Y denote independent, real-valued nonnegative random variables
with Laplace tranforms L X and L Y ,respectively. Let L X+Y denote the Laplace transform
of the random variable X + Y. Then

L X+Y (t) = L X (t)L Y (t), t ≥ 0.

Example 4.9 (Gamma distribution). Let X 1 and X 2 denote independent random variables
such that, for j = 1, 2, X j has a gamma distribution with parameters α j and β j . Let L j
denote the Laplace transform of X j , j = 1, 2. Then

α j
β
j
L j (t) = , j = 1, 2.
(β j + t) α j
Let X = X 1 + X 2 . The Laplace transform of X is therefore given by
α 1
β β 2 α 2
1
L(t) = ;
α
α
(β 1 + t) 1 (β 2 + t) 2
see Example 4.6. It follows that X has a gamma distribution if and only if β 1 = β 2 .