Page 114 - Elements of Distribution Theory

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100 Moments and Cumulants

Example 4.7 (Inverse gamma distribution). Let X denote a scalar random variable with
an absolutely continuous distribution with density function
x −3 exp(−1/x), x > 0;
this is an example of an inverse gamma distribution. The Laplace transform of this distri-
bution is given by

∞ √
L(t) = exp(−tx)x −3 exp(−1/x) dx = 2tK 2 (2 t), t ≥ 0.
0
Here K 2 denotes the modified Bessel function of order 2; see, for example, Temme
(1996).

As might be expected, the properties of the Laplace transform of X are similar to those
of the characteristic function of X;in particular, if two random variables have the same
Laplace transform, then they have the same distribution.

Theorem 4.5. Let X and Y denote real-valued, nonnegative random variables. If L X (t) =
L Y (t) for all t > 0, then X and Y have the same distribution.

Proof. Let X 0 = exp{−X} and Y 0 = exp{−Y}. Then X 0 and Y 0 are random variables
t
t
taking values in the interval [0, 1]. Since L X (t) = L Y (t), it follows that E[X ] = E[Y ] for
0 0
all t > 0; in particular, this holds for t = 1, 2,.... Hence, for any polynomial g,E[g(X 0 )] =
E[g(Y 0 )].
From the Weierstrass Approximation Theorem (see Appendix 3), we know that any
continuous function on [0, 1] may be approximated to arbitrary accuracy by a polynomial.
More formally, let h denote a bounded, continuous function on [0, 1]. Given > 0, there
exists a polynomial g such that
sup |h(z) − g (z)|≤ .
z∈[0,1]
Then

E[h(X 0 ) − h(Y 0 )] − E[g (X 0 ) − g (Y 0 )] = E[h(X 0 ) − g (X 0 )] − E[h(Y 0 ) − g (Y 0 )]
≤ E[|h(X 0 ) − g (X 0 )|] + E[|h(Y 0 ) − g (Y 0 )|]
≤ 2 .
Since E[g (X 0 ) − g (Y 0 )] = 0,

E[h(X 0 )] − E[h(Y 0 )] ≤ 2

and, since is arbitrary, it follows that E[h(X 0 )] = E[h(Y 0 )] for any bounded continuous
function h.It follows from Theorem 1.14 that X 0 and Y 0 have the same distribution. That
is, for any bounded continuous function f ,E[ f (X 0 )] = E[ f (Y 0 )]. Let g denote a bounded,
continuous, real-valued function on the range of X and Y. Since X =− log(X 0 ) and
Y =− log(Y 0 ), g(X) = f (X 0 ) where f (t) = g(− log(t)), 0 < t < 1. Since g is bounded
and continuous, it follows that f is bounded and continuous; it follows that

E[g(X)] = E[ f (X 0 )] = E[ f (Y 0 )] = E[g(Y)]
so that X and Y have the same distribution.