Page 131 - Elements of Distribution Theory

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4.4 Cumulants 117

Example 4.21 (Laplace distribution). Let X denote a random variable with a standard
Laplace distribution; see Example 4.19. Let µ and σ> 0 denote constants and let Y =
σ X + µ; the distribution of Y is called a Laplace distribution with location parameter µ
and scale parameter σ.
Using the results in Example 4.19, together with Theorem 4.14, the first four cumulants
4
2
of this distribution are µ,2σ ,0, and 12σ .
Example 4.22 (Standardized cumulants). Consider a random variable X with cumu-
√
lants κ 1 ,κ 2 ,... and consider the standardized variable Y = (X − κ 1 )/ κ 2 . The cumulants
j
2
of Y are given by 0, 1, κ j /κ , j = 3, 4,.... The cumulants of Y of order 3 and greater are
2
sometimes called the standardized cumulants of X and are dimensionless quantities. They
are often denoted by ρ 3 ,ρ 4 ,... so that
j
ρ j (X) = κ j (X)/κ 2 (X) 2 , j = 3, 4,....
We have seen that if X and Y are independent random variables, then the characteristic
function of X + Y satisfies
ϕ X+Y (t) = ϕ X (t)ϕ Y (t);

hence,
log ϕ X+Y (t) = log ϕ X (t) + log ϕ Y (t).
Since the cumulants are simply the coefficients in the expansion of the log of the character-
istic function, it follows that the jth cumulant of X + Y will be the sum of the jth cumulant
of X and the jth cumulant of Y.

Theorem 4.15. Let X and Y denote independent real-valued random variables with mth
cumulants κ m (X) and κ m (Y),respectively, and let κ m (X + Y) denote the mth cumulant of
X + Y. Then
κ m (X + Y) = κ m (X) + κ m (Y).

Proof. We have seen that ϕ X+Y (t) = ϕ X (t)ϕ Y (t). Hence, by Theorem 4.13,
m
j m
log(ϕ X+Y (t)) = log(ϕ X (t)) + log(ϕ Y (t)) = (it) (κ j (X) + κ j (Y))/j! + o(t ).
j=1
The result now follows from noting that
m
j m
log ϕ X+Y (t) = (it) κ j (X + Y)/j! + o(t )
j=1
as t → 0.

Example 4.23 (Independent identically distributed random variables). Let X 1 , X 2 ,...,
X n denote independent, identically distributed scalar random variables and let κ 1 ,κ 2 ,...
n
denote the cumulants of X 1 . Let S = X j . Then
j=1
κ j (S) = nκ j , j = 1, 2,....