Page 109 - Elements of Distribution Theory

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4.2 Moments and Central Moments 95

Example 4.2 (Cauchy distribution). Let X denote a random variable with a standard
Cauchy distribution; see Example 1.29. The same argument used in that example to show
r
that E(X) does not exist may be used to show that E(X ) does not exist when r is an odd
r
integer and E(X ) =∞ when r is an even integer.

Central moments
Let X denote a real-valued random variable such that E(|X|) < ∞ and let µ = E(X). As
noted earlier, the central moments of X are the moments of the random variable X − µ.
Clearly, the first central moment of X,E(X − µ), is 0. The second central moment of
2
X, E[(X − µ) ], is called the variance of X, which we will often denote by Var(X). The
√
standard deviation of X, defined to be the (positive) square root of its variance, Var(x),
is also often used. Note that, if µ exists, Var(X)always exists, but it may be infinite. The
following theorem gives some simple properties of the variance. The proof is straightforward
and left as an exercise.

2
Theorem 4.1. Let X denote a real-valued random variable such that E(X ) < ∞. Then
(i) Var (X) < ∞
2
2
(ii) Var (X) = E(X ) − µ where µ = E(X)
(iii) Let a, b denote real-valued constants and let Y = aX + b. Then Var (Y) =
2
a Var (X).
2
(iv) For any c ∈ R, E[(X − c) ] ≥ Var (X) with equality if and only if c = µ.
(v) For any c ∈ R,
1
Pr{|X − µ|≥ c}≤ Var (X).
c 2

Part (v) of Theorem 4.1 is known as Chebyshev’s inequality.

Example 4.3 (Standard exponential distribution). Suppose that X has a standard expo-
2
nential distribution. Then, according to Example 4.1, E(X) = 1 and E(X ) = 2. Hence,
Var(X) = 1.

r
Since (X − µ) ,r = 3, 4,..., may be expanded in powers of X and µ, clearly the central
moments may be written as functions of the standard moments. In particular, the central
r
j
moment of order r is a function of E(X ) for j = 1, 2,...,r so that if E(|X| ) < ∞, then
the central moment of order r also exists and is finite.
3
Example 4.4 (Third central moment). Let X denote a random variable such that E(|X| ) <
∞. Since

3
2
3
3
2
(X − µ) = X − 3µX + 3µ X − µ ,
it follows that
3
3
2
3
E[(X − µ) ] = E(X ) − 3µE(X ) + 2µ .