Page 314 - Elements of Distribution Theory

P1: JZP
052184472Xc10 CUNY148/Severini May 24, 2005 2:50

300 Orthogonal Polynomials

An important consequence of Lemma 10.1 is that, for any j = 0, 1,..., the function x j
has a unique representation in terms of p 0 , p 1 ,..., p j ; that is,
j
x = α 0 p 0 (x) + α 1 p 1 (x) +· · · + α j p j (x)
for some unique set of constants α 0 ,...,α j . Hence, for each n = 0, 1,...,

∞

j
x p n (x) dF(x) = 0, j = 0, 1,..., n − 1. (10.1)
−∞
Furthermore, as the following theorem shows, this property characterizes p n .
Theorem 10.1. Let {p 0 , p 1 ,...} denote a set of orthogonal polynomials with respect to a
distribution function F. Let f denote a polynomial of degree n. Then

∞

j
x f (x) dF(x) = 0, j = 0, 1,..., n − 1 (10.2)
−∞
if and only if for some α = 0
f (x) = αp n (x) a.e.(F).

j
Proof. Suppose that f = αp n for some α = 0. Fix j = 0,..., n − 1. Since x may be
written as a linear function of p 0 ,..., p n−1 ,it follows that (10.2) holds.
Now suppose that (10.2) holds for some polynomial f of degree n. Let c n denote the
n
n
coefficient of x in f (x) and let d n denote the coefficient of x in p n (x). Then
c n
f (x) − p n (x) = g(x)
d n
where g is a polynomial of degree at most n − 1. Let α = c n /d n . Then
∞ ∞ ∞

2
( f (x) − αp n (x)) dF(x) = g(x) f (x) dF(x) − α g(x)p n (x) dF(x).
−∞ −∞ −∞
By (10.2),
∞

g(x) f (x) dF(x) = 0
−∞
and, by (10.1),
∞

g(x)p n (x) dF(x) = 0.
−∞
It follows that
∞

2
( f (x) − αp n (x)) dF(x) = 0
−∞
and, hence, that
f (x) − αp n (x) = 0 a.e. (F),

proving the result.