Math 280B, Winter 2012
Feller’s method for the CLT
1. Let U1 , . . . , Un , V1 , . . . , Vn be independent random variables. We are going to compare
the distribution of U := U1 + · · · + Un to that of V := V1 + · · · + Vn . Bearing in mind
the Portmanteau Theorem, we examine E[g(U ) − g(V )] for suitable bounded continuous
functions g. First, we have the telescoping sum
(1.1)
E[g(U ) − g(V )] =
n
X
k=1
E[g(V1 + · · · +Vk−1 + Uk + · · · Un )
− g(V1 + · · · + Vk + Uk+1 + · · · Un )],
in which the k = 1 term is understood to be
E[g(U1 + · · · Un ) − g(V1 + U2 + · · · Un )]
and the k = n term is understood to be
E[g(V1 + · · · + Vn−1 + Un ) − g(V1 + · · · + Vn )].
The k th term in the above sum is equal to
(1.2)
E |g(Uk + Wk ) − g(Vk + Wk )| ,
where
Wk := V1 + · · · + Vk−1 + Uk+1 + · · · + Un ,
with the obvious interpretations in the extreme cases k = 1 and k = n. If we introduce
the notation
Ψk g(x) := E[g(Uk + x)],
Υk g(x) := E[g(Vk + x)],
the the quantity in (1.2) is equal to
(1.3)
E [Ψk g(Wk ) − Υk g(Wk )] ,
because of the mutual independence of Uk , Vk , and Wk . Since g is bounded and continuous,
both Ψk g and Υk g are bounded and continuous functions. Let us now write, for a bounded
function u : R → R,
kuk := sup{|u(x)| : x ∈ R}.
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Then the expression in (1.3) is at most
kΨk g − Υk gk.
Feeding this information into (1.1) we obtain the important estimate
(1.4)
|E[g(U ) − g(V )]| ≤
n
X
k=1
kΨk g − Υk gk.
2. We now specialize the situation of section 1 to the case of iid random variables; this
yields a direct route to the central Limit Theorem. So let X1 , X2 , . . . be iid random
variables, each with mean 0 and variance 1. Also, let Z, Z1 , Z2 , . . . be iid standard normal
random variables, independent of the Xk s. Fix n and define
Xk
Uk := √ ,
n
Zk
Vk := √ ,
n
k = 1, 2, . . . , n.
Evidently V = V1 + · · · Vn has the same distribution as Z; that is, the standard normal
distribution. Moreover, U is just the normalized sum
X + · · · + Xn
S
√n = 1 √
.
n
n
In the present case the n terms in the sum (1.4) are all the same, so we have
√
E[g(Sn / n) − g(Z)] ≤ nkΨ1 g − Υ1 gk.
(2.1)
The right side of (2.1) is dominated by
nkΨ1 g − g − g ′′ /2k + nkΥ1 g − g − g ′′ /2k,
and both of these terms converges to 0 as n → ∞. This is a consequence eof the following
lemma.
3. Lemma. Let X be a random variable with E[X] = 0 and E[X 2 ] = 1. Let g : R → R
be thrice continuously differentiable with compact support. Then
(3.1)
√
n E[g(x + X/ n)] − g(x) − g ′′ (x)/2 → 0,
n → ∞,
and the convergence is uniform in x ∈ R.
Proof. The expression in (3.1) can be written as
√
√
g(x + X/ n) − g(x) − (X/ n)g ′ (x) g ′′ (x)
2
(3.2)
E
X .
−
X 2 /n
2
2
For a fixed ω the integrand
√
√
g(x + X/ n) − g(x) − (X/ n)g ′ (x) g ′′ (x)
−
X 2 /n
2
converges to 0 as n → ∞, uniformly in x. Also, the integrand is bounded (uniformly in x
and ω) and the measure x2 P[X ∈ dx] has total mass 1, so by bounded convergence the
expectation in (3.2) converges to 0 and the convergence is uniform in x.
4. It follows from the lemma and (2.1) that
√
lim E[g(Sn / n) − g(Z)] = 0,
n→∞
√
provided g is smooth and of compact support. This in turn implies that Sn / n converges
in distribution to Z, which is the basic Central Limit Theorem.
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