Page 348 - Elements of Distribution Theory

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334 Approximation of Probability Distributions

Table 11.1. Exact probabilities in Example 11.8.

α

n 0.10 0.25 0.50 0.75 0.90
10 0.126 0.317 0.583 0.803 0.919
20 0.117 0.296 0.559 0.788 0.914
50 0.111 0.278 0.538 0.775 0.909
100 0.107 0.270 0.527 0.768 0.907
200 0.105 0.264 0.519 0.763 0.905
500 0.103 0.259 0.512 0.758 0.903

Let Q Z denote the quantile function of the standard normal distribution. Table 11.1 con-
tains probabilities of the form Pr(X n ≤ Q Z (α)) for various values of n and α; note that a
probability of this form can be approximated by α.For each value of α given, it appears that
the exact probabilities are converging to the approximation, although the convergence is, in
some cases, quite slow; for instance, for α = 0.50, the relative error of the approximation
is about 2.3% even when n = 500.

Uniformity in convergence in distribution
Convergence in distribution requires only pointwise convergence of the sequence of distri-
butions. However, because of the special properties of distribution functions, in particular,
the facts that they are nondecreasing and all have limit 1 at ∞ and limit 0 at −∞, point-
wise convergence is equivalent to uniform convergence whenever the limiting distribution
function is continuous.

Theorem 11.3. Let X, X 1 , X 2 ,... denote real-valued random variables such that
D
X n → X as n →∞.
For n = 1, 2,..., let F n denote the distribution function of X n and let F denote the distri-
bution function of X. If F is continuous, then
sup |F n (x) − F(x)|→ 0 as n →∞.
x
Proof. Fix > 0. Let x 1 , x 2 ,..., x m denote a partition of the real line, with x 0 =−∞,
x m+1 =∞, and
F(x j ) − F(x j−1 ) < /2, j = 1,..., m + 1.

Let x ∈ R, x j−1 ≤ x ≤ x j for some j = 1,..., m + 1. Since F n (x j ) → F(x j )
as n →∞,

F n (x) − F(x) ≤ F n (x j ) − F(x j−1 ) < F(x j ) + /2 − F(x j−1 )
for sufficiently large n, say n ≥ N j . Similarly,

F n (x) − F(x) ≥ F n (x j−1 ) − F(x j ) > F(x j−1 ) − /2 − F(x j )
for n ≥ N j−1 ; note that if j = 1, the F n (x j−1 ) = F(x j−1 )so that N 0 = 1.