Page 355 - Elements of Distribution Theory

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052184472Xc11 CUNY148/Severini May 24, 2005 17:56

11.3 Convergence in Probability 341

so that, for all > 0,

lim inf F n (x) ≥ F(x − ).
n→∞
That is, for all > 0,
F(x − ) ≤ lim inf F n (x) ≤ lim sup F n (x) ≤ F(x + ).
n→∞
n→∞
Suppose F is continuous at x. Then F(x + ) − F(x − ) → 0as → 0. It follows that

lim F n (x)
n→∞
D
exists and is equal to F(x)so that X n → X. This proves part (i) of the theorem.
D
Now suppose that X n → X and that X = c with probability 1. Then X has distribution
function

0if x < c
F(x) = .
1if x ≥ c
Let F n denote the distribution function of X n . Since F is not continuous at x = c,it follows
that
0if x < c

lim F n (x) = .
n→∞ 1if x > c
Fix > 0. Then
Pr(|X n − c|≥ ) = Pr(X n ≤ c − ) + Pr(X n ≥ c + )
≤ F n (c − ) + 1 − F n (c + /2) → 0as n →∞.
Since this holds for all > 0, we have

lim Pr{|X n − c|≥ }= 0 for all > 0;
n→∞
p
it follows that X n → c as n →∞, proving part (ii) of the theorem.

Convergence in probability to a constant
We now consider convergence in probability of a sequence X 1 , X 2 ,... to a constant. Without
loss of generality we may take this constant to be 0; convergence to a constant c may be
established by noting that X n converges in probability to c if and only if X n − c converges
in probability to 0.
Sinceconvergenceinprobabilitytoaconstantisequivalenttoconvergenceindistribution,
p p
by Corollary 11.2, if X n → 0 and f is a continuous function, then f (X n ) → f (0). Since,
in this case, the distribution of X n becomes concentrated near 0 as n →∞, convergence
of f (X n )to f (0) holds provided only that f is continuous at 0. The details are given in the
following theorem.

Theorem 11.7. Let X 1 , X 2 ,... denote a sequence of real-valued random variables such
p
that X n → 0 as n →∞. Let f : R → R denote a function that is continuous at 0. Then
p
f (X n ) → f (0) as n →∞.