Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

332 Consistency and Limiting Distributions

Theorem 5.2.1.IfXnconverges toXin probability, thenXnconverges toXin
distribution.

Proof:Letxbe a point of continuity ofFX(x). For every>0,

FXn(x)=P[Xn≤x]

= P[{Xn≤x}∩{|Xn−X|<}]+P[{Xn≤x}∩{|Xn−X|≥ }]

≤ P[X≤x+ ]+P[|Xn−X|≥ ].

Based on this inequality and the fact thatXn
P
→X,weseethat
lim
n→∞
FXn(x)≤FX(x+ ). (5.2.5)

To get a lower bound, we proceed similarly with the complement to show that

P[Xn>x]≤P[X≥x− ]+P[|Xn−X|≥ ].

Hence
lim
n→∞

FXn(x)≥FX(x− ). (5.2.6)

Using a relationship betweenlim and lim, it follows from (5.2.5) and (5.2.6) that

FX(x− )≤ lim

n→∞

FXn(x)≤ lim

n→∞

FXn(x)≤FX(x+ ).

Letting ↓0 gives us the desired result.

Reconsider the sequence of random variables{Xn}defined by expression (5.2.1).

Here,Xn→DXbutXn

P

→X. So, in general, the converse of the above theorem is

not true. However, it is true ifXis degenerate, as shown by the following theorem.

Theorem 5.2.2. IfXnconverges to the constantbin distribution, thenXncon-
verges tobin probability.

Proof:Let>0begiven. Then

lim

n→∞

P[|Xn−b|≤ ] = lim

n→∞

FXn(b+ )−lim

n→∞

FXn[(b− )−0] = 1−0=1,

which is the desired result.

A result that will prove quite useful is the following:

Theorem 5.2.3.SupposeXnconverges toXin distribution andYnconverges in
probability to 0. ThenXn+Ynconverges toXin distribution.

The proof is similar to that of Theorem 5.2.2 and is left to Exercise 5.2.13. We

often use this last result as follows. Suppose it is difficult to show thatXnconverges

toXin distribution, but it is easy to show thatYnconverges in distribution to

Xand thatXn−Ynconverges to 0 in probability. Hence, by this last theorem,

Xn=Yn+(Xn−Yn)→DX, as desired.
The next two theorems state general results. A proof of the first result can
be found in a more advanced text, while the second, Slutsky’s Theorem, follows
similarly to that of Theorem 5.2.1.