Robert_V._Hogg,_Joseph_W._McKean,_Allen_T._Craig

5.2. Convergence in Distribution 329

Clearly,FXn(x)=FX(x) for allxin the support ofX,sothatXn
D
→X.Onthe
other hand, the sequenceXndoes not get close toX.Inparticular,Xn →Xin
probability.

Example 5.2.1.LetXnhave the cdf

Fn(x)=

∫x

−∞

1

√

1 /n

√

2 π

e−nw

(^2) / 2
dw.
If the change of variablev=
√
nwis made, we have
Fn(x)=
∫√nx
−∞
1
√
2 π
e−v
(^2) / 2
dv.
It is clear that
lim
n→∞
Fn(x)=
⎧
⎨
⎩
0 x< 0
1
2 x=0
1 x> 0.
Now the function
F(x)=
{
0 x< 0
1 x≥ 0
is a cdf and limn→∞Fn(x)=F(x)ateverypointofcontinuityofF(x). To be
sure, limn→∞Fn(0) =F(0), butF(x) is not continuous atx= 0. Accordingly, the
sequenceX 1 ,X 2 ,X 3 ,...converges in distribution to a random variable that has a
degenerate distribution atx=0.
Example 5.2.2.Even if a sequenceX 1 ,X 2 ,X 3 ,...converges in distribution to a
random variableX, we cannot in general determine the distribution ofXby taking
the limit of the pmf ofXn. This is illustrated by lettingXnhave the pmf
pn(x)=
{
1 x=2+n−^1
0elsewhere.
Clearly, limn→∞pn(x) = 0 for all values ofx. This may suggest thatXn,for
n=1, 2 , 3 ,..., does not converge in distribution. However, the cdf ofXnis
Fn(x)=
{
0 x<2+n−^1
1 x≥2+n−^1 ,
and
lim
n→∞
Fn(x)=
{
0 x≤ 2
1 x> 2.
Since
F(x)=
{
0 x< 2
1 x≥ 2
is a cdf, and since limn→∞Fn(x)=F(x) at all points of continuity ofF(x), the
sequenceX 1 ,X 2 ,X 3 ,...converges in distribution to a random variable with cdf
F(x).