CHAPTER 1
Converges in Probability and Distribution
1. Let {Xn }, X be random variables. Then {Xn }
to X as n → ∞, written Xn →d X , if
Definition
distribution
∞
converges in
∞
lim P (Xn ≤ X) = lim FXn (x) = FX (x)
n
n
for each continuity point of the distribution function F (x).
Example 2. Let X1 , X2 , ..., Xn be a random sample from a unifrom distribution, Xi ∼ U N IF (0, 1), and let Yn = Xn:n the largest order statistic. Find limiting
distribution of Yn .
Solution. From equation
the CDF of Yn is GYn = (FX (y))n and
??


 0,
FX (x) =
x,


1,
Therefore
0≥x
0<x<1
x≥1


 0,
GYn (y) =
yn ,


1,
and
∞
lim GYn (y) =
n
0≥y
0<y<1
y≥1
∞
n∞
lim 1 = 1,
n∞




lim 0 = 0,
0≥y
0<y<1
y≥1
n
lim y n = 0,



Thus
(
∞
GY (y) = lim GYn (y) =
n
0 ,y < 1
1 ,y ≥ 1
Example 3. Suppose that X1 , X2 , ..., Xn is a random sample from a Pareto
distribution, Xi ∼ P AR (1, 1) or fX (x) = (1 + x)−2 , x > 0, and let Yn = nX1:n .
1
The CDF of Xi is FX (x) = 1 − 1+x
, x > 0, Find limiting distribution of Yn
Solution. From equation
(
GYn (y) =
1 − 1 − FX
??,
y
n
n
=1−
1
y
1+ n
n
=1− 1+
y −n
n
,0 < y
,y ≤ 0
0
and
(
∞
GY (y) = lim GYn (y) =
n
∞
lim 1 − 1 +
n
y −n
n
= 1 − e−y
,0 < y
,y ≤ 0
0
1
1. CONVERGES IN PROBABILITY AND DISTRIBUTION
Example
4. Rework Example 3 for Yn = Xn:n
Solution. From equation
(
GYn (y) =
and
2
??,
n
n
1
(FX (y)) = 1 − 1+y
,0 > y
0
,y ≥ 0

n
y
 lim
= 0 ,0 > y
GY (y) = lim GYn (y) = n∞ 1+y
n∞
0
,y ≥ 0
Thus Yn does not have limit distribution.
Definition
value y = c if
5. The function G (y) is the CDF of degenerate distribution at the
(
0 ,y < c
GY (y) =
1 ,y ≥ c
In other words, G (y) is the CDF of discrete distribution that assigns probability
one at the value y = c and zero otherwise.
Example 6. Let X1 , X2 , ..., Xn is a random sample from an Exponensial disx
tribution, Xi ∼ EXP (θ) or fX (x) = θ1 e− θ , x > 0, and let Yn = X1:n
Solution. It follows that the CDF of Yn is
(
ny
1 − e− θ
GYn (y) =
0
,y > 0
,y ≤ 0
Then
(
∞
GY (y) = lim GYn (y) =
n
∞
lim 1 − e−
n
ny
θ
= 1 ,0 < y
,y ≤ 0
which is corresponds to a degenerate distribution at the value y = 0.
0
7. A sequence of random variables Y1 , Y2 , ... is said to converto a constant c, written Yn →stochastic c, if it has a limiting
distribution that is degenerate at y = c.
Definition
gence stochastically
Example 8. For Example 2 and from Denition 5 and Denition 7, and G (y)
is the CDF of degenerate distribution at the value y = 1 and Yn →stochastic 1.
Theorem
9.
(
Continuity Theorem
) Let Y1 , Y2 , ... be a sequence of random
GY1 (y) , GY2 (y) , ... and MGF's MY1 (t) , MY2 (t) , ....
CDF GY (y), and if lim MYn (t) = MY (t) for all t in
variables with respective CDF's
If
MY (t)
is the MGF of a
open interval containing zero,
continuity points of
n
−h < t < h,
∞
lim GY
n∞
then
GY (y).
n
(y) = GY (y)
for all
In other words,
∞
∞
lim MYn (t) = MY (t) ⇒ lim GYn (y) = GY (y) ⇒ Yn →d Y
n
n
10. Let X1 , X2 , ..., Xn be a random sample from a Bernoulli distrin
P
bution, Xi ∼ BIN (1, p), and consider Yn = Xi .
Example
i=1
1. CONVERGES IN PROBABILITY AND DISTRIBUTION
3
Solution. Let np = µ for xed µ > 0 then p → 0 as n → ∞. Thus, from
Theorem ??,
n
t
+ q) = (pe
n
t
µe
= n + 1 − nµ
n
µ(et −1)
= 1+
n
MYn (t)
And
∞
∞
lim MYn (t) = lim
n
1+
n
µ (et − 1)
n
n
t
= eµ(e
−1)
t
Since MY (t) = eµ(e −1) is the MGF of the Poisson distribution with mean µ. Thus,
Yn →d Y ∼ P OI (µ = np).
Example
and Zn =
11. Let X1 , X2 , ..., Xn be a random sample and consider Yn =
Solution. Let σn =
a2
2
σn
Xi
i=1
Yn −np
√
npq .
M Yn − np (t) = e
n
P
− npt
σn
σn
√
M Yn
npq , Zn =
t
σn
Yn
σn
− σnpn . Since Zn =
Yn
σn
− σnpn , then MZ n (t) =
. Therefore, using the series expansion ea = 1 + a +
+ ...,
n
t
npt
= e− σn pe σn + q
h pt t
in
= e− σn pe σn + q
h
in
2 2
t2
= 1 − σptn + p2σt2 + ... p(1 + σtn + 2σ
2 + ...) + q
n
n
h
in
2 2
pt2
= 1 − σptn + p2σt2 + ... 1 + σptn + 2σ
2 + ...
hn
in n
d(n)
t2
= 1 + 2n + n
MZ n (t)
where d (n) → 0 as n ∞. Thus,
∞
∞
lim MZn (t) = lim
n
In other words, Zn =
Example
n
Yn −np
√
npq
1+
t2
d (n)
+
2n
n
n
t2
=e2
→d Z ∼ N (0, 1).
12. Let Z1 , Z2 , ..., Zn be a random sample and Zi ∼ N (0, 1), Find
n
P
the limiting distribution of Zn =
1
Zi+ n
√
n
i=1
.
1. CONVERGES IN PROBABILITY AND DISTRIBUTION
4
Solution. Since MGF of Zi is MZ (t) = e 2 t , then from Theorem
Zn is,
Zn t
1 2
MZn (t)
=E
e
 n

P


= E e

Z +1
n
 i=1√i

n



t 



√t + √t
n
n
n
P
Zi
!
=E e
t
√t
√ Z
√t Z
√t Z
= e n E e n 1 e n 2 ...e n n
n
√t
= e n MZn √tn
i=1
Therefore,
∞
∞
lim MZn (t) = lim e
n
Thus, Z n →d Z ∼ N (0, 1).
n
√t
n
t2
e 2n
n
∞
= lim e
n
√t
n
t2
t2
e2 =e2
??
MGF of