1 1.1- random walk on z: 1.1.1- simple random walk : let be independent identically distributed...
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1
1.1- Random walk on Z:
1.1.1- Simple random walk :
Let be independent identically distributed random variables (iid) with only two possible values {1,-1}, such that: ,
or
Define the simple random walk process , by:
,
Define:
To be waiting time of the first visit to state 1 .
)1,1,0( qpqp
qpw
ppwX k .1
.1
0, nSn
1,1
nXSn
kkn0S
1:1inf nSnT
}1,{ nn
qXpX 1111 11
2
A state (i) is called recurrent if the chain
returns to (i) with probability 1 in a finite
number of steps, other wise the state is
transient.
If we define the waiting time variables as ,
then state i is recurrent if .
That is the returns to the state (i) are sure
events , and the state (i) is transient if:
i
1 ii
0 ii
In this case there is positive probability of never
returning to state (i), and the state (i) is a positive
recurrent if .
Hence if then the state (i)
is a null recurrent .
iXE i |
iXTEE iii |
3
We say that i leads to (j) and write if
for some
We say i communicates with j and write if both and .
ji jXP ni ( .0)0 n
ji ji
ij
A Markov chain with state space S is said to be irreducible if , ji Sji ,
If x is recurrent and then y is recurrent.yx
4
If the Markov chain is irreducible, and one of the states is recurrent then all the states are recurrent .
The simple random walk on is recurrent
iff
2
1qp
1) If and , then .
2) If and , then .
yx zy zx yx zy zx
5
We define the state or position of random walk at time zero as , and the position at time n as0S
n
kkn XS
1, then by strong law of large number
saEXn
Sn .1
Because:are independent identically distributed and
from the definition ,then .sX i '
11 X 11 XEHence
saqpEXn
Sn
n.lim 1
And
saqpn
Sn
n.1lim
This means that qpnSn ~ and consequence of that we have
qpif
qpifSn
nlim
The state 0 is transient if the number of visit to zero is finite almost surly, which mean that with probability 1 the number of visit is finite .
6
Hence zero is transient state.Now if p=q=1/2, we claim recurrence.It is enough to show that 0 is recurrent (from theorem 1.2.4), then the state 0 is recurrent iff
)0,0(G
Define
1
)(00)0,0(
n
nPG
Where 0|0)(
00 SSpp nn
1
2
1
)2(00
2
1
!!
)!2(
2)0,0(
n
n
n
nnn
nn
n
qpn
nPG
7
Now using sterling formula n
e
nnn
2~!
Hence
1
2
12
2
21
11
2
1
2
24
)0,0(
n
n
nn
n
n
e
nn
e
nn
G
However
)0,0(G and 0 is recurrent, and consequently the simple random walk on Z is recurrent .Now we will show that symmetric simple random walk is null recurrent, which means that NEWhere
0|0:1inf
0
SSn
totimereturnN
n
8
The probability generating function is defined as:
1
)(k
kN kNpssEsF
Then
1
)()()1(k
NPkNPF
Also..........)2()1()( 2 NPsNPssF
Hence........)2(2)1()( NPsNPsF
And
1
)()1(k
ENkNPkF
2411)( sqpsF
2
11
411)1()(
qp
qpFNP
9
Put:
2
1qp
Hence:
21)(
s
ssF
and
NESFs
)(lim1
Hence zero is a null recurrent, and the simple symmetric random walk on Z is null recurrent.
We proved that simple random walk on Z is
recurrent if .2
1qp
We have:2411)( sqpsF
Then:
2412
8)(
sqp
sqpsF
10
If is an infinite sequence of events such that:
1.
2. , provided
that are independent events.
,........, 21 AA
0suplim, nn
nn APthenAP
1suplim, nn
nn APthenAP
sAi '
11
It is well known that if are
independent random variables , then
, but this may fail in
the case of infinite product , To show this we
can introduce this counter example .
n ,.......,, 21
n
ii
n
ii EE
11
12
* Consider to be independent random variables such that:
That is we have infinite sequence of independent identically distributed random variables, then:
,...., 21 YY
2
102 ii yPyP
1)( 1 yE
We define anew sequence as follows:
then
n
iin y
1
: nn 2,0
Now define the waiting time
that means the first
time the sequence equal to zero.
Then :
0:1inf nXnT
11
n
kkn EYE
n
13
.
:)2(&)1(
)2(1
11
1
n
kk
kk
n
kk
EE
fromhence
E
On the other hand
112
1)()(
21
21
k
ikIK
kTPTP
k
kk yyyPkTP
2
1)0,2,......,2()( 11
hence: T is finite almost surly.
)1(0)(
0)lim(
:
.0
1
kk
nn
n
YE
E
then
saX
14
i. X is finite random variable iff ii. X is bounded iff
sa.
M
Now we introduce the notion of lebesgue measure, and Borel -field
15
The -field is a collection of sub-sets of such that
1.
2.
3.
The elements of are called measurable sets or events.
CAA
sAforA ii
i ',1
* The intersection of all -fields that contain all open
intervals is called Borel -field and denoted by .
It is known that a lebesgue measure on (Borel -field)
is the only measure that assigns for every interval
(a,b] a measure , and .
baabba ,,
Provided that EX and EY are finite.
)()()( YEXEYXE
16
If is finite then: Note : If are independent identically distributed random variables, and N is positive integer valued random variable independent of then:
iEX ni ,......2,1
n
i
n
iii EXXE
1 1
sX i '
sX i '
ENEXXEN
ii
1
1.
N
i
n
iii EXnnNXEnNXE
1 11.)/()/(
11
.))/(( EXNNXEN
ii
ENEXEXNENXEEN
ii .).())/(( 11
1
and
Hence
17
If , and
Then
0nX saXX n .
EXEX n
If random variables, then
,....)2,1(0 nX n
1 1i i
ii EXXE
We define a new sequence
then:
now
11
:
n
n
kkn YXY
1
limk
knn
XY
)1()(
lim)lim(
11
ii
ii
nn
nn
EXXE
EYYE
18
But is this theorem true if ?
The following example shows that this
theorem may fail if all .
0)0( nXp
0nX
Define are independent identically
distributed random variables, such that:
sYi '
2
1)1()1( ii YpYp
and define :
and
n
kkn YS
1
}1:1inf{: nSnT
19
)1(11
.
nTn
nn
n IYX then:
On the other hand :
)2(.1:1
saXEhencen
n
saSYIY T
T
nnnT
nn .1.
11
1. nTnn IEYEX
Because we have symmetric simple random walk, then recurrence implies that
and
then: Define also:
saST .1
1)( TSE
)()1(: nTnnTnn IYIYX
saT .
20
the event does occur or does not
depending only on , which means
that .
Hence are independent random
variables.
}1{ nT),....,,{ 121 nYYY
),.....,,(}1{ 121 nYYYnT )1(& nTn IY
then
and
Hence from (2) & (3) we conclude that:
.
0. )1( nTnn IEYEXE)3(.0 saXE
nn
n
nn
n XEXE
The sigma-field generated by a random variable
X is: )}(:)({)( 1 BBXXwhere is the Borel . field
21
A common error is repeated in some books, about linear correlation which is
But this relation must be written as:)1(1 xybaXY
)2(.1 sabaXYxy
The following example shows that the formula (1) may not be true.
* Consider a random variable .
We define the triple where
and and
P=Lesbgue measure on .
)1,0(~ U),,( P
)1,0( }:)1,0{( B
is the Borel -field restricted on (0,1) .
22
Now define
otherwiseO
irrationalisXifXY :
)()( rationalXPXYP
0))1,0(( Qmeasurelebesgue
1
.1)(
xythusand
saXYthenXYP
From (1) & (2) we get a = 0 , y = b
Suppose for the contrary that
)2........(..........003
1
)1.......(..........002
1
,
31
21
baY
X
baY
Xset
basomeforbaXY
This Contradiction because the assumption is not true.
23
It is already known that if the moment
generating function do exist ,then all the
moment do exist ,but it could happen that
all the moments do exist but the moment
generating function does not exist, the
following counterexample explains this
We know if the moment generating
function is defined , then
Let )1,0(~ NX
tM
0kk MXE
24
22
1
)(
)(kkX
thk
eeE
YofmomentkYE
And define
, that is Y
follows the lognormal distribution.
XYeY X log
Hence all the moment of y do exist, whereas the moment generating function of Y does not exist, as we show now .
dxe
dxee
eE
eEtM
Xet
Xet
et
ytY
X
X
X
0
221
221
2
1
2
1
)(
)()(
Then the moment generating function does not exist.
25
The sequence of moments does not determine uniquely the distribution, the following example show this:
We have the following two distributions with identical sequence of moments
,......2,1;3
23
3
10)2
,.......2,1;3
23)1
12
1
2
kk
YP
YP
kk
XP
k
k
k
k
26
Now:
362
12
3
23)()1
22
12
12
kkk
k
kkXEa
kkk
k
kk
k
kkXEa 12
32
3
23)()2
14
1
2
22
Obviously . 2,)( nXE n
2,)()2
3
12
3
230)()1
2
1212
1
nYEb
kkYEb
n
kk
k Also
27
It is known from the literature that the sequence of moments of a random variable uniquely determines the distribution if it satisfies one of the following equivalent conditions:
),0()()3
),()()2
.2
)(suplim)1
1
2
1
1
2
1
2
2
1
2
xifm
xifm
finiteisk
m
k
kk
k
kk
kk
n
28
The joint distribution of random variables determines uniquely the marginal distributions, but the converse is not necessarily true .
2
1)1(
2
1
4
1
4
1),0()0(
XP
YfXPY
This is a joint probability function
4
1;
1
0
10Y
X
4
1
4
1
4
1
4
1
29
These two marginal functions are
corresponding to the joint function fir
any
Conclusion:
The marginal distributions don’t
determine the joint distribution.
4
1
10X
)(xf2
1
2
1
The marginal distribution of X is:
10Y
)(Yf2
1
2
1
The marginal distribution of Y is:
30
The “positive correlated” is not transitive property.
A matrix is said to be positive
definite if it satisfies one of the
following equivalent conditions:
i.
ii. All the eignvalues are positive.
iii. The determinants of all left upper
sub-matrices are positive.
nnA
.00` XXAX
31
Any positive definite matrix could be
variance-covariance matrix of some
random vector.
Now Consider the matrix
1
1
1
21
41
21
21
41
21
We can make sure that is variance-covariance
matrix of a random variable .
Z
Y
X
04
1
02
1
02
1
zx
zxzx
zy
zyzy
yx
yxyx
then
Then X and Y are positively correlated, and Y and Z as well, but X and Z are negatively correlated
32
We say that converges weakly to X if:
At all continuity points of F.
This is denoted by
or
nX
)()(lim xFxFnn
XX Wn XX D
n
The following example shows that a sequence of continuous random variable does not necessary converge weakly to a continuous random variable.
33
Let Then X is a degenerate random variable with cumulative distribution function
Consider a sequence of cumulative density functions such that:
Then
saX .1
11
10)()(
x
xxXPXf
)(xFn
11
10
00
)(
x
xx
x
xF nn
11
10)(lim
x
xxFn
n
)(xFObviously, is the cumulative distribution function of a continuous random variable . Whereas, F is the cumulative distribution function of a discrete random variable X. Never the less, . XX W
n
nFnX
34
The median is any x such that
If F is continuous random variable, then
the median is unique .
bxa
2
1)(:sup
2
1)(:inf
xFXb
xFXa
Let X be a random variable such that
35
Consider a random variable X, such that
P(X)
210-1X
4
1
4
1
3
1
6
1
thus the median is any
Which means that a=0, and b=1.
That is, the median is not unique.
This is a disadvantage of the median as a
measure of location as the mean .
)1,0(x
21
216
5
102
1
014
1
10
)(
x
x
x
x
x
xF
The cumulative density function of X
36
If
Then :
Let
then
The median does not satisfy the relation
Mod(X+Y) = Mod(X) + Mod(Y)
),(~ nX
0,)(
)( 1
xeXn
xf xnn
)1,1(~ X
0,)( texf t
)0( t
t Ie
Assume that Y is an independent copy of X, so we can find Med(X) and Med(Y) as follows
37
this implies that
Med(X) =Med(Y)=log2
now to calculated Med (Z) suppose for contrary
that
Med(X+Y) = Med(X) + Med(Y)
This means that
Med(Z) = 2 log 2
)1,2(~ YXZ
But
2log
2
11)(
0
thence
edtetF tx
tX
With probability density function
38
tttx
z
tz
etedxextF
and
tettf
0
1
0,)(
.If the median of Z were 2log 2, then
2
1)2log2( ZF
However,
2
14034.0
2log21)2log2( 2log22log2
eeFZ
Hence
)()()( YMedXMedZMed
39
The mod is the value of X that makes the density function is maximum.
The mode is not unique, it is 0 and 1
P(X)
210X
9
4
9
4
9
1
A random variable X has the following distribution
40
Suppose that X&Y follow the distribution above and they are independent.Let Z=X+Y
P(X)
10X
5
3
5
2
P(Z)
210Z
25
9
25
12
25
4
000)()(1)( YModXModZMod
We show in this example that the mode is not a linear operator.This is a disadvantage of the mode.
41
and
from (1) & (2) we conclude that the Mode is not linear operator.
Let
0,)(
)1,1(~
texf
Xt
and Y is an independent copy of X, Since f (0)=1 is the maximum value of f(x), hence
)1()(0)( YModXMod
Now Z = X + Y, then , the probability density function of Z
)1,2(~ Z
0,)( tettf tz
,)( ttz eettf
)2()(10 ZModteet tt
42
We defined a simple random walk in chapter (1), and we know that the following Markov Chain
qkkppkkp )1,(,)1,(
is recurrent iff 2
1qp
This walk is called simple symmetric random walk (SSRW).
43
Consider a state i and a state j such that:
egersomeforkijij int k,
221,1 iiiiii( Transitive property) .
)1(ijjithen By the same way
kegersomeforkijijIf int,
)2(, ijjithen Hence simple symmetric random walk (SSRW) on Z is irreducible.
Simple symmetric random walk (SSRW) on Z is irreducible
44
A sequence 0, nn of random variables is Martingale if:
saE nnn .,......,,| 11
and sub-martingale if
saE nnn .,......,,| 11
and a super-martingale if :
saE nnn .,......,,| 11
45
Consider independent random variables
0, nn, such that 0)( iE
We claim that
n
kknS
0is a martingale, for
n
nn
nnnn
nnnnn
S
ES
ESE
SESSSE
1
1
11
||
|,,|
46
Consider independent random variables 1, ndn
Define a sequence k
n
kn dZ
1
. We show that
1, nEZ
ZW
n
nn
is a martingale.
n
nnnn
nnnnn
nnnnn
nnn
nn
nnn
W
EdZEdEZ
dEZEdEZ
dZEEdEZ
ZEEZ
EZ
ZEWE
11
11
11
11
1
11
1
|1
|1
|1
||
47
1Z
’s independent random
Consider a family tree where , and let
1nZ the number of children at thn )1( generation , and let
nkd the number of children of the thk
individual of the thn generation then:
n
n
Z
kZnnnnkn ddddZ
1211 ......
1,0, ZwherenZn
and 1,0, kndnk
,then nkd is a doubly indexed family .
nkdVariables, for every n and k and for fixed n they are independent identically distributed random variables.
We assume that
48
nZ
knkn dZif
11
1
11
n
nn EZ
ZW
Consider 1,0, kndnk , we Show that
, then
is a martingale.
sX i '
nz
knkn dZ
11
We use the fact that: If
Independent identically distributed random variables and
N random variable and independent of sX i '
then:
11
EXNEXEN
ii
are
Integer-valued
49
Since variables, and for fixed n they are independent identically distributed. Also
sd nk ' are independent random
sd nk ' are independent of nZ Then: nnn ZEdEZE 11)(
And
nn
n
nn
nn
nn
nnn
nn
n
Z
knk
nn
n
WEZ
Z
EdEZ
EdZ
EdEZ
dEZ
EdEZ
dE
EZ
ZE
n
1
1
1
1
50
(Not every martingale has an almost sure limit).In a symmetric simple random walk on Z, we have
2111 XPXp , then
nSis a martingale since
n
nn
nnn
nnnnn
n
kkn
S
XES
XES
XSESE
XS
1
1
11
1
and nn
Slim
simple random walk (SSRW), is recurrent this means that it keeps oscillating between all the integers.
does not exist, because this symmetric
51
If 0, nX nis a sub-martingale such that: n
nXEsup
Then for every 0
nXE
ini
p X
0max
52
.
is greater than or equal tofor the first time at time k .
We define kiforXXA iKK ,:
which means that KA is the event that the process
n
kkAA
0
, where A is the event that the process
is greater than or equal to by time n.
We want to show that:
n
kk
n
APAp
XEAp
0
Since on KA , kX
n
kA
n
kk
kIE
APAp
0
0
But since sAK ' are disjoint.
Hence
53
Then
nAn
An
An
n
kAn
n
kAn
n
kknn
n
kAnn
n
kAk
XEIXE
IXE
IXE
IXE
IXE
IXEE
IXEE
IXEAp
n
kk
k
k
k
k
0
0
0
0
0
0
In the following example we can see if the sequence is not a sub-martingale, then the last theorem fail.
54
Consider independent identically distributed random variables such that
2
102 ii XpXp
It is clear that the sequence nX
is not a sub-martingale since obviously, 1nEX
Hence:
)1(max1max00
ini
ini
XpXp
Doob’s inequality requires that for 2
then (1) will be
211
211 n
which is not true for large n.
This validity because the sequence is not a martingale.
55
In this chapter we describe convergence of sequences of random variables. Almost sure convergence, because of its relationship to pointwise convergence, and convergence in distribution, because of its being the easiest to establish. Convergence in probability is significant for weak laws of large numbers, and in statistics, for certain forms of consistency. Mean convergence is used to establish convergence of moments.
56
be random variables on .Let ,....,, 21 XXX p,,
We have these modes of convergence
1. Almost sure convergence 2. Convergence in probability3. Convergence in distribution4. Convergence in mean
Almost sure convergence, also known as convergence with probability one, is the probabilistic version of pointwise convergence.
The sequence nX converges to X almost surly, denoted by
XX san . , if
1: XXp n
57
The sequence nX converges to X in probability,
XX pn
0lim
XXp nn
for every .0
, ifdenoted by
0lim
XXE nn
The sequence nX converges to X in mean,
XX mn , ifdenoted by
58
In distribution
In mean
In probability
Almost sure
Defining conditionMode
1: XXp n
0lim
XXp nn
for all 0 0lim
XXE n
n
tFtF XXn n
lim
at continuity points t ofXF
Table 5.1. Definition of convergence for random variables
This convergence is some times called weak convergence. We summarize these modes in the following table.
The sequence nX converges to X in distribution,
XX Dn , if
XXXn
FoftspocontinuityattFtFn
intlim
denoted by
59
Almost sure convergence.
Convergence in mean.
Convergence in probability.
Convergence in distribution.
Figer 5.2- Implications Among Forms of Convergence
60
The last figure depicts the implications that are always valid. None of the other implications holds in general.
XX san . , then XX p
n
Suppose that 0 , and for each n, let XXA nn
Then, convergence in probability follows:
0..,limlim
oiApApAp nmk
mm
mm
If
61
If XX mn , then .XX p
n
By Chebyshev’s inequality, for each 0
XXEXXp nn
1
Therefor, 0 XXE n
implies that 0 XXp n
If XX pn , then XX D
n
The reader can see the proof in the book of Alan F. Karr p.(141).
62
The implication convergence in probability does not always lead to almost sure convergence and this counter example shows us that.
of independent random variables such that ;
nnnn XpandXp 11 110
We can see that for ,0 0
1
nXp n
1, nX nConsider a sequence
Hence
XX pn
Now claim that nX does not converge to zero almost surly,
n n
n nXp
1)1(
And since nX are independent, then from Borel-Cantelli lemma,
1.1)(: oiXp n This means that nX does not converge to zero almost surly.
Then
63
If is a sequence of independent random variables and
1, nX n
yprobabilitinSXSn
KKn
1
then:
SS san .
This theorem suggests that a convergence in probability is equivalent to almost sure convergence in case of martingales. But, this is false in general as we show in the following example.
64
Set
211
nforand
ZX
0
0
11
1
nnn
nn
n XifZXn
XifZX
Where nn XX ,......,1
Consider a sequence 1, nZn
of independent random variables such that:
npw
npw
npw
Zn
2
1.1
2
11.0
2
1.1
Define a new sequence as follows:nX
65
We claim that nX is a martingale which means that:
11 nnn XXENow
nnXnnnX
nXnnXn
nXnnXn
nXnXnnn
ZEIXnZEI
IZXnIZE
IXEIXE
IXIXEXE
nn
nn
nn
nn
0110
010
00
00
11
11
11||
Since nZ is independent of 1n
Then
1
01
010
11
11
n
Xn
nXnnXnn
Xn
IXn
ZEIXnZEIXE
n
nn
66
Notice that
.00 definitionfromZX nn
n
ZpXp nn
1100
For 0 0
10
nXpXp nn
But
nn
n nXp
10
This means that
1.0 oiXp n
Which is equivalent to saying does not converge to zero almost surly.That is, convergence in probability does not imply almost sure convergence even for martingales.
nX
67
Define
0011 nn XandX
And
1:0&0:1 XXHence the cumulative density function of is
nX
.11
102
1
00
xF
x
x
x
xXpxF nn
ThenXX D
n on the other hand,
11 XXp n does not converge to 0.
Hence does not converge to X in probability.nX
This example shows that convergence in distribution does not imply convergence in probability.Let each of 0 & 1 has mass = ½ 1,0
68
We can see that
01
n
XE n
Hence
0 mnX
And in example (13) we show that does not converge to 0 almost surly.
nX
This example shows that convergence in mean does not imply almost sure convergence.Define
npw
npw
X n 11.0
1.1
69
Hence does not converge to 0 in mean.nX
On the other hand
c
n
n
nif
ifnX
1
1
,00
,0
It is clear that
0. sanX
This example shows the converse implication of the last example, is not true in general.Define on the probability space a sequence of random variables as follows:
P,,
nX
n
InX n 1,0:
Where and P=lebesgue measure 1,0
Now
11
nn
nXpnEXXE nnn
70
XX n in the pth mean if
0,0 pnasxXEp
n
This example shows that convergence in probability does not imply convergence in the pth mean.Define
npw
npwn
X n
log
11.0
log
1.
For every ,0
0log
10
nXpXp n
p
n
71
This means
0 pnX
And
n
nXE
pp
n log
nXHence for every p>0, does not converge to 0 in the pth mean.
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6- The Integrals of Lebesgue Measure, by Denjoy, Perron, and Henstock/Russell A. Gordan.
1- Adventures in Stochastic Processes, by Sidney I. Resnick.
2- Introduction to Probability Theory, by Paul G.Hoel, Sidney C. Port and Charles J.Stone.
3- Markov Chains, by J.R.Norries.
4- Probability by Alan F. Karr.
5- Random Walks and Electric Networks, by Peter G.Doyle and J.Laurie Snell.