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Stochastic Processes
(Lecture #4)
Davy Paindaveine
Universite Libre de Bruxelles
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Today
Today, our goal will be
to finish with conditional expectations,
to have a quick look at conditional variances,
and
to discuss limits of sequences of r.v.s, and
to study famous limiting results.
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Conditional expectation
Conditional expectationallows for exploiting some information(e.g., the information that some event occurred) in order to
improve the (unconditional) best guessE[X].
This available information can take various forms:
(the occurrence of) an event.
(the value of )a r.v.
(the occurrence of some event in) a sigma-algebra.
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Conditional expectation (w.r.t. a r.v.)
LetXbe an integrable r.v. on(,A, P).LetY be adiscreter.v. on(,A, P), say with distribution
distribution ofY
values y1 y2 . . .
probabilities p1 p2 . . .
Then we defineE[X|Y]as the r.v.
E[X|Y] : R
E[X|Y =Y()].The last conditional expectation is w.r.t. an event, and hence is
well defined.
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Conditional expectation (w.r.t. a r.v.)
LetXbe an integrable r.v. on(,A, P).LetYbe anabsolutely continuousr.v. on(,A, P), say with pdff.
Then we defineE[X
|Y]as the r.v. such that
(i)E[X|Y]is(Y)-measurable.(ii)
A
E[X|Y]() dP() = A
X() dP(), for allA (Y).
In practice:
IfX is discrete,E[X|Y =y] =i xiP[X=xi|Y =y].
IfXis absolutely continuous,
E[X|Y =y] = 1fY(y)
R
x f(X,Y)(x, y) dx=
R
x f(X|Y)(x, y) dx,
wheref(X
|Y)
(x, y)denotes the pdf ofXconditionally onY =y.
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Conditional expectation (w.r.t. a r.v.)
Same properties as in the discrete case:
E[X|Y]is(Y)-measurable E[X
|Y] = E[X]iffX
Y.
E[X|Y] =X iffX is(Y)-measurable. E
E[X|Y]
= E[X].
Remark: E[X]= E
E[X|Y]
=R
E[X|Y =y] fY(y) dy.(//total probability formula).
We still define: P[A|Y] = E[IA|Y]]. P[A] = E[IA] = E
E[IA|Y]
=
R
E[IA|Y =y] fY(y) dy=
RP[A|Y =y] fY(y) dy.
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Conditional expectation (w.r.t. a r.v.)
Example: letX Unif(0, 1). IfX=x, you flipm times a coin suchthatP[Head] =x. LetNbe the number of heads.
E[N] =? Distribution ofN?
E[N|X=x] =mn=0
n P[N=n|X=x] =mn=0
n
m
n
xn(1 x)mn =
=
mn=1
m
m 1n 1
xn(1x)mn =mx
mn=1
m 1n 1
xn1(1x)(m1)(n1) =mx.
E[N|X] =mX ((X)-measurable).
Consequently, E[N] =E
E[N|X]
= E[mX] =m E[X] = m2.
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Conditional expectation (w.r.t. a r.v.)
Distribution ofN?
P[N=n] = R
P[N=n
|X=x] fX(x) dx=
1
0
P[N=n
|X=x] dx
=
10
m
n
xn(1x)mn dx=
m
n
10
xn(1x)mn dx=. . .= 1m + 1
,
for alln= 0, 1, . . . , m.
Hence,Nis uniformly distributed on {0, 1, . . . , m}.(which now makes clear whyE[N] = m2).
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Conditional expectation
Conditional expectationallows for exploiting some information(e.g., the information that some event occurred) in order to
improve the (unconditional) best guessE[X].
This available information can take various forms:
(the occurrence of) an event.
(the value of ) a r.v.
(the occurrence of some event in)a sigma-algebra.
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Conditional expectation (w.r.t. a-algebra)
LetXbe an integrable r.v. on(,A, P).Let Fbe a sigma-algebra A.
Then we defineE[X
|F]as the r.v. such that
(i)E[X|F]is F-measurable.(ii)
A
E[X|F]() dP() = A X() dP(), for allA F.
Same kind of properties as forE[X|Y]: E[X|F]is F-measurable E[X
|F] = E[X]iffX
F.
E[X|F] =X iffX is F-measurable. E
E[X|F]
= E[X].
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Conditional expectation (w.r.t. a-algebra)
Let F1 F2be two-algebras in A.Extra properties:
EE[X|F2
]F1= E[X
|F1].
E
E[X|F1]F2
= E[X|F1].
Conditional expectation w.r.t. to a-algebra also allows for
defining conditional expectationw.r.t. a collection of r.v.s.
More specifically, we define
E[X|Y1, . . . , Y n] := E[X|(Y1, . . . , Y n)],
where(Y1, . . . , Y n)is the smallest-algebra containing
{Y1i (B)
|B
B, i= 1, . . . , n
}.
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Conditional expectation (w.r.t. a-algebra)
To understand how one can computeE[X|F]in practice,see exercise 6, in homework 1.
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Conditional variances
LetXandYbe r.v.s on(,A, P).As we have seen, one can exploit the information sitting inY
in order to improve the (unconditional) best guessE[X].
the improved guess is the conditional expectationE[X|Y].Similarly, onceYis observed, the knowledge of the
(unconditional) dispersion ofX, namely
Var[X] = E
(X E[X])2
,
can be improved into theconditional variance
Var[X|Y] = E
(X E[X|Y])2Y.
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Conditional variances
In time series analysis, many models are designed to explain thedynamics of the conditional varianceVar[Xt|Xt1, Xt2, . . .](e.g., the so-called stochastic volatility models).
Remarks:
Similarly as forVar[X],
Var[X|Y] = E[X2
|Y] E[X|Y]2
(exercise; better for computations, not for the interpretation).
Var[X]
= E[Var[X
|Y]], but we have
Var[X] = E
Var[X|Y]+ VarE[X|Y].Indeed, E[Var[X|Y]] = E[X2] E[
E[X|Y]
2
] = Var[X] +
E[X]
2
E[E[X|Y]2
] = Var[X] +
E
E[X|Y]2
E[E[X|Y]2
].
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Today
Today, our goal will be
to finish with conditional expectations,
to have a quick look at conditional variances,
and
to discuss limits of sequences of r.v.s, and
to study famous limiting results.
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Convergence of sequences of r.v.s
LetX1, X2, . . .bei.i.d.r.v.s, that is, r.v.s that areindependent andidenticallydistributed. AssumeX1 is square-integrable, and
write:= E[X1]and2 := Var[X1].
Let Xn := 1nn
i=1 Xi. Then
E[Xn] = 1nn
i=1E[Xi] = E[X1] =, and
Var[Xn] =
1n2 Var[
ni=1 Xi] =
1n2
ni=1Var[Xi] =
1nVar[X1] =
2
n,
which converges to0asn .
Consequently, we feel intuitively that Xn X, whereX=a.s.
How to make this convergence precise?
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Convergence of sequences of r.v.s
Consider a sequence of r.v.s(Xn)and a r.v. X, defined on(,A, P).How to defineXn X(asn )?
Xna.s.
X(almost surely) P[{ |Xn() X()}] = 1.Xn
P X(in probability) For all >0,P[|Xn X| > ] 0.
Xn
Lr
X(inLr
,r >0) E[|Xn X|r
] 0.Xn
D Xin distribution (or in law) FXn(x) FX(x)for allxatwhichFX is continuous.
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Convergence of sequences of r.v.s
Example:
LetY1, Y2, . . .be i.i.d. r.v.s, with common distribution
distribution ofYi
values 0 2
probabilities 12
1
2
DefineXn=ni=1 Yi.
The distribution ofXn is
distribution ofXn
values 0 2n
probabilities 1 12n
1
2n
We feel thatXn X, whereX= 0a.s... But in what sense?Stochastic Processes (Lecture #4) p. 19/
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Convergence of sequences of r.v.s
We feel thatXn X, whereX= 0a.s...But in what sense?
In probability?
For all >0,P[|Xn X| > ] =P[Xn> ] P[Xn >0] = 12n 0,asn .
Xn P X, asn .
InL1?
E[
|Xn
X
|] = E[Xn] = 0
1
12n + 2n
12n = 1,
which does not go zero, asn . the convergence does not hold in theL1 sense.
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Convergence of sequences of r.v.s
It can be shown that:
Xn a.s. X Xn P X Xn Lr
X
Xn D X
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C f f
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Convergence of sequences of r.v.s
Some arrows can sometimes be reverted:
Xn P X there exists a subsequence(Xnk)for whichXnk
a.s. X. Xn P Xand theXns are uniformly integrable Xn L
r X. Xn D a Xn P a(a is some constant).
Definition: theXns are uniformly integrable (i)supnE[|Xn|]< .(ii)P[A]
0
supn A |
Xn()
|d
0.
Recall: ifX is integrable,P[A] 0 A |X()| d 0(hence,only thesupnbrings some extra condition in (ii) above).
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Li iti th
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Limiting theorems
Here are the two most famouslimiting resultsin probability and
statistics...
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Li iti th
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Limiting theorems
Thelaw of large numbers(LLN):
LetX1, X2, . . .be i.i.d. integrable r.v.s.
Write:= E[X1].
Then
Xn := 1
n
n
i=1
Xia.s. .
Remarks:
Basically no assumption. Interpretation for favourable/fair/defavourable games of
chance.
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Limiting theorems
An example...
LetX1, X2, . . .be i.i.d., withXi Bern(p).Then= E[Xi] =p, so that
Xn := 1
n
ni=1
Xia.s. p.
Interpretation:
The empirical proportion of successes converges to the
theoretical proportion of successes (that is, the probability of
success).
Remark: this also shows that, if(Yn)is a sequence of r.v.s with
Yn Bin(n, p), thenYn/na.s.
p.Stochastic Processes (Lecture #4) p. 25/
Limiting theorems
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Limiting theorems
... and some applets:
http://hadm.sph.sc.edu/COURSES/J716/a01/stat.html
http://www.stat.berkeley.edu/stark/Java/Html/lln.htm ...
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Limiting theorems
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Limiting theorems
Thecentral limit theorem(CLT):
LetX1, X2, . . .be i.i.d. square-integrable r.v.s.
Write:= E[X1]and2 := Var[X1].
ThenXn /
n
D Z, withZ N(0, 1).
Remarks:
Xn/n
=XnE[Xn]
Var[Xn].
It says sth about the speed of convergence in Xn a.s. . It allows for computingP[Xn B]for largen...
It is valid whatever the distribution of theXis!Stochastic Processes (Lecture #4) p. 27/
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Limiting theorems
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Limiting theorems
(B) Throw 9 dices and denote by Xthe sum of the results.Repetition of this yields i.i.d. r.v.sX1, X2, . . .
Note thatXi =Yi1+ . . . + Yi9, whereYij is the result of the jth
dice in theith roll. Then we have
= E[Xi] = E[Yi1] + . . . + E[Yi9] = 9 3.5 = 31.5
and
2 = Var[Xi] = Var[Yi1] + . . . + Var[Yi9] = 9 2.91666...= 26.25,
so that
n(Xn 31.5)
26.25
D Z, withZ N(0, 1).
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Limiting theorems
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Limiting theorems
... and some applets:
http://www.math.csusb.edu/faculty/stanton/probstat/clt.html
http://www.stat.vt.edu/sundar/java/applets/(then go to Statistical Theory Central Limit Theorem
Main page).
http://www.jcu.edu/math/isep/Quincunx/Quincunx.html http://bcs.whfreeman.com/ips4e/cat 010/applets/
CLT-Binomial.html
...
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