Statistics - Probability and Stochastic Processes.pdf
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Pobierz
Chapter 34 - probability and Stochastic Process
Poularikas A. D. “Probability and Stochastic Processes”
The Handbook of Formulas and Tables for Signal Processing.
Ed. Alexander D. Poularikas
Boca Raton: CRC Press LLC,1999
© 1999 by CRC Press LLC
Probability and
34
Stochastic Processes
34.1 Axioms of Probability
34.2 Conditional Probabilities—Independent Events
34.3 Compound (Combined)Experiments
34.4 Random Variable
34.5 Functions of One Random Variable (r.v.)
34.6 Two Random Variables
34.7 Functions of Two Random Variables
34.8 Two Functions of Two Random Variables
34.9 Expected Value, Moments, and Characteristic
Function of Two Random Variables
34.10 Mean Square Estimation of R.V.'s
34.11 Normal Random Variables
34.12 Characteristic Functions of Two Normal Random
Variables
34.13 Price Theorem for Two R.V.'s
34.14 Sequences of Random Variables
34.15 General Concepts of Stochastic Processes
34.16 Stationary Processes
34.17 Stochastic Processes and Linear
Deterministic Systems
34.18 Correlation and Power Spectrum of Stationary
Processes
34.19 Linear Mean-Square Estimation
34.20 The Filtering Problem for Stationary Processes
34.21 Harmonic Analysis
34.22 Markoff Sequences and Processes
References
34.1 Axioms of Probability
34.1.1 Axioms of Probability
I.
P
()
0
,II.
P
()
1
, III. If
AB
0
then
PA B PA PB
+ =
)
( )
( )
[. S = a set of elements of out-
comes {
} of an experiment (certain event), 0 = empty set (impossible event). {
} = elementary event
if {
} consists of a single element. A + B =
union
of events, AB =
intersection of events
, event = a
subset of S, P(A) = probability of event A.
© 1999 by CRC Press LLC
(
34.1.2 Corollaries of Probability
PPA PA A
(), ( )
0
= −
1
( ) ,(
1
complement set of
A
)
PA B PA PB PA B PA PB PAB PA PB
(
+ ≠
)
( )
( ),
+ =
(
)
( )
( )
( )
( )
( )
Example
S = {
hh,ht,th,tt
} (tossing a coin twice),
A
= {heads at first tossing} = {
hh,h
t},
B
= {only one head came
up} = {
ht,th
},
G
= {heads came up at least once} = {
hh,ht,th
},
D
= {tails at second tossing} = {
ht, tt
}
34.2 Conditional Probabilities—Independent Events
34.2.1 Conditional Probabilities
PAM
(
)
PAM
PM
(
)
()
probability of event
probabilty of event
AM
M
conditional probaqbility of given
A
M
.
1.
PAM
(
)
0
if
AM
0
2.
PAM
(
)
PA
PM
()
()
P A
()
if
AM A A M
(
)
3.
PAM
(
)
PM
PM
()
()
1f
MA
4.
P A BM
(
)
P AM P BM
(
)
(
)
if
AB
0
Example
Pf
16
/ ,
i
1 6
,
L
.
M
{
odd}
{ , , },
f f f A f AM f PM
135
{ },
1
{ }, ( )
1
36
/ , (
PAM
)
16
/
,then
(|
even)
PAM PM
( )/( ) /
13
34.2.2 Total Probability
() ( )( )
1
1
+ +
L
PBA PA
n
( )( )
n
arbitrary event,
AA
i j
0
i j
,, ,
LL
n A
+ +
1
A
n
= S = certain event.
34.2.3 Baye's Theorem
PA B
(
)
PBA PA
PBA PA
(
i
) ( )
i
i
(
) ( )
+ +
L
PBA PA
(
) ( )
1
1
n
n
AA
i j
0
,
i j
1 2
, ,
L
n A A
,
+ +
1
2
= =
L
A S
n
certain event, B=arbitrary
34.2.4 Independent Events
() ()()
implies A and B are independent events.
34.2.5 Properties
1.
PAB P A
( )
( )
© 1999 by CRC Press LLC
( )
i
Pf
1
PB PBA PA
≠ =
1 2
≠ =
PAB PAPB
2.
3.
PBA PB
() ()
L L= =
PA B PA PB PAPB
(
12
n
)
( )
1
( ),
n
A
independent events
4.
+ = + −
AB AB PAB
(
) () () ()()
5.
If A nad B are independent. Overbar means complement set.
6. If
= +
(
), (
+ = −
)
1
PABP AB PAPB
(
),
( )
( ) ( )
AA A
i
, ,
23
A
1
A
23
PAAA PA PA PA PA PAA
are independent and is independent of
then
(
) ()( )() ()( )
123
1
2
3
1
23
. Also
PA A A
[(
] ( ) ( ) (
PAA PAA PAAA PA
) ()
12 3
12
13
123
1
[( ) ( ) ( )( ] ( )(
PA PA PA PA
2
3
2
3
PA PA A
1
2
3
)
34.2.6
PA B C PA PB PC PAB PAC PBC PABC
(
+ +
)
( )
( )
( )
( )
( )
( )
(
)
34.3 Compound (Combined, Experiments
34.3.1
S=S
1
S
2
=
Cartesian product
Example
S
1
={1,2,3},
S
2
={heads, tails},
S
=
S
1
S
2
= {(1 heads),(1 tails),(2 heads),(2 tails),(3 heads),(3 tails)}
34.3.2
If
ASA S
1
1
,
2
2
then
AA ASAS
1
2
(
1
2
)(
2
1
)
(see
Figure 34.1
)
S
2
A
1
S
2
A
2
A
2
A
1
A
2
S
1
A
1
S
1
FIGURE 34.1
34.3.3 Probability in Compound Experiments
PA PA S
() (
1
1
2
)
where
1
A
1
and
2
A
2
34.3.4 Independent Compound Experiments
PA A P A P A
(
1
2
)
1
( ) ( )
1 2
2
Example
P
(heads) =
p
,
P
(tails) =
q
,
p+ q=
1,
E
= experiment tossing the coin twice =
E
1
E
2
(
E
1
= experiment
of first tossing),
E
2
= experiment of second tossing),
S
1
={
h,t
}
P
1
{
h
}=
p
P
2
{
t
}=
q
,
E
2
=
E
1
= experiment
of the second tossing,
S S S
= × =
[ , , , }, { } {} {}
hh ht th tt P hh P h P h
p
2
assume independence,
1
2
1
2
P ht
{} , {} , [,}
pq P th qp P t t q
2
. For heads at the first tossing,
H h t
1
{,}
PH Phh
or
() {}
1
P ht
{}
p
2
pq p
© 1999 by CRC Press LLC
PAA A PA PA
i
34.3.5 Sum of more Spaces
SS SS
= +
1
2
,
1
outcomes of experiment and = outcomes of experiment
E
1
S
2
E
2
S
.
space of
the experiment
EE E
= +
1
2
AA A
;
= +
1
where and are events of and
A
1
A
2
E
1
E
2
AS
:
,
2
1
1
AS
2
;
PA PA PA
() ( ) ( )
.
2
1
2
34.3.6 Bernoulli Trials
P
()
probability of event A, E
E
E
...
E = perform experiment n times = combined experiment.
pk
()
n
k
pq
knk
probability that events occurs k times in any order
PA pPA qp q
() , () ,
+ =
1
n
Example
5
522
!
1
6
2
5
6
52
A fair die was rolled 5 times.
p
5
()
2
= probability that "four" will come up
(
)! !
twice.
Example
Two fair dice are tossed 10 times. What is the probability that the dice total seven points exactly four
times?
Solution
1
6
2
1
6
5
6
Event
B
{( , ),( , ),( , ),( , ),( , ),( , )}, ( )
16 2 5 34 43 52 61
P B
= ⋅
6
= =
p P
, ( )
81
p
. The
10
4
1
6
4
5
6
6
probability of B occuring four times and six times is
B
0 0543
.
.
34.3.7
Pk k k
1
≤ ≤
2
}
probability of success of A (event) will lie between and
k
1
k
2
k
2
k
n
k
Pk k k
{
≤ ≤
}
p k
( )
pq
knk
1
2
n
kk
1
kk
1
k
2
n
k
1
k
1. Approximate value:
pq
k n k
e
(
k np
) /
2
npq
,
npq
1
2
npq
kk
1
kk
1
34.3.8 DeMoivre-Laplace Theorem
pk
()
n
k
pq
k n k
1
e
(
k np
) /
2
npq
,
npq
1
n
2
npq
34.3.9 Poisson Theorem
n
kn k
pq
!
()
!
np
k
k
a
k
k
knk
e
p
e
a
!
,
→ ∞ →
n
,
p
0
,
np a
!(
)!
© 1999 by CRC Press LLC
= − =
{
2
2
− −
− −
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