Extension of the concept of independence to an arbitrary class of events utilizes
the product rule.
Definition. A class of events is said to be (stochastically) independent
iff the product rule holds for every finite subclass of two or more events in the class.
A class {A,B,C}{A,B,C} is independent iff all four of the following product rules hold
P
(
A
B
)
=
P
(
A
)
P
(
B
)
P
(
A
C
)
=
P
(
A
)
P
(
C
)
P
(
B
C
)
=
P
(
B
)
P
(
C
)
P
(
A
B
C
)
=
P
(
A
)
P
(
B
)
P
(
C
)
P
(
A
B
)
=
P
(
A
)
P
(
B
)
P
(
A
C
)
=
P
(
A
)
P
(
C
)
P
(
B
C
)
=
P
(
B
)
P
(
C
)
P
(
A
B
C
)
=
P
(
A
)
P
(
B
)
P
(
C
)
(3)If any one or more of these product expressions fail, the class is not independent.
A similar situation holds for a class of four events: the product rule must hold for every pair,
for every triple, and for the whole class. Note that we say “not independent” or
“nonindependent” rather than dependent. The reason for this becomes clearer in dealing with
independent random variables.
We consider some classical exmples of nonindependent classes
- Suppose {A1,A2,A3,A4}{A1,A2,A3,A4} is a partition, with each P(Ai)=1/4P(Ai)=1/4. Let
A=A1⋁A2B=A1⋁A3C=A1⋁A4A=A1⋁A2B=A1⋁A3C=A1⋁A4
(4)P(AB)=P(A1)=1/4=P(A)P(B)andsimilarlyfortheotherpairs,butP(AB)=P(A1)=1/4=P(A)P(B)andsimilarlyfortheotherpairs,but
(5)P(ABC)=P(A1)=1/4≠P(A)P(B)P(C)P(ABC)=P(A1)=1/4≠P(A)P(B)P(C)
(6) - Consider the class {A,B,C,D}{A,B,C,D} with AD=BD=∅AD=BD=∅, C=AB⋁DC=AB⋁D,
P(A)=P(B)=1/4P(A)=P(B)=1/4, P(AB)=1/64P(AB)=1/64, and P(D)=15/64P(D)=15/64. Use of a minterm maps shows
these assignments are consistent. Elementary calculations show the product rule applies
to the class {A,B,C}{A,B,C} but no two of these three events forms an independent pair.
As noted above, the replacement rule holds for any pair of events. It is easy to show,
although somewhat cumbersome to write out, that if the rule holds for any finite number k
of events in an independent class, it holds for any k+1k+1 of them. By the principle
of mathematical induction, the rule must hold for any finite subclass. We may
extend the replacement rule as follows.
If a class is independent, we may replace any of the sets by its complement, by a
null event, or by an almost sure event, and the resulting class is also independent.
Such replacements may be made for any number of the sets in the class.
One immediate and important consequence is the following.
If {Ai:1≤i≤n}{Ai:1≤i≤n} is an independent class and the the class
{P(Ai):1≤i≤n}{P(Ai):1≤i≤n} of individual probabilities is known, then
the probability of every minterm may be calculated.
Suppose the class {A,B,C}{A,B,C} is independent with respective probabilities
P(A)=0.3P(A)=0.3, P(B)=0.6P(B)=0.6, and P(C)=0.5P(C)=0.5. Then
{Ac,Bc,Cc}{Ac,Bc,Cc} is independent and P(M0)=P(Ac)P(Bc)P(Cc)=0.14P(M0)=P(Ac)P(Bc)P(Cc)=0.14
{Ac,Bc,C}{Ac,Bc,C} is independent and P(M1)=P(Ac)P(Bc)P(C)=0.14P(M1)=P(Ac)P(Bc)P(C)=0.14
Similarly, the probabilities of the other six minterms, in order, are 0.21, 0.21, 0.06,
0.06, 0.09, and 0.09. With these minterm probabilities, the probability of any Boolean
combination of A,BA,B, and C may be calculated
In general, eight appropriate probabilities must be specified to determine the minterm
probabilities for a class of three events. In the independent case, three appropriate probabilities are sufficient.
Suppose {A,B,C}{A,B,C} is independent with P(A∪BC)=0.51P(A∪BC)=0.51, P(ACc)=0.15P(ACc)=0.15,
and P(A)=0.30P(A)=0.30. Then P(Cc)=0.15/0.3=0.5=P(C)P(Cc)=0.15/0.3=0.5=P(C) and
P
(
A
)
+
P
(
A
c
)
P
(
B
)
P
(
C
)
=
0
.
51
so
that
P
(
B
)
=
0
.
51
-
0
.
30
0
.
7
×
0
.
5
=
0
.
6
P
(
A
)
+
P
(
A
c
)
P
(
B
)
P
(
C
)
=
0
.
51
so
that
P
(
B
)
=
0
.
51
-
0
.
30
0
.
7
×
0
.
5
=
0
.
6
(7)With each of the basic probabilities determined, we may calculate the minterm probabilities, hence
the probability of any Boolean combination of the events.
Frequently we have a large enough independent class {E1,E2,⋯,En}{E1,E2,⋯,En} that
it is desirable to use MATLAB (or some other computational aid) to calculate the probabilities
of various “and” combinations (intersections) of the events or
their complements. Suppose the independent class {E1,E2,⋯,E10}{E1,E2,⋯,E10} has respective
probabilities
0
.
13
0
.
37
0
.
12
0
.
56
0
.
33
0
.
71
0
.
22
0
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43
0
.
57
0
.
31
0
.
13
0
.
37
0
.
12
0
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56
0
.
33
0
.
71
0
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22
0
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43
0
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57
0
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31
(8)It is desired to calculate (a) P(E1E2E3cE4E5cE6cE7)P(E1E2E3cE4E5cE6cE7), and (b) P(E1cE2E3cE4E5cE6cE7E8E9cE10)P(E1cE2E3cE4E5cE6cE7E8E9cE10).
We may use the MATLAB function prod and the scheme for indexing a matrix.
>> p = 0.01*[13 37 12 56 33 71 22 43 57 31];
>> q = 1-p;
>> % First case
>> e = [1 2 4 7]; % Uncomplemented positions
>> f = [3 5 6]; % Complemented positions
>> P = prod(p(e))*prod(q(f)) % p(e) probs of uncomplemented factors
P = 0.0010 % q(f) probs of complemented factors
>> % Case of uncomplemented in even positions; complemented in odd positions
>> g = find(rem(1:10,2) == 0); % The even positions
>> h = find(rem(1:10,2) ~= 0); % The odd positions
>> P = prod(p(g))*prod(q(h))
P = 0.0034
In the unit on MATLAB and Independent Classes, we extend the use of MATLAB in the calculations
for such classes.
Catalogue of Useful Matlab Files
