Independence of events (Page 3/3)

Applied probability Page 3 / 3

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)

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 withindependent random variables.

We consider some classical exmples of nonindependent classes

Some nonindependent classes

Suppose ${A_{1}, A_{2}, A_{3}, A_{4}}$ is a partition, with each $P (A_{i}) = 1 / 4$ . Let
$A = A_{1} ⋁ A_{2} B = A_{1} ⋁ A_{3} C = A_{1} ⋁ A_{4}$
Then the class ${A, B, C}$ has $P (A) = P (B) = P (C) = 1 / 2$ and is pairwise independent, but not independent, since
$P (A B) = P (A_{1}) = 1 / 4 = P (A) P (B) and similarly for the other pairs, but$

$P (A B C) = P (A_{1}) = 1 / 4 \neq P (A) P (B) P (C)$
Consider the class ${A, B, C, D}$ with $A D = B D = \emptyset$ , $C = A B ⋁ D$ , $P (A) = P (B) = 1 / 4$ , $P (A B) = 1 / 64$ , and $P (D) = 15 / 64$ . Use of a minterm maps shows these assignments are consistent. Elementary calculations show the product rule appliesto the class ${A, B, C}$ but no two of these three events forms an independent pair.

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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 $k + 1$ of them. By the principle of mathematical induction, the rule must hold for any finite subclass. We mayextend the replacement rule as follows.

General replacement rule

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.

Minterm probabilities

{A_{i} : 1 \leq i \leq n}

is an independent class and the the class

{P (A_{i}) : 1 \leq i \leq n}

of individual probabilities is known, then the probability of every minterm may be calculated.

Minterm probabilities for an independent class

Suppose the class ${A, B, C}$ is independent with respective probabilities $P (A) = 0.3$ , $P (B) = 0.6$ , and $P (C) = 0.5$ . Then

${A^{c}, B^{c}, C^{c}}$ is independent and $P (M_{0}) = P (A^{c}) P (B^{c}) P (C^{c}) = 0.14$

${A^{c}, B^{c}, C}$ is independent and $P (M_{1}) = P (A^{c}) P (B^{c}) P (C) = 0.14$

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 Booleancombination of $A, B$ , and C may be calculated

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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.

Three probabilities yield the minterm probabilities

Suppose ${A, B, C}$ is independent with $P (A \cup B C) = 0.51$ , $P (A C^{c}) = 0.15$ , and $P (A) = 0.30$ . Then $P (C^{c}) = 0.15 / 0.3 = 0.5 = P (C)$ and

P (A) + P (A^{c}) P (B) P (C) = 0.51 so that P (B) = \frac{0.51 - 0.30}{0.7 \times 0.5} = 0.6

With each of the basic probabilities determined, we may calculate the minterm probabilities, hence the probability of any Boolean combination of the events.

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Matlab and the product rule

Frequently we have a large enough independent class ${E_{1}, E_{2}, \dots, E_{n}}$ that it is desirable to use MATLAB (or some other computational aid) to calculate the probabilitiesof various “and” combinations (intersections) of the events or their complements. Suppose the independent class ${E_{1}, E_{2}, \dots, E_{10}}$ has respective probabilities

0.13 0.37 0.12 0.56 0.33 0.71 0.22 0.43 0.57 0.31

It is desired to calculate (a) $P (E_{1} E_{2} E_{3}^{c} E_{4} E_{5}^{c} E_{6}^{c} E_{7})$ , and (b) $P (E_{1}^{c} E_{2} E_{3}^{c} E_{4} E_{5}^{c} E_{6}^{c} E_{7} E_{8} E_{9}^{c} E_{10})$ .

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.

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Source: OpenStax, Applied probability. OpenStax CNX. Aug 31, 2009 Download for free at http://cnx.org/content/col10708/1.6

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