4.1 Matlab and independent classes  (Page 4/4)

Verification of this far reaching result rests on the minterm expansion and two elementary facts about the disjoint subclasses of an independent class. We state these facts andconsider in each case an example which exhibits the essential structure. Formulation of the general result, in each case, is simply a matter of careful use of notation.

1. A class each of whose members is a minterm formed by members of a distinct subclass of an independent class is itself an independent class.

   Consider the independent class $\{A_1,A_2,A_3,B_1,B_2,B_3,B_4\}$ , with respective probabilities 0.4, 0.7, 0.3, 0.5, 0.8, 0.3, 0.6. Consider M ₃ , minterm three for the class $\{A_1,A_2,A_3\}$ , and N ₅ , minterm five for the class $\{B_1,B_2,B_3,B_4\}$ . Then

   $$P\left(M_3\right)=P\left(A_1^c{A}_2{A}_3\right)=0.6\cdot 0.7\cdot 0.3=0.126\text{and}P\left(N_5\right)=P\left(B_1^c{B}_2{B}_3^c{B}_4\right)=0.5\cdot 0.8\cdot 0.7\cdot 0.6=0.168$$

   Also

   $$\begin{array}{c}P\left(M_3{N}_5\right)=P\left(A_1^c{A}_2{A}_3{B}_1^c{B}_2{B}_3^c{B}_4\right)=0.6\cdot 0.7\cdot 0.3\cdot 0.5\cdot 0.8\cdot 0.7\cdot 0.6\\ =(0.6\cdot 0.7\cdot 0.3)\cdot (0.5\cdot 0.8\cdot 0.7\cdot 0.6)=P\left(M_3\right)P\left(N_5\right)=0.0212\end{array}$$

   The product rule shows the desired independence.

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   Again, it should be apparent that the result holds for any number of A _i and B _j ; and it can be extended to any number of distinct subclasses of an independent class.

2. Suppose each member of a class can be expressed as a disjoint union. If each auxiliary class formed by taking one member from each of the disjoint unions isan independent class, then the original class is independent.

   Suppose $A=A_1\bigvee A_2\bigvee A_3$ and $B=B_1\bigvee B_2$ , with $\{A_i,B_j\}$ independent for each pair $i,j$ . Suppose

   $$P\left(A_1\right)=0.3,P\left(A_2\right)=0.4,P\left(A_3\right)=0.1,P\left(B_1\right)=0.2,P\left(B_2\right)=0.5$$

   We wish to show that the pair $\{A,B\}$ is independent; i.e., the product rule $P\left(AB\right)=P\left(A\right)P\left(B\right)$ holds.

   COMPUTATION

   $$P\left(A\right)=P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)=0.3+0.4+0.1=0.8\text{and}P\left(B\right)=P\left(B_1\right)+P\left(B_2\right)=0.2+0.5=0.7$$

   Now

   $$AB=(A_1\bigvee A_2\bigvee A_3)(B_1\bigvee B_2)=A_1{B}_1\bigvee A_1{B}_2\bigvee A_2{B}_1\bigvee A_2{B}_2\bigvee A_3{B}_1\bigvee A_3{B}_2$$

   By additivity and pairwise independence, we have

   $$\begin{array}{c}P\left(AB\right)=P\left(A_1\right)P\left(B_1\right)+P\left(A_1\right)P\left(B_2\right)+P\left(A_2\right)P\left(B_1\right)+P\left(A_2\right)P\left(B_2\right)+P\left(A_3\right)P\left(B_1\right)+P\left(A_3\right)P\left(B_2\right)\\ =0.3\cdot 0.2+0.3\cdot 0.5+0.4\cdot 0.2+0.4\cdot 0.5+0.1\cdot 0.2+0.1\cdot 0.5=0.56=P\left(A\right)P\left(B\right)\end{array}$$

   The product rule can also be established algebraically from the expression for $P\left(AB\right)$ , as follows:

   $$\begin{array}{c}P\left(AB\right)=P\left(A_1\right)[P\left(B_1\right)+P\left(B_2\right)]+P\left(A_2\right)[P\left(B_1\right)+P\left(B_2\right)]+P\left(A_3\right)[P\left(B_1\right)+P\left(B_2\right)]\\ =[P\left(A_1\right)+P\left(A_2\right)+P\left(A_3\right)][P\left(B_1\right)+P\left(B_2\right)]=P\left(A\right)P\left(B\right)\end{array}$$

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   It should be clear that the pattern just illustrated can be extended to the general case. If

   $$A=\underset{i=1}{\overset{n}{\bigvee }}{A}_i\text{and}B=\underset{j=1}{\overset{m}{\bigvee }}{B}_j,\text{with}\text{each}\text{pair}\{A_i,B_j\}\text{independent}$$

   then the pair $\{A,B\}$ is independent. Also, we may extend this rule to the triple $\{A,B,C\}$

   $$A=\underset{i=1}{\overset{n}{\bigvee }}{A}_i,B=\underset{j=1}{\overset{m}{\bigvee }}{B}_j,\text{and}C=\underset{k=1}{\overset{r}{\bigvee }}{C}_k,\text{with}\text{each}\text{class}\{A_i,B_j,C_k\}\text{independent}$$

   and similarly for any finite number of such combinations, so that the second proposition holds.

3. Begin with an independent class $\mathcal{E}$ of n events. Select m distinct subclasses and form Boolean combinations for each of these. Use of the minterm expansion for each of these Boolean combinations and the two propositions just illustrated shows that the class of Boolean combinations is independent

   To illustrate, we return to [link] , which involves an independent class of ten events.

   A hybrid approach

   Consider again the independent class $\{E_1,E_2,\cdots ,E_{10}\}$ with respective probabilities $\{0.130.370.120.560.330.710.220.430.570.31\}$ . We wish to calculate

   $$P\left(F\right)=P\left(E_1,\cup ,E_3,(E_4\cup E_7^c),\cup ,E_2,(E_5^c\cup E_6{E}_8),\cup ,E_9,E_{10}^c\right)$$

   In the previous solution, we use minprob to calculate the $2^{10}=1024$ minterms for all ten of the E _i and determine the minterm vector for F . As we note in the alternate expansion of F ,

   $$F=A\cup B\cup C,\text{where}A=E_1\cup E_3(E_4\cup E_7^c)B=E_2(E_5^c\cup E_6{E}_8)C=E_9{E}_{10}^c$$

   We may calculate directly $P\left(C\right)=0.57\cdot 0.69=0.3933$ . Now A is a Boolean combination of $\{E_1,E_3,E_4,E_7\}$ and B is a combination of $\{E_2,E_5,E_6,E_8\}$ . By the result on independence of Boolean combinations, the class $\{A,B,C\}$ is independent. We use the m-procedures to calculate $P\left(A\right)$ and $P\left(B\right)$ . Then we deal with the independent class $\{A,B,C\}$ to obtain the probability of F .

   >>p = 0.01*[13 37 12 56 33 71 22 43 57 31];>>pa = p([1 3 4 7]); % Selection of probabilities for A>>pb = p([2 5 6 8]); % Selection of probabilities for B>>pma = minprob(pa); % Minterm probabilities for calculating P(A)>>pmb = minprob(pb); % Minterm probabilities for calculating P(B)>>minvec4;>>a = A|(B&(C|Dc)); % A corresponds to E1, B to E3, C to E4, D to E7>>PA = a*pma' PA = 0.2243>>b = A&(Bc|(C&D)); % A corresponds to E2, B to E5, C to E6, D to E8>>PB = b*pmb' PB = 0.2852>>PC = p(9)*(1 - p(10)) PC = 0.3933>>pm = minprob([PA PB PC]);>>minvec3 % The problem becomes a three variable problem>>F = A|B|C; % with {A,B,C} an independent class>>PF = F*pm' PF = 0.6636 % Agrees with the result of [link] Got questions? Get instant answers now!