Random variables for all i in the codomain of Ri and r2 in the codomain of R2. In words, the probability that R, takes on a particular value is unaffected by the value of r As an example are C and M independent? Intuitively the answer should be"no The number of heads, C, completely determines whether all three coins match; that is whether M= l. But to verify this intuition we must find some 21, 2 E R such that: Pr(C=x1∩M=x2)≠Pr(C=x1)·Pr(M=x2) One appropriate choice of values is T1=2 and 2=1. In that case, we have Pr(c=2nM=1)=0 but Pr(C=2). Pr(M=1) ≠0 The notion of independence generalizes to a set of random variables as follows. Ran dom variables R1, R2, .. Rn are mutually independent if Pr(R1=x1∩R2=x2∩…∩Rn1=xn) Pr(R1=a1). Pr(R2=T2). Pr(Rn=an) for all x1,…, Inin the codomains of R1,…,Rn A consequence of this definition of mutual independence is that the probability of ssignment to a subset of the variables is equal to the product of the probabilites of the individual assignments. Thus, for example, if R1, R2,..., R1oo are mutually independent random variables with codomain n, then it follows that Pr(R1=9∩R7=84∩R23=13)=Pr(R1=9)·Pr(R7=84)Pr(R23=13) (This follows by summing over all possible values of the other random variables; we omit the details. 1.5 An Example with dice Suppose that we roll two fair, independent dice. The sample space for this experiment consists of all pairs(r1, r2)where r1, r2 E(1,2, 3, 4, 5, 6. Thus, for example, the outcome (3, 5)corresponds to rolling a 3 on the first die and a 5 on the second. The probability of each outcome in the sample space is 1 /6 1 6=1 36 since the dice are fair and indepen We can regard the numbers that come up on the individual dice as random variables D1 and D2. So D1( 3, 5)=3 and D2(3, 5)=5. Then the expression D1+ D2 random variable; lets call it T for"total". More precisely, weve defined T(w)=D1(w)+ D(w) for every outcome w Thus, T(3, 5)=D1( 3, 5)+ D2 (3, 5)=3+5=8. In general, any function of random variables is itself a random variable. For example,VDi+cos(D2)is a strange, but well defined random variable� 4 Random Variables for all x1 in the codomain of R1 and x2 in the codomain of R2. In words, the probability that R1 takes on a particular value is unaffected by the value of R2. As an example, are C and M independent? Intuitively, the answer should be “no”. The number of heads, C, completely determines whether all three coins match; that is, whether M = 1. But to verify this intuition we must find some x1, x2 ∈ R such that: Pr (C = x1 ∩ M = x2) = Pr (C = x1) · Pr (M = x2) One appropriate choice of values is x1 = 2 and x2 = 1. In that case, we have: 3 1 Pr (C = 2 ∩ M = 1) = 0 but Pr (C = 2) · Pr (M = 1) = = 0 8 · 4 � The notion of independence generalizes to a set of random variables as follows. Ran￾dom variables R1, R2, . . . , Rn are mutually independent if Pr (R1 = x1 ∩ R2 = x2 ∩ · · · ∩ Rn = xn) = Pr (R1 = x1) · Pr (R2 = x2)· · ·Pr (Rn = xn) for all x1, . . . , xn in the codomains of R1, . . . , Rn. A consequence of this definition of mutual independence is that the probability of an assignment to a subset of the variables is equal to the product of the probabilites of the individual assignments. Thus, for example, if R1, R2, . . . , R100 are mutually independent random variables with codomain N, then it follows that: Pr (R1 = 9 ∩ R7 = 84 ∩ R23 = 13) = Pr (R1 = 9) · Pr (R7 = 84) · Pr (R23 = 13) (This follows by summing over all possible values of the other random variables; we omit the details.) 1.5 An Example with Dice Suppose that we roll two fair, independent dice. The sample space for this experiment consists of all pairs (r1, r2) where r1, r2 ∈ {1, 2, 3, 4, 5, 6}. Thus, for example, the outcome (3, 5) corresponds to rolling a 3 on the first die and a 5 on the second. The probability of each outcome in the sample space is 1/6 1· /6 = 1/36 since the dice are fair and indepen￾dent. We can regard the numbers that come up on the individual dice as random variables D1 and D2. So D1(3, 5) = 3 and D2(3, 5) = 5. Then the expression D1 + D2 is another random variable; let’s call it T for “total”. More precisely, we’ve defined: T(w) = D1(w) + D2(w) for every outcome w Thus, T(3, 5) = D1(3, 5) + D2(3, 5) = 3 + 5 = 8. In general, any function of random variables is itself a random variable. For example, √D1 + cos(D2) is a strange, but well￾defined random variable