Let's also define an indicator random variable s for the event that the total of the two dice is seven: 1 if T( ifT()≠7 So S is equal to 1 when the sum is seven and is equal to 0 otherwise. For example, S(4,3)=1,butS(5,3) Now lets consider a couple questions about independence. First, are D1 and T inde- pendent? Intuitively, the answer would seem to be"no" since the number that comes up on the first die strongly affects the total of the two dice. But to prove this, we must find integers c1 and t, such that: Pr(D1 2)+Pr(D1=.1). Pr(T= 2) For example, we might choose T1=2 and 2=3. In this case, we have Pr(T=2|D1=3)=0 since the total can not be only 2 when one die alone is 3. On the other hand, we have (T=2).Pr(D1=3)=Pr({1,1})·Pr({(3,1),(3,2),…,(3,6)}) ≠0 So, as we suspected, these random variables are not independent. Are S and D independent? Once again, intuition suggests that the answer is"no The number on the first die ought to affect whether or not the sum is equal to seven But this time intuition turns out to be wrong! These two random variables actua Proving that two random variables are independent takes some work.(fortunately this is an uncommon task; usually independence is a modeling assumption. Only rarely do random variables unexpectedly turn out to be independent. In this case, we must show that Pr(S=rinD=T2)=Pr(S=a1). Pr(D1=x2) for all 1 E 0, 1) and all 2 E(1, 2, 3, 4, 5, 6. We can work through all these possibilities Suppose that 1=1. Then for every value of z2 we have Pr(S=1)=Pr(.6,2.),…,(6,1)=6 Pr(D1=2)=P(x21,(22,…,(x26)=6 Pr(S=1nD1=x2)=Pr(x2,7-x2)) Since 1/6. 1/6=1 /36, the independence condition is satisfied� Random Variables 5 Let’s also define an indicator random variable S for the event that the total of the two dice is seven: � 1 if T(w) = 7 S(w) = 0 if T(w) =� 7 So S is equal to 1 when the sum is seven and is equal to 0 otherwise. For example, S(4, 3) = 1, but S(5, 3) = 0. Now let’s consider a couple questions about independence. First, are D1 and T inde￾pendent? Intuitively, the answer would seem to be “no” since the number that comes up on the first die strongly affects the total of the two dice. But to prove this, we must find integers x1 and x2 such that: Pr (D1 = x1 ∩ T = x2) = Pr (D1 = x1) · Pr (T = x2) For example, we might choose x1 = 2 and x2 = 3. In this case, we have Pr (T = 2 | D1 = 3) = 0 since the total can not be only 2 when one die alone is 3. On the other hand, we have: Pr (T = 2) · Pr (D1 = 3) = Pr ({1, 1}) · Pr ({(3, 1), (3, 2), . . . , (3, 6)}) 1 6 = = 0 36 · 36 � So, as we suspected, these random variables are not independent. Are S and D1 independent? Once again, intuition suggests that the answer is “no”. The number on the first die ought to affect whether or not the sum is equal to seven. But this time intuition turns out to be wrong! These two random variables actually are independent. Proving that two random variables are independent takes some work. (Fortunately, this is an uncommon task; usually independence is a modeling assumption. Only rarely do random variables unexpectedly turn out to be independent.) In this case, we must show that Pr (S = x1 ∩ D1 = x2) = Pr (S = x1) · Pr (D1 = x2) (1) for all x1 ∈ {0, 1} and all x2 ∈ {1, 2, 3, 4, 5, 6}. We can work through all these possibilities in two batches: • Suppose that x1 = 1. Then for every value of x2 we have: 1 Pr (S = 1) = Pr ((1, 6), (2, 5), . . . , (6, 1)) = 6 1 Pr (D1 = x2) = Pr ((x2, 1), (x2, 2), . . . , (x2, 6)) = 6 1 Pr (S = 1 ∩ D1 = x2) = Pr ((x2, 7 − x2)) = 36 Since 1/6 1· /6 = 1/36, the independence condition is satisfied