Libro Probabilidad

Continuous Distributions
MomentGenerating Function etθ2 − etθ1 t (θ2 − θ1 )

Distribution

Probability Function f (y) = 1 ; θ ≤ y ≤ θ2 θ2 − θ1 1

Mean θ1 + θ2 2

Variance (θ2 − θ1 )2 12

Uniform

Normal

f (y) =

1 1 (y − µ)2 √ exp − 2σ 2 σ 2π −∞ < y < +∞

µ

σ2

exp µt +

t 2σ 2 2

Exponential

f (y) =

1 −y/β e ; β>0 β 0 0 and P(B) > 0. Are A and B independent?Prove your answer. Suppose that A and B are mutually exclusive events, with P(A) > 0 and P(B) < 1. Are A and B independent? Prove your answer. Suppose that A ⊂ B and that P(A) > 0 and P(B) > 0. Show that P(B|A) = 1 and P(A|B) = P(A)/P(B). If A and B are mutually exclusive events and P(B) > 0, show that P(A) . P(A|A ∪ B) = P(A) + P(B)

2.8 Two Laws of Probability
The following two laws give theprobabilities of unions and intersections of events. As such, they play an important role in the event-composition approach to the solution of probability problems. THEOREM 2.5 The Multiplicative Law of Probability The probability of the intersection of two events A and B is P(A ∩ B) = P(A)P(B|A) = P(B)P(A|B). If A and B are independent, then P(A ∩ B) = P(A)P(B). Proof The multiplicative lawfollows directly from Definition 2.9, the definition of conditional probability. Notice that the multiplicative law can be extended to find the probability of the intersection of any number of events. Thus, twice applying Theorem 2.5, we obtain P(A ∩ B ∩ C) = P[(A ∩ B) ∩ C] = P(A ∩ B)P(C|A ∩ B) = P(A)P(B|A)P(C|A ∩ B). The probability of the intersection of any number of, say, k events can be obtained inthe same manner: P(A1 ∩ A2 ∩ A3 ∩ · · · ∩ Ak ) = P(A1 )P(A2 |A1 )P(A3 |A1 ∩ A2 ) · · · P(Ak |A1 ∩ A2 ∩ · · · ∩ Ak−1 ). The additive law of probability gives the probability of the union of two events.

58

Chapter 2

Probability

THEOREM 2.6

The Additive Law of Probability The probability of the union of two events A and B is P(A ∪ B) = P(A) + P(B) − P(A ∩ B). If A and B are mutuallyexclusive events, P(A ∩ B) = 0 and P(A ∪ B) = P(A) + P(B).

Proof

The proof of the additive law can be followed by inspecting the Venn diagram in Figure 2.10. Notice that A ∪ B = A ∪ (A ∩ B), where A and (A ∩ B) are mutually exclusive events. Further, B = (A ∩ B) ∪ (A ∩ B), where (A ∩ B) and (A ∩ B) are mutually exclusive events. Then, by Axiom 3, P(A ∪ B) = P(A) + P(A ∩ B) and P(B) = P(A ∩ B) +P(A ∩ B). The equality given on the right implies that P( A ∩ B) = P(B) − P(A ∩ B). Substituting this expression for P(A ∩ B) into the expression for P(A ∪ B) given in the left-hand equation of the preceding pair, we obtain the desired result: P(A ∪ B) = P(A) + P(B) − P(A ∩ B). The probability of the union of three events can be obtained by making use of Theorem 2.6. Observe that P(A ∪ B ∪ C) =P[A ∪ (B ∪ C)] = P(A) + P(B ∪ C) − P[A ∩ (B ∪ C)] = P(A) + P(B) + P(C) − P(B ∩ C) − P[(A ∩ B) ∪ (A ∩ C)] = P(A) + P(B) + P(C) − P(B ∩ C) − P(A ∩ B) − P(A ∩ C) + P(A ∩ B ∩ C) because (A ∩ B) ∩ (A ∩ C) = A ∩ B ∩ C. Another useful result expressing the relationship between the probability of an event and its complement is immediately available from the axioms of probability.

F I G U R E 2.10 Venndiagram for the union of A and B

A

B

Exercises

59

THEOREM 2.7

If A is an event, then P(A) = 1 − P(A). Observe that S = A ∪ A. Because A and A are mutually exclusive events, it follows that P(S) = P(A) + P(A). Therefore, P(A) + P(A) = 1 and the result follows. As we will see in Section 2.9, it is sometimes easier to calculate P(A) than to calculate P(A). In such cases, it iseasier to find P(A) by the relationship P(A) = 1 − P(A) than to find P(A) directly.

Proof

Exercises
2.84
If A1 , A2 , and A3 are three events and P(A1 ∩ A2 ) = P(A1 ∩ A3 ) = 0 but P(A2 ∩ A3 ) = 0, show that P(at least one Ai ) = P(A1 ) + P(A2 ) + P(A3 ) − 2P(A1 ∩ A2 ).

2.85 2.86

If A and B are independent events, show that A and B are also independent. Are A and B independent? Suppose...