By Howard Raiffa

ISBN-10: 047138349X

ISBN-13: 9780471383499

"In the sphere of statistical choice conception, Raiffa and Schlaifer have sought to advance new analytic thoughts in which the fashionable concept of application and subjective chance can truly be utilized to the industrial research of normal sampling problems."

—From the foreword to their vintage paintings *Applied Statistical selection Theory*. First released within the Sixties via Harvard college and MIT Press, the publication is now provided in a brand new paperback version from Wiley

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**Sample text**

I. for i ~ j. then 11 P(b) = P(b & a l ) + ... + P(b & a,). for any proposition b. Proof: h entails (b & (1 , ) v ... v (b & an) v [b & -(a, v ... va,)}. Furthermore, all the disjuncts on the right-hand side arc mutually exclusive. Let a = (/1 V . . v a". Hence by (10) we have that P(h) = P(b & a l ) + ... + P(b & a,) + P(h & -a). But P(h & -0) $ P(-a). by (8), and P(-a) = I - P(a) = 1 - 1 = 0. Hence P(b & -a) = 0 and (11) follows. Coro/fan) I. Ifa l v ... I entails -aJ. j, then P(b) = 'LP(b & (Ii)' ~ Corollary 2.

Also if a entails -- b so P(a v - h) = P(a) + P(-b). But by (5) P(-b) = I - P(h) , whence P(a) = P(h). ¢? b then a We can now prove the important property of probability functions that they respect the entailment relation; to be precise, the probability of any consequence of a is at least as great as that of a itself: (8) If a entails b then pro) :s; P(b). Proof: If 0 entails b then [a v (h & -a)] ¢ ? b. Hence by (7) P(b) = pra v (b & -a)}. But a entails -(b & -a) and so pra v (b & -a)] = pray + P(b &-a).

N(n - l)(n - 2) ... 1, and O! is set equal to 1). By the independence and constant probability assumptions, the probability of each conjunct in the sum is p'"(I - pr ',since P(Xi = 0) = 1 - p. ~II) is said to possess the binomial distrihution. The mean of ~11) is np, as can be easily seen from the facts that 41 THE PROBABILITY CALCULUS and that E(X) = P . I + (1 - p) . 0 = p. The squared standard deviation, or variance of Yin)' is E(Y(n) - npj2 = E(Y(n/) + E(npj2- E(2~n)np) = E(~n/) + (npj2- 2npE(Y(n) = E(~n/) - (npj2.

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