By Howard Raiffa

ISBN-10: 047138349X

ISBN-13: 9780471383499

"In the sphere of statistical determination conception, Raiffa and Schlaifer have sought to boost new analytic innovations in which the fashionable conception of application and subjective likelihood can truly be utilized to the commercial research of average sampling problems."

—From the foreword to their vintage paintings *Applied Statistical determination Theory*. First released within the Nineteen Sixties via Harvard collage and MIT Press, the booklet is now provided in a brand new paperback variation from Wiley

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**Sample text**

Prove that the following properties are equivalent: (i) Γ is independent of G under P , (ii) for every probability Q on (Ω, F), equivalent to P , with able, Q(Γ) = P (Γ). dQ dP G measur- 2. Let Q be a probability on (Ω, F) which is equivalent to P and consider the following properties: is G measurable (j) dQ dP (jj) for every set Γ ∈ F independent of G under P , Q(Γ) = P (Γ). Prove that (j) implies (jj). 3. Prove that in general, (jj) does not imply (j). Hint: Let G = {∅, Ω, {a}, {a}c }, assuming that {a} ∈ F, and P ({a}) > 0.

1. Let Γ ∈ F. Prove that the following properties are equivalent: (i) Γ is independent of G under P , (ii) for every probability Q on (Ω, F), equivalent to P , with able, Q(Γ) = P (Γ). dQ dP G measur- 2. Let Q be a probability on (Ω, F) which is equivalent to P and consider the following properties: is G measurable (j) dQ dP (jj) for every set Γ ∈ F independent of G under P , Q(Γ) = P (Γ). Prove that (j) implies (jj). 3. Prove that in general, (jj) does not imply (j). Hint: Let G = {∅, Ω, {a}, {a}c }, assuming that {a} ∈ F, and P ({a}) > 0.

Furthermore, since Zα is BA -measurable, EQα (EQα (Zα | B)1A ) = αEP (EQα (Zα | B)) = αEP (Zα ) . On the other hand, we have: Q(A) = EQα (Zα 1A ) 1A P (A| B) = αEP (Zα ) . = αEP Zα Put Z = E(Zα | B), then from the above, we verify that EP (Z) = 1. s. s. We have proven that Z satisﬁes the required conditions. Now, let Q and Qα be two probabilities which satisfy (a) and (b) together with ˆ | . This implies that (c), (d) and (e). By (c) and (e), it is obvious that Q|BA = Q BA ˆ dQ ˆ on the σ-ﬁeld F.

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