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kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

kniga 7 - Probability and Statistics 1 - Sheynin, Oscar

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P( ≤ ≤ |, 1 , 2 , …, s ) (57)<strong>and</strong> assuming that the prior distribution of the parameters has density (, 1 , 2 , …, s ), wewill obtain the following expression for the unconditional probability (56):P( ≤ ≤ ) = … P( ≤ ≤ |, 1, …, s )(, 1 , …, s )dd 1 …d s . (58)A particular case of such rules, when the conditional probability (57) remains constant atall possible values of , 1 , …, s , is especially important for practice. If this conditionalprobability is constant <strong>and</strong> equals ", then, on the strength of (59),P( ≤ ≤ ) = … "(, 1, …, s )d d 1 …d s = ".This means that the unconditional probability (56) does not depend on the unconditionaldistribution of the parameters 10 .We have already indicated in §1 that the very hypothesis on the existence of a priordistribution of the parameters is not always sensible. However, if the conditional probability(57) does not depend on the values of the parameters <strong>and</strong> is invariably equal to one <strong>and</strong> thesame number " then it is natural to consider that the unconditional probability (56) exists <strong>and</strong>is equal to " even in those cases in which the hypothesis on the existence of a priordistribution of the parameters is not admitted 11 .If the conditional probability (57) is equal to " for all the possible values of the parameters(so that, consequently, the same is true with respect to the unconditional probability (56) forany form of the prior distribution of the parameters), we shall say, following Fisher [1; 2],that our rule has a certain confidence probability equal to ". It is easy to see that for Problem1 the rule that recommends to assume that a is situated within the boundariesa ≤ a ≤ a (59)wherea = x + c/hn, a = x + c/hn (60)has a certain confidence probabilityc" = (1/) ′′c′exp (– 2 ) d. (61)Indeed, for any a <strong>and</strong> h,P(a ≤ a ≤ a|a; h) = P( x + c/hn ≤ a ≤ x + c/hn|a; h) =−−c′′cP(a – c/hn ≤ x ≤ a – c/hn|a; h) = (1/) ′c(1/) ′′c′exp (– 2 ) d.exp (– 2 ) d =For example, if c = – 2 <strong>and</strong> c = 2,

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