By Rueda R.
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Put up yr be aware: First released January 1st 1988
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Extra info for A Bayesian Alternative to Parametric Hypothesis Testing
Let f be a measurable function such that A f dµ is ﬁnite. Then ∞ f dµ = A f dµ. i=1 Ai To justify this statement we can ﬁrst consider f to be a non-negative simple function. Then the σ-additivity follows from the fact that in an inﬁnite series with non-negative terms the terms can be re-arranged. For an arbitrary non-negative measurable f we use the deﬁnition of the integral as a limit of integrals of simple functions. 7. If f is a non-negative function, the σ-additivity of the integral implies that the function η(A) = A f dµ is itself a measure.
Let there be two sequences fn (1) (1) (2) (2) fn+1 ≥ fn and fn+1 ≥ fn for all n, and Ω fn dµ from the (2) and fn such that lim fn(1) (ω) = lim fn(2) (ω) = f (ω) for every ω. 5 that for any k, (1) Ω fk dµ ≤ lim n→∞ fn(2) dµ, Ω and therefore, fn(1) dµ ≤ lim lim n→∞ Ω n→∞ fn(2) dµ. Ω We obtain fn(1) dµ ≥ lim lim n→∞ (1) by interchanging fn Ω n→∞ fn(2) dµ Ω (2) and fn . Therefore, fn(1) dµ = lim lim n→∞ Ω n→∞ fn(2) dµ. 6. Let f be a non-negative measurable function and fn a sequence of non-negative simple functions which converge monotonically to f from below.
Let A1 × B1 , A2 × B2 , ... be a sequence of non-intersecting rectan∞ gles such that A × B = n=1 An × Bn . Consider the sequence of functions n fn (ω1 ) = i=1 χAi (ω1 )µ2 (Bi ), where χAi is the indicator function of the set Ai . Similarly, let f (ω1 ) = χA (ω1 )µ(B). Note that fn ≤ µ2 (B) for all n and limn→∞ fn (ω1 ) = f (ω1 ). Therefore, the Lebesgue Dominated Convergence Theorem applies. We have n m(Ai × Bi ) = lim lim n→∞ n i=1 n→∞ µ1 (Ai )µ2 (Bi ) = lim i=1 n→∞ fn (ω1 )dµ1 (ω1 ) Ω1 52 3 Lebesgue Integral and Mathematical Expectation f (ω1 )dµ1 (ω1 ) = µ1 (A)µ2 (B) = m(A × B).
A Bayesian Alternative to Parametric Hypothesis Testing by Rueda R.