Stat 310A/Math 230A Theory of Probability Practice Final Solutions
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چکیده
Solution : Assume, without loss of generality |x2 − x1| ≥ 2−n+1. Then there exists an integer k ∈ {1, . . . , 2 − 1}, such that x1 < k · 2−n < (k + 1)2−n < x2. Of course PX((x1, x2)) ≥ P(k · 2−n ≤ X(ω) ≤ (k + 1)2−n) . (4) The integer k admits the unique binary expansion k = ∑n i=1 ki2 n−i. Then P(k · 2−n ≤ X(ω) ≤ (k + 1)2−n) = P(Cn,(k1...,kn)) = p 1(1− p)0 , (5) with n0(k) and n1(k) the number of zeros and ones in (k1, . . . , kn). For p ∈ (0, 1) the above probability is strictly positive.
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Stat 310A/Math 230A Theory of Probability Midterm Solutions
with a1, a2 > 0. (Indeed any rank 2 matrix can be written as RA with A as above, and R orthogonal. Hence any affine transformation can be written as h ◦ f with f as claimed and g a rigid motion.) Hence, by Caratheodory uniqueness theorem it is sufficient to show that λ2(ASu(α1, α2)) = |det(A)|λ2(Su(α1, α2)) for all u = (u1, u2) ∈ R, αi ∈ R+ (3) Su(α, β) ≡ { x = (x1, x2) ∈ R : x1 ∈ (u1, u1 + α1]...
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Solution : Throughout the solution we will use the fact that λ2 = λ1 × λ1 whence we obtain the action of λ2 on rectangles: λ2(A1 × A2) = λ1(A1)λ1(A2). Also, for J1, J2 ⊆ R two intervals, let TJ1,J2 be any triangle with two sides equal to J1 (parallel to the first axis) and J2 (equal to the second axis). from the addittivity of λ2 it follows immediately that λ2(TJ1,J2) = |J1| · |J2|/2. (We use h...
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[lo] H. Cramer and M. Leadbetter, “The moments of the number of [13] L. Ehrman, “Analysis of a zero-crossing frequency discriminator crossings of a level by a stationary normal process,” Ann. Math. with random inputs,” IEEE Trans. Aerospace and Navigational Stat., vol. 36, pp. 165661663, 1965. Electronics, vol. ANE-12, pp. 113-119, June 1965. [II] M. Loeve, Probability Theory. New York: Van Nos...
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