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This implies that U(x) = 0 ae Lebesgue measure on 1,∞) and Z 1 0 U(x)xn−1dx = 0 Consider T = h(X(n)) To have E(TU) = 0, we must have Z 1 0 h(x)U(x)xn−1dx = 0 Thus, we may consider the following function h(x) = c 0 ≤ x ≤ 1 bx x > 1, where c and b are some constants From the previous discussion, Eh(X(n))U(X(n)) = 0, θ ≥ 1 2. Solve your math problems using our free math solver with stepbystep solutions Our math solver supports basic math, prealgebra, algebra, trigonometry, calculus and more. U I W i ́A ړI ͂Ȃ A g ̉ ɂ 鉽 \ B t Ɉ˗ č ꍇ ́A Ԃ̘g g ܂߂āA ݂ \ ̒ ŋ@ \ ̂ v Z X ˁB ƁA Ă 邱 Ƃ́A N C A g ͌ ĉ ߂邩 m 邱 Ƃ͂Ȃ Ă ƁB ނ ͉ K v Ȃ̂ ŏ Ō ܂ł킩 Ȃ B v ̎ ق A ͔ނ Ɏ M ^ Ȃ Ƃ Ȃ B N C A g ̌ Ƃ͕ Ȃ Ƃ ƂˁB ŁA A ꂪ C ɓ Ȃ ̂ł ΁A w ̍ ȉƂɗ ݂Ȃ x Č v.

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