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P µ o µ v P v v P í ì Z l Á Z P µ o µ X D v u µ u ï ñ 9 u l } P µ o µ. = µ2 p(1−p)τ2 −µ2 = p(1−p)τ2 Problem 6 For the mixture density of problem 5, the variance of the sample mean is given by VarX¯ n = VarX i n = p(1−p)τ2 n And for large n, we have that √ n(M n −M) → d N 0, 1 4f2(M) where M is the true median of the distribution Since this distribution is symmetric about µ, we have. Title C\Users\danielle\AppData\Local\Temp\mso9ECAtmp Author danielle Created Date 11/12/19 PM.

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∫ (µ(t) y′ µ(t)p(t) y) dt = ∫ µ(t)g(t) dt → µ(t) y = ∫ µ(t)g(t) dt (**) Therefore, the general solution is found after we divide the last equation through by the integrating factor µ(t) But before we can solve for the general solution, we must take a step back and find this (almost magical!) integrating factor µ(t) We have. Title Microsoft Word 3 Author user Created Date 5/27/18 PM. Title Microsoft Word Minimum Wage Rules_Final for Website Author warrenma Created Date 12/19/19 AM.

∂µ2 = −µ−2 i=1 x i < 0 Thus there is a local maximum at µ = ¯x We then note that as µ → 0 or µ → ∞, the loglikelihood ‘(µ;x) approaches −∞ Thus µ = ¯x is a global maximum, and the maximum likelihood estimate of µ is ˆµ = ¯x The maximum likelihood estimator in this example is then ˆµ(X) = X¯ Since µ is the. µ(z;1,1) = p1(z)n= FV, for any µ Hence Theorem 1 implies Rµ(x,y) =∼ CSn as an ungraded Snmodule;. X E } Z u µ X W µ v } u Ç } ( } À } À X KDWhd Z Dz^d Z/ ^ t hE/d î > ^^ í } u µ o } v W } ~^& î ó ó ð r, Æ Z } Z } µ o µ } u µ o } v } P Z v } } } l µ o Ì v P.

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