Answerkey2

# X n 1 n p n i 1 x i by denition the weight for each x

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X n = 1 n P n i =1 X i X i is same as 1 n across all X i , i.e., w i = 1 n for all i = 1 ;:::;N: sample average X n as E ² X n ³ = and V ar ² X n ³ = 2 n : 1

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Given these, the law of large numbers indicates that the sample average X n converges in prob- ability to , (or equivalently, X n ! p where " ! p " denotes the "convergence in probability"). In other words, as we have more observations, our sample average is a consistent estimator for the population mean. Furthermore, the central limit theorem provides the asymptotic distribution of the sample av- erage X n : As n ! 1 ; the distribution of p n ( X n ) ± becomes arbitrarily well approximately by the standard normal distribution under some mild conditions (or, equivalently X n is approximately normally distributed with mean E X n ± = and the variance V ar X n ± = ± 2 n ). Z 1 ;Z 2 ;Z 3 ;:::: Suppose the random variables are continuous and that the n th random variable, Z n , has pdf given by, f Z n ( z ) = 8 < : ne nz for z & 0 0 for z < 0 (Ans) (a) In order to verify R 1 f Z n ( z ) dz = 1 for each n; we start by partitioning the entire domain into two sub-intervals as follows, R 1 f Z n ( z ) dz = R 0 f Z n ( z ) dz + R 1 0 f Z n ( z ) dz = R 0 &1 0 dz + R 1 0 ne nz dz = n ± ² ² 1 n e nz j 1 z =0 ³ = 0 ² ( ² 1) = 1 where the above pdf indicates that f Z n ( z ) = 0 In the above calculation, it is worth noting that R e z dz = e z + C for some constant C: In this problem, we have R e nz dz = ² 1 n ± ± e nz + C for some constant C: The easiest way to check this is to di/erentiate the right hand side of ² 1 n ± ± e nz with respect to z and see if your answer corresponds to e nz : 2
(b) Given " > 0 ; it becomes, P ( Z n > " ) = R 1 " ne

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X n 1 n P n i 1 X i by denition the weight for each X i is...

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