And v ar ² x n ³ 2 n 1 given these the law of large

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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 ² X n ³ = ° and the variance V ar ² X n ³ = ° 2 n ). 2. Suppose we have an in°nite sequence of random variables 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 ² 0 0 for z < 0 (Ans) (a) In order to verify R 1 °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 °1 f Z n ( z ) dz = R 0 °1 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 ³ ° ´ 1 n e ° nz j 1 z =0 ± = 0 ´ ( ´ 1) = 1 where the above pdf indicates that f Z n ( z ) = 0 for the °rst sub-interval in the second equality. 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 ³ ³ 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 ³ ³ e ° nz with respect to z and see if your answer corresponds to e ° nz : 2
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(b) Given " > 0 ; it becomes, P ( Z n > " ) = R 1 " ne ° nz dz = n ³ ° ´ 1 n ± ³ e ° nz j 1 z = " = e ° n" (c) In part (b), we °nd that P ( Z n > " ) = e ° n" for some " > 0 and each n: As n ! 1 ; it becomes, lim n !1 P ( Z n > " ) = lim n !1 e ° n" = 0 which, in turn, indicates lim n !1
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