be independent identically distributed Gaussian random variables with mean zero

Be independent identically distributed gaussian

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be independent, identically distributed Gaussian random variables with mean zero and variance 1. Let X - 1 = 0 and X n = rX n - 1 + W n n 0 , (1) where 0 < r < 1. a) Show that X n as n + converges in distribution to a random variable X . What is the distribution of the random variable X ? b) It is not that hard to show that (please do not try to show that and just believe me) P {| X n - X n - 1 | ≥ 2 } ≥ P { W n 2 } ≥ 0 . 02 n 1 . (2) Then show that for X in part (a) we have (It turns out that this inequality is true for any X .) P {| X n - X | ≥ 1 } + P {| X n - 1 - X | ≥ 1 } ≥ P {| X n - X n - 1 | ≥ 2 } ∀ n 1 . (3) Knowing that, then does X n as n + converge in Probability to our random variable X from part a)? Solution1 a) X 0 = W 0 X 1 = rX 0 + W 1 is Gaussian (0 , 1 + r 2 ) X 2 = rX 1 + W 2 is Gaussian (0 , 1 + r 2 + r 4 ) So on X n = rX n - 1 + W n is Gaussian (0 , n i =0 r 2 i ) CDF of X n converges to Gaussian (0 , 1 1 - r 2 ) b) For the second part P {| X n - X | ≥ 1 } + P {| X n - 1 - X | ≥ 1 } ≥ P {| X n - X | ≥ 1 } orP {| X n - 1 - X | ≥ 1 } ≥ P {| X n - X n - 1 | ≥ 2 } ≥ 0 . 02 n 1 . As n + P {| X n - X | ≥ 1 } does not converge to 0 so then X n does not converge in probability to X . Page 2 of 5
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ELEC 533 (Behnaam Aazhang ): Midterm Examination 2017 Problem 1 Problem 2 Suppose we have a random process X t with t T . We take three samples of the process at t = 1 , 2 , and 3 seconds to form a three dimensional vector X = X 1 X 2 X 3 . Assume that X is a Gaussian random vector with mean μ = 1 1 0 and covariance matrix denoted by C X or Σ | X = 1 ρ 0 ρ 1 ρ 0 ρ 1 with some | ρ | ≤ 1 / 2.
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  • Spring '14
  • Aazhang,Behnaam
  • Probability theory, random process Xt, Behnaam Aazhang, different points of time t1

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