32. The Renyi representation. Suppose E₁,..., En are iid exponentially dis-tributed random variables with parameter λ> 0 so thatP[E₁ ≤ x] = 1 - e-ix, x>0.LetE1.n ≤ E2.n ≤ ≤ En.n<.be the order statistics. Prove the n spacingsE1.n, E2.n-E1.n...., En.n-En-1,nare independent exponentially distributed random variables where Ek+1,n-Ek,n has parameter (n - k)λ. Intuitively, this results from the forgetfulnessproperty of the exponential distribution.

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Answered: 32. The Renyi representation. Suppose…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Suppose that X;; i = 1,2, 3, .. n are independent random variables with a common uniform distribution over (0, 1) and Yg is distributed over (0, Yg -1) with Y, =1,k = 1,2, ..., n. Let Z = k = 1,2,3, .n R.D.R EGALA STAT 203 March 4, 2021 Show by mathematical induction induction or any other method that Zg has probability density function (-In z)*-1 fR(z) = k = 1,2,3, . n (k – 1)! Hence deduce that Zg and Y, have the same distribution (k = 1,2,3, ..n)
Page 353 5.4.4
Let {e} be a zero-mean, unit-variance white noise process. Consider a process {Y} which starts at time t = 0, and is defined recursively as follows: Let Yo = cieo, Y₁ = c₂Yo + e₁, and Y₁ = ¢₁Yt-1 + 2Yt-2 + et for t≥ 2 (as in an AR(2) process). a) Show that the process has zero mean. 1 b) For values of 01, 02 in the stationarity region for an AR (2) process, show how to choose C₁, C₂ such that both Var(Y) = Var(Y₁) and the lag 1 autocorrelation cov(Yo, Y₁) equals that of a stationary AR(2) process with parameters 01, 02. c) Once the process {Y} has been generated, show how one can transform in into a new process which has any desired mean and variance. (Remark. This gives a method for simu- lating stationary AR(2) processes.)
4.2 The probability distribution of the discrete random variable X is f(x) = 3 (*)*** Find the mean of X. 3-x (¹)ª (²³)³—ª, x = 0, 1, 2, 3.
Suppose X is a discrete random variable. Let the pmf of X be equal to 5 - x f(x) = x = 1,2,3,4 10 Find the cdf of X, that is F(X).
2. Let X₁, X2, X3,..., be independent identically distributed Unif(0, 1) random variables. Let Sn be the largest gap between any two of the points {X1, X2,..., Xn} (e.g. if n = 3 and (X₁, X2, X3) = (0.1, 0.5, 0.7) then S3 = 0.4). Show that for any x > 0 lim P(nSn > x) = 1. n→∞ Hint: to lower bound nSn > x, if suffices to demonstrate that there is an empty interval of some length x/n which contains none of the points. Make a list of some candidate intervals and then use Cheby- shev's inequality to show one of them is empty with probability going to 1.
Q 6.1. Suppose Z = (Z₁, Z2, Z3) is a standard multi-variate Gaussian random variable i.e., for i ≤ 3, Zi~ N(0, 1) are i.i.d. random variables. Each of the random variables, (a)–(d), on the left is equal in distribution to exactly one random variables, (1)–(4), on the right. Pair up according to "equal in distribution" and explain briefly your reasoning. X₁ (a) X₂ (b) (x₂) - (2) (c) (x²) - (¹/1² 1/√²) (2₁) (3/√2 = √2 = (V2) Z₁ (d) (x) - (1) (²) X2 Y₁ (9)-(-) (2) (1=1) Y₂ (1) Y₁ (²) (4) - (1/² (3) (1/√√2 1/√2-1/√2) (x₁) = (²₁ ² ✓/³²) (3) 2 2 √2 1 Z3 Y₁ → ()-()) (4) = Y₂ 1/2) (2) 1) (2)
2. Suppose one has n non-degenerate random variables X1, X2,, X, so that X1+ X2 + · · · + Xn = L for some constant L. (Recall that a random variable is non-degenerate if it is not a constant in disguise.) Show that there must be at least one pair of indices i j so that p(X;, X;) < 0.
Q 6.1. Suppose Z = (Z1, Z2, Z3) is a standard multi-variate Gaussian random variable i.e., for i ≤ 3, Zi~ N(0, 1) are i.i.d. random variables. Each of the random variables, (a)-(d), on the left is equal in distribution to exactly one random variables, (1)-(4), on the right. Pair up according to "equal in distribution" and explain briefly your reasoning. (a) (X₂) (¹) (X) = (2) 1/√2 (Ⓒ) (x₂) - (3/1² ¹1/1²) (2) (c = 2 0 = 2 Z₁ 1) (²) 3 (4¹) (X) = (²¹) (1) (2) (3) Y₁ 1 (29) - ()) (2) = Y₂ 1 Y₁ (121) = (1/² √₂) (2) (1/√2 1/√2 /2 −1/√√2, Y₂ X₁ X₂ = 22 1 1 2 0 Y₁ 2 1 (4) (2)) = (²) (²) 1 1 Z₁ Z₂ Z3
3

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