In this problem we refer to Example 8.6. Suppose the random variables (X₁, Y₁), (X2, Y2),..., (Xn, Yn)are all independent and that for each i we have X₁, Yį distributed as N(μi, 0²).a) Set v = and express the likelihood function in terms of the parameters1,,n, and v.b) Show that the log likelihood isn(Xì - Hi) + (Yi - Hi)21 (μ₁,..., fln, v) = 2n log. + n logv - v1√2πi=1c) Calculate the MLEs for the parameters 1,...,fln, V. EXAMPLE 8.6 The Neyman-Scott ParadoxHere is an example of a distribution where the MLE is not consistent. Youwill be asked to prove most of the claims here in Problem 8.7. Suppose randomvariables (X₁, Y₁), (X2, Y2),..., (Xn, Yn) are all independent and that for eachi we have X₁, Y₁ distributed as N(i, 02). Note that the variables have meansindexed by i but they all have the same variance. In this case, we can form thelog-likelihood function and show that the MLEs for the parameters are22==Xi + Yi2n1-n(Xi - Yi)²4These are in fact the unique solutions of the likelihood equation.Because of the last expression, it is easy to see that222and therefore the MLE is not consistent (it is biased).(8.4)(8.5)

We have two independent random variables Xi,Yi i=1,2,....,n and for each i, they are distributed as…

Answered: 8.7 In this problem we refer to Example…

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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Find the regression equation, letting overhead width be the predictor (x) variable. Find the best predicted weight of a seal if the overhead width measured from a photograph is 2 cm. Can the prediction be correct? What is wrong with predicting the weight in this case? Use a significance level of 0.05.
1. The random variables §, §1, §2,... are independent and identically distributed with 1/4 and P(§ = j) = c/j for j = 1,2,3. Let X₁ = 0 and Xn = max(§₁, ..., Èn) for n = 1, 2,.... distribution P(§ = 0) = (a) What value must c take? (b) Explain why {Xn, n = 0, 1, 2,...} is a Markov chain. (c) Write down the transition matrix. (d) Draw the transition diagram and classify the states (aperiodic, transient, re- current, eorgodic, etc). (e) Calculate P(Xn = 0). (f) Calculate P(X₁ = 3, X₂ = 1|X₁ = 3).
Suppose that X, Y, and Z are jointly distributed random variables, that is, they are defined on the same sar Suppose that we also have the following. E(X)=-5 E(Y)=8 E(Z)=6 Var (X)=24 Var (Y) = 16 Var (Z) = 33 Compute the values of the expressions below. E (4Y+4)= 36 E(-4Z+2X)- -3 = 11.33 Var (4Y)-2= 254 E(-x²) = 49 0|0 X
Question 3. Let X₁,, Xn be a random sample from a distribution with the pdf given by f₁₁x(x) = exp(-ª) if x ≥ A, otherwise f(x) = 0, where > 0. Find the MLE's of 0 and X. Start by writing the likelihood function and note the constraint involving A. Question 4. #4.2.9 of the textbook.
Let X ∼ N (0, 1) and Y ∼ Ber(p) be two independent random variables. find the law of S = X + Y
3. If X;~N(0,1), i = 1,2,3,4 are independent Normal random variables and Y is given as Y =(X,+X,)' +(X, + X,)' . If aY~x², . find the constant a. 2
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)
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
7. Suppose that the joint p.d.f. of two random variables X and Y is as follows: (4 – 2x – y) for x > 0, y > 0, f(x, y) = and 2x + y 2|X = 0.5).
Example 17.37. (a) Let x1, x2, ..., Xn be a random sample from the uniform distribution with p.d.f. : 1,0 0 f(x, 0) 0, elsewhere Obtain the maximum likelihood estimator for 0.

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