Suppose thatY₁ = B₁X₁ + €i, i = 1,..., nN(0,0²), σ > 0 holds true. We only have the observed versions of(Yi, X₁), which are (x₁, y₁), . (Xn, Yn).(a) Find the Maximum Likelihood Estimator (MLE) for ₁.(b) Show that MLE B₁ is unbiased.(c) Show that B₁ has a normal distribution.(d) Find Var(8₁) (treating the X;'s as fixed).iidwhere ti~

Answered: Suppose that Y₁ = B₁ X₁ + €į, i=1,...,…

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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Question

Suppose that
Y₁ = B₁X₁ + €i, i = 1,..., n
N(0,0²), σ > 0 holds true. We only have the observed versions of
(Yi, X₁), which are (x₁, y₁), . (Xn, Yn).
(a) Find the Maximum Likelihood Estimator (MLE) for ₁.
(b) Show that MLE B₁ is unbiased.
(c) Show that B₁ has a normal distribution.
(d) Find Var(8₁) (treating the X;'s as fixed).
iid
where ti~

Expert Solution

Step 1: Determine the maximum likelihood estimator

Given: $Y subscript i equals beta subscript 1 X subscript i plus epsilon subscript i$ where $ε_{i}$ is normally distributed with mean 0 and variance $σ^{2}$ .

a) The likelihood function for the normal distribution is:

$L open parentheses beta subscript 1 close parentheses equals product from i equals 1 to n of fraction numerator 1 over denominator square root of 2 pi sigma squared end root end fraction e to the power of negative fraction numerator open parentheses Y subscript i minus beta subscript 1 X subscript i close parentheses squared over denominator 2 sigma squared end fraction end exponent$

To maximize the likelihood, it's easier to work with the log-likelihood:

$l open parentheses beta subscript 1 close parentheses equals sum from i equals 1 to n of open square brackets negative 1 half log open parentheses 2 pi sigma squared close parentheses minus fraction numerator open parentheses Y subscript i minus beta subscript 1 X subscript i close parentheses squared over denominator 2 sigma squared end fraction close square brackets$

Differentiate with respect to $β_{1}$ and set the derivative equal to zero to find the MLE:

$fraction numerator partial differential l open parentheses beta subscript 1 close parentheses over denominator partial differential beta subscript 1 end fraction equals sum from i equals 1 to n of fraction numerator X subscript i open parentheses Y subscript i minus beta subscript 1 X subscript i close parentheses over denominator sigma squared end fraction equals 0$

Solving for $β_{1}$ :

$beta with hat on top subscript 1 equals fraction numerator sum subscript i equals 1 end subscript superscript n X subscript i Y subscript i over denominator sum subscript i equals 1 end subscript superscript n X subscript i superscript 2 end fraction$

b) $E open parentheses beta with hat on top subscript 1 close parentheses equals E open parentheses fraction numerator sum subscript i equals 1 end subscript superscript n X subscript i Y subscript i over denominator sum subscript i equals 1 end subscript superscript n X subscript i superscript 2 end fraction close parentheses$

$equals fraction numerator sum subscript i equals 1 end subscript superscript n X subscript i E open parentheses Y subscript i close parentheses over denominator sum subscript i equals 1 end subscript superscript n X subscript i superscript 2 end fraction$

$equals fraction numerator sum subscript i equals 1 end subscript superscript n X subscript i open parentheses beta subscript 1 X subscript i close parentheses over denominator sum subscript i equals 1 end subscript superscript n X subscript i superscript 2 end fraction$

$equals beta subscript 1$