3.STATISTICAL INFERENCE Let Y₁, Y₂, ..., Yn be a random sample whose probability density function is given by-{oPro,e6840,0 < y < ∞ and ß > 0elsewhere200and suppose that n = 200, Σ²00 y₁ = 20, Σ²ºy² = 100, 200³ = 250 and Ây = 0.025.i=1i) Derive the standard error of ß, se() = 0.0009, using MLE approach.f(y; B) =

Given density function is, fy=y36β4e-yβ, y>0 Consider, the likelihood function, L=∏i=1nfy…

Answered: Let Y₁, Y2,..., Yn be a random sample…

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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3.STATISTICAL INFERENCE

Let Y₁, Y₂, ..., Yn be a random sample whose probability density function is given by
-{oPro,
e
684
0,
0 < y < ∞ and ß > 0
elsewhere
200
and suppose that n = 200, Σ²00 y₁ = 20, Σ²ºy² = 100, 200³ = 250 and Â
y = 0.025.
i=1
i) Derive the standard error of ß, se() = 0.0009, using MLE approach.
f(y; B) =

Expert Solution

Step 1

Given density function is,

$f (y) = \frac{y^{3}}{6 β^{4}} e^{- \frac{y}{β}}, y > 0$

Consider, the likelihood function,

$L = \prod_{i = 1}^{n} f (y) = \prod_{i = 1}^{n} \frac{y_{i}^{3}}{6 β^{4}} e^{- \frac{y_{i}}{β}} = \frac{y_{1}^{3}}{6 β^{4}} e^{- \frac{y_{1}}{β}} * \frac{y_{2}^{3}}{6 β^{4}} * e^{- \frac{y_{2}}{β}} * . . . . * \frac{y_{n}^{3}}{6 β^{4}} * e^{- \frac{y_{n}}{β}} = \frac{\prod_{i = 1}^{n} {(y_{i})}^{3}}{{(6 β^{4})}^{n}} * e^{- \frac{\sum_{i = 1}^{n} y_{i}}{β}}$

Apply 'ln' on both sides,

$\ln L = \ln [\frac{\prod_{i = 1}^{n} {(y_{i})}^{3}}{{(6 β^{4})}^{n}} * e^{- \frac{\sum_{i = 1}^{n} y_{i}}{β}}] = \ln [\prod_{i = 1}^{n} {(y_{i})}^{3}] - \ln {(6 β^{4})}^{n} - \frac{\sum_{i = 1}^{n} y_{i}}{β} = \sum_{i = 1}^{n} \ln y_{i}^{3} - n \ln 6 - 4 n \ln β - \frac{\sum_{i = 1}^{n} y_{i}}{β}$

MLE of $β$ obtained by $\frac{d}{d β} \ln L = 0 .$

$\frac{d \ln L}{d β} = 0 \frac{d}{d β} [\sum_{i = 1}^{n} \ln y_{i}^{3} - n \ln 6 - 4 n \ln β - \frac{\sum_{i = 1}^{n} y_{i}}{β}] = 0 0 - 0 - \frac{4 n}{β} + \frac{\sum_{i = 1}^{n} y_{i}}{β^{2}} = 0 \frac{\sum_{i = 1}^{n} y_{i}}{β^{2}} = \frac{4 n}{β} \hat{β} = \frac{\sum_{i = 1}^{n} y_{i}}{4 n}$