UCD STAT20100: Inferential Statistics Assignment 2, Semester 2, 2020

Verified

Added on Β 2022/09/14

|5
|903
|11
Homework Assignment
AI Summary
This document presents a complete solution to an Inferential Statistics assignment, focusing on a random sample from a Gaussian population. The solution begins by determining the method of moments estimator for the parameter Ο„, utilizing the second non-central moment. It then derives the likelihood and log-likelihood functions for Ο„. The assignment proceeds to demonstrate that nX is a sufficient statistic for Ο„, followed by a proof that the maximum likelihood estimator for Ο„ is Ο„Μ‚ = 1/(2X). Finally, the solution calculates a 95% confidence interval based on a given sample realization, providing the necessary calculations and Excel output to support the results. The assignment comprehensively addresses key concepts in inferential statistics, including estimation, likelihood, sufficiency, and confidence intervals.
Document Page
1
INFERENTIAL STATISTICS
Name of the Student
Name of the University
Author’s Note.
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
2
Q. Consider a random sample X1, X2, …, Xn from a Gaussian Population with mean 0 and
unknown variance 𝜎 > 0. The parameter of interest is 𝜏 = 1
2𝜎2. To simplify the notation, you can
use 𝑋̿= βˆ‘ 𝑋𝑖
21
𝑛
𝑛
𝑖=1 .
Solution.
From the above question, then making 𝜎 the subject give;
𝜎2 = 1
2𝜏
The pdf of the Gaussian distribution will be given as;
𝑓(π‘₯, 𝜎) = 1
√2πœ‹πœŽ2 βˆ— exp {βˆ’1
2𝜎2 ( 𝑋𝑖 βˆ’ 0)2} 1
Changing the parameterization yields;
𝑓(π‘₯, 𝜏) = √𝜏
βˆšπœ‹ exp{βˆ’πœ(𝑋𝑖)2}
a) The Moments Estimator of 𝜏 is given by;
πœ‡2
β€² = 𝑀2
β€² 2
By MGF technique the second moment for the gaussian distribution will be given by;
𝑀π‘₯(𝑑) = exp (πœ‡π‘‘ +
𝜎2𝑑2
2 ) 3
Which on substitution and finding the second derivative yields the second-row population
moments as;
πœ‡2
β€² = 𝜎2 = 1
2𝜏
So that πœΜ‚= 1
2𝜎2 which should be equated to the sample raw moments XΜΏ = βˆ‘ Xi
21
n
n
i=1 .
Since 𝜎2Μ‚ = XΜΏ
Then πœΜ‚= 1
2XΜΏ is the MoM estimator.
1 Takeyuki Hida, β€˜5. Gaussian Processes’, Stationary Stochastic Processes. (MN-8) (2015).
2 Courtney Taylor, β€˜Moment Generating Function for Binomial Distribution’ (2020)
<https://www.thoughtco.com/moment-generating-function-binomial-distribution-3126454> accessed 8
April 2020.
3 ibid.
Document Page
3
b) The likelihood function of 𝜏 is given by
𝐿(𝜏) = ∏ 𝑓(π‘₯𝑖, 𝜏)
𝑛
𝑖=1
4
Which on substitution yields
= (𝜏)𝑛
2 + (πœ‹)βˆ’π‘›
2 + exp {βˆ’πœ βˆ‘ π‘₯𝑖
2
𝑛
𝑖=1
} 5∎
The log likelihood function is given by
ln 𝐿(𝜏)
Or simply on substitution yields
= 𝑛
2 ln(𝜏) βˆ’ 𝑛
2 ln(πœ‹) βˆ’ 𝜏 βˆ‘ π‘₯𝑖
2
𝑛
𝑖=1 ∎
6
c) To find if the statistic T is sufficient 𝜏, then by factorization theorem 7 the MLE should be
a function of T.
I.e. 𝑔(𝑇, 𝜏)β„Ž(π‘₯)and should only depend on 𝜏.
From the previous equation 5 on the likelihood function, the factorized form can be written as;
( 𝜏
πœ‹
)
𝑛
2 exp{βˆ’πœ βˆ‘ π‘₯𝑖
2
𝑛
𝑖=1
} βˆ— 1
So that 𝑔(𝑇, 𝜏) = (
𝜏
πœ‹)
𝑛
2 exp{βˆ’πœβˆ‘ π‘₯𝑖
2𝑛
𝑖=1 }
And β„Ž(π‘₯) = 1
From 𝑔(𝑇, 𝜏) we only have one minimal sufficient statistic 𝜏 =βˆ‘ π‘₯𝑖
2𝑛
𝑖=1
To test whether it is unbiased, then
𝜏
𝑛 = XΜΏ
So that 𝜏 = 𝑛XΜΏis a sufficient statistic.
d) The maximum likelihood estimator will be given by minimizing the log likelihood
4 Mike Allen, β€˜Maximum Likelihood Estimation’, The SAGE Encyclopedia of Communication Research
Methods (2017).
5 ibid.
6 ibid.
7 John A Rice, Mathematical Statistics and Data Analysis (Cengage Learning/Brooks/Cole 2013).
Document Page
4
function 8.
πœ•π‘™(𝜏)
πœ•πœ = 𝑛
2𝜏 βˆ’ βˆ‘ π‘₯𝑖
2𝑛
𝑖=1 equating to 0 and a bit of algebra gives;
𝑛
2𝜏 = βˆ‘ π‘₯𝑖
2𝑛
𝑖=1 or dividing both side of the equation by 2
𝑛 and taking reciprocal yields;
1
𝜏 = 2βˆ‘ π‘₯𝑖
2𝑛
𝑖=1
𝑛 π‘œπ‘Ÿ πœΜ‚= 1
2XΜΏ
β„Žπ‘’π‘›π‘π‘’ π‘π‘Ÿπ‘œπ‘£π‘’π‘‘βˆŽ
e) The 95% 𝐢. 𝐼is given by the formula9
Pr{πœ’π‘›βˆ’1
2 (1 βˆ’π›Ό
2) ≀ π‘›πœΜ‚
𝜎2 ≀ πœ’π‘›βˆ’1
2 (𝛼
2)} 10 where 𝜎2 = 1
2𝜏
Or
( π‘›πœΜ‚
πœ’π‘›βˆ’1
2 (𝛼
2) , π‘›πœΜ‚
πœ’π‘›βˆ’1
2 (1βˆ’π›Ό
2)) 11
From the excel output below, πœΜ‚= 3.5461, 𝑛 = 6, πœ’π‘›βˆ’1
2 (𝛼
2) = 12.8325and πœ’π‘›βˆ’1
2 (1 βˆ’π›Ό
2) =
0.8312.
On substituting yields
( 6(3.5461)
12.8325,6(3.5461)
0.8312 )
And thus, the value of πœΜ‚ lies between 1.6580 and 25.5971.
8 ibid.
9 ibid.
10 ibid.
11 ibid.
tabler-icon-diamond-filled.svg

Paraphrase This Document

Need a fresh take? Get an instant paraphrase of this document with our AI Paraphraser
Document Page
5
x_i x_i^2 Column1 labels Chisquare Values
-0.36 0.1296 1-alpha/2 0.8312
0.25 0.0625 alpha/2 12.8325
0.57 0.3249 confidence levels
-0.4 0.16
-0.13 0.0169 lower 1.658
0.39 0.1521 upper 25.5974
0.846
0.141
3.5461
Works Cited.
Allen M, β€˜Maximum Likelihood Estimation’, The SAGE Encyclopedia of Communication
Research Methods (2017)
Hida T, β€˜5. Gaussian Processes’, Stationary Stochastic Processes. (MN-8) (2015)
Rice JA, Mathematical Statistics and Data Analysis (Cengage Learning/Brooks/Cole 2013)
Taylor C, β€˜Moment Generating Function for Binomial Distribution’ (2020)
<https://www.thoughtco.com/moment-generating-function-gaussian-distribution-3126454>
accessed 8 April 2020
chevron_up_icon
1 out of 5
circle_padding
hide_on_mobile
zoom_out_icon
[object Object]