Question

Let X1,.. ,X be a random sample from an N(p,02) distribution, where both and o are unknown. You will use the following facts for this ques- tion: Fact 1: The N(u,) pdf is J(rp. σ)- exp Fact 2 If X,x, is a random sample from a distribution with pdf of the form I-8, f( 0,0) = for specified fo, then we call and 82 > 0 location-scale parameters and (6,-0)/ is a pivotal quantity for 8, where 6, and ô, is the MlE's of and , respectively LaX = ΣΧ/m md S3ΣΧ- X), (n-) . (a) Show that p and o are location-scale paramet ers for the N (4,02) pdf. (b) The MLE's of p and o are X and EX, - X} /n, respectively. Using this fact, together with Facts 1 and 2, find a pivotal quantity for p (c) Using the result found in part (b), show that (X-)/(S/m is a pivotal quantity for i (d) Using the fact that (X-)/(S//m) t(n- 1), show (providing all the details) that (-a/()S/Vn, X +ts,a/2{n = 1)S/wm is a1-a confidence interval for i. Here, the quantile to(m) is defined a for T~ t(n- 1) by P(T t(m))

Answer 1

(a) If X1, X2,...X_n be a random sample from having pdf:

Nσ) Νίμ. σε

(Fact 1)

f (ε: μ, σ') Feupl-(π-μ) V2Τσ

Every normal distribution is a version of the standard normal distribution i.e.,

π.μ f (α: μ, σ ) -φ(Ε ΕφΕ σ b

where $\phi(.)$ is the pdf of standard normal distribuiton

2 Ф(г) -т" Еегр- V2т 2

Now comparing it with fact 2

$f(x,\theta_{1},\theta_{2})=\frac{1}{\theta_{2}}f_{0}(\frac{x-\theta_{1}}{\theta_{2}})$

for specified f₀ = $\phi(.)$ and $\\ \theta_{1} and \ \theta_{2}$ are location -scale parameters

we get,

$\\ \theta_{1}=\mu\\ \theta_{2}=\sigma$

Hence, using fact 2
_{μαnd σ}
are location-scale parameters for
_{Nσ) Νίμ. σε}

(b) MLE of
_{μαnd σ}

_{$\\\widehat{X}=\sum\frac{Xi}{n}\\ S^{2}=\sum\frac{(Xi-\widehat{X})}{(n-1)}$}

Likelihood function of
_{Nσ) Νίμ. σε}

$\\L=\prod_{i=1}^{n}f(x:\mu,\sigma)=\ \prod_{i=1}^{n}f(x;\mu,\sigma^{2})=\prod_{i=1}^{n}(\frac{1}{\sqrt{2\pi\sigma^{2}}}exp[-\frac{1}{2\sigma^{2}}(x-\mu)^{2}])\\ logL=log(\frac{1}{2\pi}^{\frac{n}{2}})-\frac{n}{2}log\sigma^{2}-\frac{1}{2\sigma^{2}}\sum_{i=1}^{n}(xi-\mu)\\ now\ differentiating\ with\ respect\ to\ \mu\ and\ \sigma^2\ and\ equating\ it\ equal\ to\ 0\\ \frac{\partial logL}{\partial \mu}=-\frac{1}{\sigma^{2}}\sum(xi-\mu)=0\\ \Rightarrow \widehat{\mu}=\sum\frac{Xi}{n}\\ \Rightarrow \widehat{\mu}=\widehat{X}\\ \frac{\partial logL}{\partial \sigma^{2}}=-\frac{n}{2}\frac{1}{\sigma^{2}}+\sum\frac{(xi-\mu)^{2})}{2(\sigma^{2})^{2}}=0\\ \Rightarrow \widehat{\sigma^{2}}=\sum\frac{(Xi-\mu)}{n}\\ \Rightarrow \widehat{\sigma}=\sqrt{\sum\frac{(Xi-\widehat{X})^{2}}{n}}$

using Fact 1 and Fact 2 pivotal quantity for $\mu$ is:

$\frac{(\widehat{X}-\mu)}{\widehat{\sigma}}=\frac{(\widehat{X}-\mu)}{\sqrt{\sum\frac{(Xi-\widehat{X})^{2}}{n}}}$

Nouw NI amd -1) S CHu tnd Pivatal quantity onfideuc intuvel f dFor = /-d PC KI xSk ニーd n42 omd ki= -tn1(/) n )cantidence ntwal