A geometric distribution has a pdf given by P(X=x) = p(1-p)^x, where x = 0, 1, 2, \u2026, and 0 < p=””>< 1.=”” this=”” form=”” of=”” the=”” geometric=”” starts=”” at=”” x=”0,” not=”” at=”” x=”1.” given=”” are=”” the=”” following=””><\/p>\n\n\n\n
E(X) = (1-p)\/p, and Var(X) = (1-p)\/p^2<\/p>\n\n\n\n
A random sample of size n is drawn; the data are X1, X2, \u2026, Xn.<\/p>\n\n\n\n
A. Derive the Fisher information function for the parameter p.<\/p>\n\n\n\n
B. Find the Crame\u0301r-Rao lower bound (CRLB) for the variance of an unbiased estimator for p.<\/p>\n\n\n\n
C. Find a sufficient statistic for the parameter p<\/p>\n\n\n\n
D. Show that the sample mean, xbar, is an unbiased estimate of E(X) = (1-p)\/p Find the variance of xbar.<\/p>\n\n\n\n
E. Argue whether or not the sample mean is a minimum variance unbiased estimate (MVUE) of population mean, mew.<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The Fisher information function ( I(p) ) is given by:<\/p>\n\n\n\n [ Where ( L(p) ) is the likelihood function of the sample. For a random sample of size ( n ), the likelihood function is:<\/p>\n\n\n\n [ Taking the logarithm of the likelihood:<\/p>\n\n\n\n [ Now, we compute the first and second derivatives with respect to ( p ):<\/p>\n\n\n\n [ [ The Fisher information is:<\/p>\n\n\n\n [ Since ( \\mathbb{E}[X_i] = \\frac{1 – p}{p} ), the Fisher information function is:<\/p>\n\n\n\n [ The CRLB is given by:<\/p>\n\n\n\n [ From part A, we know the Fisher information ( I(p) ). Therefore, the CRLB is:<\/p>\n\n\n\n [ This is the minimum variance for an unbiased estimator of ( p ).<\/p>\n\n\n\n By the factorization theorem, a statistic is sufficient for ( p ) if the likelihood can be factored into a product of two terms, one of which depends only on the data through the statistic. The likelihood function is:<\/p>\n\n\n\n [ This can be factored as:<\/p>\n\n\n\n [ Thus, the statistic ( T(X) = \\sum_{i=1}^{n} X_i ) is sufficient for ( p ).<\/p>\n\n\n\n The sample mean ( \\overline{X} ) is an unbiased estimator of ( E(X) = \\frac{1 – p}{p} ) because:<\/p>\n\n\n\n [ To find the variance of ( \\overline{X} ), recall that:<\/p>\n\n\n\n [ Since ( \\text{Var}(X_1) = \\frac{1 – p}{p^2} ), we have:<\/p>\n\n\n\n [ To determine if ( \\overline{X} ) is the MVUE, we need to check if it is unbiased and has the minimum variance among all unbiased estimators. From parts D and B, we know that ( \\overline{X} ) is an unbiased estimator of ( \\frac{1 – p}{p} ), and its variance is given by:<\/p>\n\n\n\n [ From the CRLB, the minimum variance for an unbiased estimator of ( p ) is ( \\frac{p^2}{n(1 – p)} ). Since the sample mean does not achieve this lower bound, it is not the MVUE.<\/p>\n\n\n\n Thus, while ( \\overline{X} ) is an unbiased estimator, it is not the MVUE because it does not attain the Cram\u00e9r-Rao lower bound. The MVUE for ( p ) could be a more complex estimator that incorporates the sufficient statistic ( T(X) = \\sum_{i=1}^{n} X_i ).<\/p>\n","protected":false},"excerpt":{"rendered":" A geometric distribution has a pdf given by P(X=x) = p(1-p)^x, where x = 0, 1, 2, \u2026, and 0 < p=””>< 1.=”” this=”” form=”” of=”” the=”” geometric=”” starts=”” at=”” x=”0,” not=”” at=”” x=”1.” given=”” are=”” the=”” following=””> E(X) = (1-p)\/p, and Var(X) = (1-p)\/p^2 A random sample of size n is drawn; the data […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-191369","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191369","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=191369"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/191369\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=191369"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=191369"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=191369"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}A. Fisher Information Function for Parameter ( p )<\/h3>\n\n\n\n
I(p) = – \\mathbb{E} \\left[ \\frac{\\partial^2}{\\partial p^2} \\log L(p) \\right]
]<\/p>\n\n\n\n
L(p) = \\prod_{i=1}^{n} P(X_i = x_i) = \\prod_{i=1}^{n} p(1 – p)^{x_i}
]<\/p>\n\n\n\n
\\log L(p) = \\sum_{i=1}^{n} \\log \\left[ p(1 – p)^{x_i} \\right]
= n \\log p + \\sum_{i=1}^{n} x_i \\log(1 – p)
]<\/p>\n\n\n\n\n
\\frac{\\partial}{\\partial p} \\log L(p) = \\frac{n}{p} – \\sum_{i=1}^{n} \\frac{x_i}{1 – p}
]<\/p>\n\n\n\n\n
\\frac{\\partial^2}{\\partial p^2} \\log L(p) = – \\frac{n}{p^2} – \\sum_{i=1}^{n} \\frac{x_i}{(1 – p)^2}
]<\/p>\n\n\n\n
I(p) = – \\mathbb{E} \\left[ \\frac{\\partial^2}{\\partial p^2} \\log L(p) \\right]
]<\/p>\n\n\n\n
I(p) = \\frac{n}{p^2} + \\frac{n(1 – p)}{p^3}
]<\/p>\n\n\n\nB. Cram\u00e9r-Rao Lower Bound (CRLB) for the Variance of an Unbiased Estimator of ( p )<\/h3>\n\n\n\n
\\text{CRLB} = \\frac{1}{I(p)}
]<\/p>\n\n\n\n
\\text{CRLB} = \\frac{p^2}{n(1 – p)}
]<\/p>\n\n\n\nC. Sufficient Statistic for ( p )<\/h3>\n\n\n\n
L(p) = p^n (1 – p)^{\\sum_{i=1}^{n} x_i}
]<\/p>\n\n\n\n
L(p) = \\left( p^n \\right) \\left( (1 – p)^{\\sum_{i=1}^{n} x_i} \\right)
]<\/p>\n\n\n\nD. Unbiased Estimate of ( E(X) = \\frac{1 – p}{p} ) and Variance of ( \\overline{X} )<\/h3>\n\n\n\n
\\mathbb{E}[\\overline{X}] = \\mathbb{E}[X_1] = \\frac{1 – p}{p}
]<\/p>\n\n\n\n
\\text{Var}(\\overline{X}) = \\frac{\\text{Var}(X_1)}{n}
]<\/p>\n\n\n\n
\\text{Var}(\\overline{X}) = \\frac{1 – p}{n p^2}
]<\/p>\n\n\n\nE. Is the Sample Mean the Minimum Variance Unbiased Estimator (MVUE)?<\/h3>\n\n\n\n
\\text{Var}(\\overline{X}) = \\frac{1 – p}{n p^2}
]<\/p>\n\n\n\n