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Estimation of Parameters in a Gamma Distribution for Life Span Model, Study Guides, Projects, Research of Statistics

A problem about estimating the parameters  and checking the appropriateness of a gamma distribution model for life spans using a given pdf. The problem involves calculating the likelihood function, finding its maximum likelihood estimator, and comparing the log likelihood values for different parameter values.

Typology: Study Guides, Projects, Research

Pre 2010

Uploaded on 09/17/2009

koofers-user-0ei
koofers-user-0ei 🇺🇸

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Download Estimation of Parameters in a Gamma Distribution for Life Span Model and more Study Guides, Projects, Research Statistics in PDF only on Docsity! Project #5 STAT 380 Fall 2005 1) (18 total points) Actuaries use PDFs to help model the life spans of people. These models then can be used to help determine rates to charge for life insurance. Suppose the random variable X denotes life span in years of an individual where X is constrained to be between 0 and 100 years. Suppose the PDF below is used to model life span for a particular population. 1 1 1 ( ) 1 1 x 1 x 0 x 100 f(x) 100 ( ) ( ) 100 100 0 otherwise                            where the numerical constant  is greater than 0, the numerical constant  is greater than 0, and ( ) is the gamma function. Throughout this problem, assume  = 2. Do not assume the value of  is a particular number! Thus, the PDF becomes, 1 2 1 1 1 ( 2) 1 1 1 1 1 f(x) x 1 x ( 1) x 1 x 100 ( ) (2) 100 100 100 100 100                                        for 0 < x < 100 and f(x) = 0 otherwise. The purpose of this problem is to estimate  and determine if the PDF really is appropriate or not for this setting. Suppose a random sample of life spans from this particular population is obtained. Answer the questions below. a) (2 points) Using by-hand calculations, find the likelihood function. b) (2 points) Using by-hand calculations, show the natural log of the likelihood function can be expressed as     n 1 n i i i 1 lnL(x ,...,x | ) n( 1)ln(100) nln( 1) nln( ) ( 1)ln x ln 100 x                   Make sure to show each step of your calculations and provide justification. It will help to remember from an algebra class the following way to reexpress items in a natural log: 2 2 1 2 1 2 1 2ln(a a ) ln(a ) ln(a ) 2ln(a ) ln(a )    . c) The actual observed values for a sample of size n = 1,000 are included in the actuary.xls file on the projects web page of the course website. Using this information, answer the questions below. Note that the ln( ) function in Excel can be used to find a natural log value. i) (1 point) If  = 12, what is the value of the log likelihood function? ii) (1 point) If  = 5, what is the value of the log likelihood function? iii) (1 point) Which value of , 5 or 12, is more plausible for this observed data. Explain. iv) (2 point extra credit) Plot the log likelihood function. What do you think the maximum likelihood estimate of  should be approximately. You can use any software package of your choice to answer this question. d) (2 points) Using by-hand calculations, show the derivative with respect to  of the natural log likelihood function can be expressed as n1 n i i 1 lnL(x ,...,x | ) 2 1 1 n ln x ( 1) 100                 . 1
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