Question

5. Assume that the number of cases of tetanus reported in the United States during a single month in 2005 has a Poisson distribution with parameter u. The number of cases reported in January and February are 1 and 3, respectively. [10+2+3 points] (a) Find and plot the likelihood function over the space of potential values for μ. (b) What is the maximum likelihood estimate (MLE) of u? (c) Give an estimate of the probability that there is no case of tetanus reported for a given month.

Answer 1

Solution :

$\mathrm{Y\:\:be\:the\:number\:of\:cases\:of\:tetanus\:reported\:in\:the\:United\:States\:during\:a\:single\:month\:in\:2005.}$

$\mathrm{Z\:\:has\:a\:Poisson\:distribution\:with\:unknown\:parameter\:\mu }$

$\mathrm{Probability\:mass\:function\:of\:Z\:is\:given\:by\:\:P(Z}=x)=e^{-\mu }\frac{\mu ^{x}}{x!},\quad\:x=0,1,2,...$

( a )

$\mathrm{We\:have\:two\:observations,\:\:Z_{1}\:\:and\:\:Z_{2}\:\:from\:the\:above\:distribution\:with\:\:Z_{1}=1\:\:and\:\:Z_{2}=3.}$

$\mathrm{\:The\:likelihood\:function\:is\:given\:by}$

$\mathrm{L(\mu \mid Z_{1}=1,\:\:Z_{2}=3)=P(Z_{1}=1)*P(Z_{2}=3)}$

$=e^{-\mu }\frac{\mu }{1}*e^{-\mu }\frac{\mu^{3} }{3!}$

$=e^{-2\mu }\frac{\mu^{4} }{3!}$

$=e^{-2\mu }\frac{\mu^{4} }{3*2}$

$=e^{-2\mu }\frac{\mu^{4} }{6}$

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( b )

$\mathrm{The\:maximum\:likelihood\:estimate\:of\:\:\mu \:\:is\:\:\:2.}$

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( c )

$\mathrm{The\:probability\:that\:there\:is\:no\:case\:of\:tetanus\:reported\:for\:a\:given\:month\:is\:}$

$\mathrm{P(Z}=0)=e^{-\mu }$

$\mathrm{\:The\:estimated\:value\:of\:the\:probability\:that\:there\:is\:no\:case\:of\:tetanus\:reported\:for\:a\:given\:month\:is\: \:exp(-\hat{\mu} )}$

$=e^{-2}$

$=\frac{1}{e^2}$

$=0.13534$