This preview shows page 110 - 112 out of 261 pages.
, and so the posterior distribution is inverse gamma2f&*)°+00with and . The Bayesian premium is the predictive mean. Since $,00cc(~ *~ &*)I´[ ` µ ~(is exponential given ), it follows that the predictive mean is equal to the posterior mean,[0which is,$ccf%*f%&*)~~ )' ÀThe variance of the predictive distribution isI´[ `B~ ($Á B~ )$Á B~ +)µ f ²I´[ `B~ ($Á B~ )$Á B~ +)µ³(&&%&'(%&'.We have found the predictive mean to be .I´[ `B~ ($Á B~ )$Á B~ +)µ ~ )'(%&'I´[ `B~ ($Á B~ )$Á B~ +)µ ~I´[ ` µ i² `B~ ($Á B~ )$Á B~ +)³ 1((&&%&'%&'$^(04 00.Since is exponential given , we have [I´[ ` µ ~ &000(&&(the second moment of an exponential distribution is 2 times the square of the mean).Therefore,I´[ `B~ ($Á B~ )$Á B~ +)µ ~&i² `B~ ($Á B~ )$Á B~ +)³ 1(&&%&'%&'$^(04 00,which is (2nd moment of the posterior distribution). Since the posterior is inverse gamma, its& g2nd moment is . The predictive variance is²³²&*)³²f%³²f&³²)³²(³,$$c &&cc~~ 'Á )%%À&)&²'Á )%%À&)³ f ²)'³~ (Á &%'À&)&. ¡Example CR-25:Tom has a coin, but he doesn't know the probability of tossing a head. Heassumes a prior distribution for the probability of tossing a head to be=4²=³ ~ &=$ ¡ = ¡ %for . Tom tosses the coin 10 times and observes 4 heads.Find the posterior distribution of the probability of tossing a head. Suppose that Tom tosses thecoin another 5 times. Find the expected number of heads in those next 5 tosses.Solution:We first observe that , which is a beta4²=³ ~ &= ~i =²% f =³"""²'³²&³ ²%³&f%%f%distribution with , . The number of heads tossed in 10 tosses has a binomial. ~ &/ ~ %distribution with parameters and , and probability function; ~ %$=3²B`=³ ~= ²% f =³B=45%$BB%$fB. The joint density of and is3²(Á =³ ~ 3²(`=³ i²=³ ~= ²% f =³ii =²% f =³445%$²'³²&³ ²%³((%$f(&f%%f%""", which is proportionalto . Therefore, the posterior is also a beta distribution, with and .=²% f =³. ~ */ ~ +*f%+f%ccThe predictive mean for the number of heads in 5 more tosses isI´`µ ~I´`=µ i²=`³ 1=no. heads in 5 more tosses 4 headsno. heads in 5 more tosses4 heads($%4~)= i²=`³ 1= ~ ) g= ~ ) i~ ) i~ &À'($%44 heads(posterior mean of ). .*. e/*e+ccc¡