Quiz5_570统计.doc

AMS570 Quiz 5Name: _ID: _This is a close-book exam, due 1:40pm. Please do NOT leave after the quiz lecture will start after the quiz.Let X1,Xn be a random sample from an exponential distribution exp with pdf:f x;=1exp-x, where 0 and x0,Please derive(1) The maximum likelihood estimator (MLE) for . (2) Is the above MLE unbiased for ?(3) The method of moment estimator (MOME) for . (4) Please derive the distribution of the first order statistic X1, where X1= min(X1,Xn), and further show whether Y=nX1 is an unbiased estimator of or not.Solution:(1) The likelihood function is L=i=1nf xi;=i=1n1exp-xi=1nexp-i=1nxiThe log likelihood function is l=lnL=ln1nexp-i=1nxi=-nln-i=1nxiSolving l=-n+i=1nxi2=0We obtain the MLE for :=X(2) SinceEX=EX=0x1exp-xdx=We know the MLE =X is an unbiased estimator for .(3) Now we derive the method of moment estimator (MOME) for . Since we have only one parameter, , to estimate, the equation of the first population moment and sample moment will suffice. That is, solving: EX=X, we obtain the MOME: =X(4) Now we derive the general formula for the pdf of the first order statistic as follows:PX1x=PX1x,Xnx=i=1nPXixTherefore we have 1-FX1x=i=1n1-Fx=1-FxnDifferentiating with respect to x, and then multiplying by (-1) at both sides leads to:fX1x=nfx1-Fxn-1Now we can derive the pdf of the first order statistic for a random sample from the given exponential family. First, the population cdf is:Fx=0x1exp-udu=-exp-u0x=1-exp-xNow plugging in the population pdf and cdf we have: fX1x=n1exp-xexp-xn-1=nexp-nx,when 0 and x0. Thus we know that X1exp n, and its mean shoul