Consider a random sample of size n from a distribution with

Consider a random sample of size n from a distribution with pdf f(x;theta)=(theta)(x^(theta-1)) if 0<x<1 and zero otherwise; theta>0.

Find the UMVUE of 1/(theta). Hint: E[-ln(X)]=1/(theta)

Solution

pdf is f(x:theta)=theta*(x^theta-1)   0<x<1

                     =0               otherwise       theta>0

we have a random sample of size n

so the joint pdf is- f(x1,x2,...,xn:theta)=thetaⁿ(x1*x2*x3*.....xn)^theta-1        0<x1,x2,...,xn<1

                                                      =exp[ln{thetaⁿ(x1*x2*x3*.....xn)^theta-1 }]

                                                      =exp[n(lntheta)+(theta-1){lnx1+lnx2+...+lnxn}]

                                                      =exp[n(lntheta)+theta{lnx1+lnx2+...+lnxn}-{lnx1+lnx2+..+lnxn}]

now range of xi\'s are independent of the parameter theta. and the range is an open space and the joint pdf can be written in the form exp[u(theta)*v(x1,x2,...,xn)+w(theta)+z(x1,x1,...xn)]

where u(theta)=theta z(x1,x1,...xn)=-{lnx1+lnx2+..+lnxn}    w(theta)=n(lntheta) and v(x1,x2,...,xn)={lnx1+lnx2+...+lnxn}

hence the distribution belongs to ONE PARAMETER EXPONENTIAL FAMILY.

HENCE T={lnx1+lnx2+...+lnxn} IS A COMPLETE SUFFICIENT STATISTIC.

now we know that E[-ln(x)]=1/theta

now E[-T/n]=E[-{lnx1+lnx2+...+lnxn}/n]=1/n[E[-ln(x1)]+E[-ln(x2)]+....+E[-ln(xn)]]=(n/theta)/n=1/theta

hence -T/n is an unbiased estimator of 1/theta

and T is a complete sufficient statistic.

hence by LEHMANN-SCHEFFE THEOREM -{lnx1+lnx2+...+lnxn}/n is the UMVUE of 1/theta