The geometric mean of a sequence of positive numbers x1 xn

The geometric mean of a sequence of positive numbers x1,. .. , xn is nth root x1. . . xn. Let x1,. .. , Xn be a random sample on an exponential distribution with mean theta. We would like to use the geometric mean G = nth root X1 .. . Xn to estimate theta. All results may be expressed in terms of the gamma function and the Euler constant Show that G is a biased estimator of theta. Deduce an estimator of theta based on G that is unbiased. Call this estimator T. Let Y = in X where X is exponential with mean theta. Obtain the mean of Y. Let Y1,. .. , Yn be a random sample on the distribution of in X. Show that Y is not a consistent estimator of ln theta. Is G a consistent estimator of theta? Explain.

Solution

Given X₁,X₂,……,X_n are n i.i.d random variables (r.v.) from exponential distribution with mean i.e. their common pdf is

f(x) = exp(-x/ )/        , if x>0 , >0

Now G= (X₁ X₂ …..X_n)^1/n

(a)

E[G]   = E(X₁ X₂ …..X_n)^1/n

                ⁼ {E(X₁^1/n)E(X₂^1/n)…..E(X_n^1/n)}   Since they are i.i.d

          = {E(X₁^1/n)}ⁿ

          = { ^1/n gamma(1+1/n)}ⁿ

          = {gamma(1+1/n)}ⁿ

Hence G is biased for .

(b)

An unbiased estimator T of is given by

          T= G/{gamma(1+1/n)}ⁿ

(c)

Y= lnX

E[Y] = E[ln X]

          = Integration(from=0,to=, lnx*f(x)dx)

          =int(0, ,(lnx*exp(-x/ ))/ )

          = -.577216            if =1

If not equal to 1 E[Y] cannot have a closed form.

(d)

E[Ybar]

=E[SUM(Y_i)/n]

=SUM(E[Y_i])/n

=n*E[Y₁]/n                              Since Ys are i.i.d

=E[Y₁]   

Now lim(n tends to ,E[Ybar]) not equal to ln.

Hence Ybar is not consistent for ln.

(e)

E[G] = {gamma(1+1/n)}ⁿ   which tends to as n tends to .

Now E[G²] = E(X₁ X₂ …..X_n)^2/n

                ⁼ {E(X₁^2/n)E(X₂^2/n)…..E(X_n^2/n)}   Since they are i.i.d

          = {E(X₁^2/n)}ⁿ

          = { ^2/n gamma(1+2/n)}ⁿ

          = ² {gamma(1+2/n)}ⁿ

Therefore,

Var(G) = E[G²] – E²[G]

          = ² [ {gamma(1+2/n)}ⁿ – {gamma(1+1/n)}ⁿ ]

tends to 0 as n tends to .

Hence G is consistent for .