From: thompson@atlas.socsci.umn.edu (T. Scott Thompson)
Newsgroups: sci.math.stat
Subject: Re: Question on ratio estimate
Date: 3 Sep 92 04:39:08 GMT
Reply-To: thompson@atlas.socsci.umn.edu
Organization: Economics Department, University of Minnesota

wangj@eniac.seas.upenn.edu (Jie Wang ) writes:

>Hi, there, could someone out there help me with the following
>ratio estimate problem ? Any comments and/or references are
>appreciated. Now the problem:

>Suppose {X(i)} and {Y(i)} are i.i.d. sequences respectively, but
>X(i) and Y(i) may be correlated. Let m(x) and m(y) be the expected
>values of X(i) and Y(i) repectively. We would like to estimate 
>	    m(x)
>	R = ----. 	We know that
>	    m(y)


>	^      [ sum_1_to_n X(i) ] / n
>	R(n) = -----------------------  
>               [ sum_1_to_n Y(i) ] / n

>is a consistent but biased estimator for R. Does anyone know of
>any RIGOROUS treatment on the variance of this estimator ?
>Particularly, I am interested in some sufficient conditions of X(i),Y(i),
>that imply
>                           ^
>lim_n_to_infinity   n var( R(n) ) 

>         E(X**2)    E(Y**2)      E(X*Y)
>= R**2 { -------  + ------- - 2--------- }.
>         m(x)**2    m(y)**2    m(x)*m(y)

You can derive a closely related approximation using the "delta
method."  Let {Z_i} be a sequence of i.i.d. random Kx1 vectors, with
mean vector m, and finite covariance matrix V.  Let Zbar(n) be the
vector of sample means of the first n elements of the {Z_i} sequence.
Let g map K-dimensional space to J-dimensional space with J .LE. K,
and let g be continuously differentiable on a neighborhood of m with G
= g'(m).  Then by a Taylor expansion

	n^(1/2)*[ g(Zbar(n)) - g(m) ] = G*d + e

where   d = n^(1/2)[ Zbar(n) - m ]   and  e = o[ n^(-1/2)*|d| ].

By standard central limit theory d ---> Normal( 0, V ).  This in turn
implies that e --> 0 in probability and G*d ---> Normal(0,W), where

         W = G*V*transpose(G)

Thus   n^(1/2)* [ g(Zbar(n)) - g(m) ] ---> Normal(0,W).

To apply this to your problem let K = 2, J = 1 and set 
Z_i = (X_i,Y_i).  Set

	g(z) = z(1) / z(2)  so that G = [ 1 / m(y),  -m(x)/m(y)^2 ].

[Implicitly m(y) != 0.]  Then

     W = Var(X)/m(Y)^2 + Var(Y)*m(x)^2/m(y)^4 - 2*Cov(X,Y)*m(x)/m(y)^3

       = R^2 * [ Var(X)/m(x)^2 + Var(Y)/m(y)^2 - 2*Cov(X,Y)/(m(x)*m(y)) ]

which coincides (almost) with your formula.  (You incorrectly used the
absolute second moments of X and Y instead of the variances and
covariances.)

Applying the preceeding argument gives:

                      ^
(*)        n^(1/2)* [ R(n) - R ] ---> Normal( 0, W ) in distribution

as n -> infinity.  Most likely this is the result that you really
want.  It is all that is needed for construction of an asymptotically
valid confidence interval or test statistic, for example.

The result that you asked for, namely

                   ^
            n Var( R(n) ) ---> W  as n -> infinity

does not follow from (*) unless you make stronger assumptions.  Indeed
(*) will be true in cases where the variance of your estimator does
not exist for any value of n.

I have said enough.  Who wants to fill in the rest?
--
T. Scott Thompson              email:  thompson@atlas.socsci.umn.edu
Department of Economics        phone:  (612) 625-0119
University of Minnesota        fax:    (612) 624-0209