2
$\begingroup$

I am reading a paper on dual frame surveys which has confused me.

Without going too deep into the notation, Hartley has an estimator for dual frame surveys which we will denote $\hat{Y}(\theta)$. This takes both samples from the two frames and performs a weighted average estimate for the overlap domain $Y_{ab}$.

Now Fuller and Burmeister (retrieved here) suggests that if we take additional measurements in our survey such as $x_i$ then incorporating the difference $\hat{X}_{ab}^A-\hat{X}_{ab}^B$ into this estimate could reduce the variance, where $\hat{X}_{ab}^A$ is the estimated total $X$ for the overlap of the two frames using the sample from frame $A$, etc.

My question first of all is whether or not this is true? and if so, why?

I can't understand why adding additional terms which do not measure $y_i$ would help reduce the variance of our estimate of the population total $Y$.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

Consider two variables $U$ and $V$. From basic properties of variances,

$\text{Var}(U-V) = \text{Var}(U) + \text{Var}(V) - 2 \text{Cov}(U,V)$

So if $ \text{Cov}(U,V)>Var(V)/2$ then $\text{Var}(U-V) < \text{Var}(U)$

So if $E(V)=0$ but $\text{Cov}(U,V)$ is reasonably large, $U-V$ has the same expectation but smaller variance than $U$.

The trick is to construct some $V$ that is sufficiently positively related (or equivalently a $W=-V$ that is sufficiently negatively related to add) to the quantity we're using as our original estimator.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.