3 Conservative bias correction for the Horvitz-Thompson variance estimator under non-measurability

Peter M. Aronow and Cyrus Samii

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The case where $A$ may be less than zero for a non-measurable design suggests the need for some adjustment that will guarantee a bias that is weakly bounded below by zero. We first develop a general bias correction that is guaranteed to be conservative, later providing a special simplified case for practical usage.

3.1  General formulation

Consider the following variance estimator:

$${\widehat{\text{Var}}}_C(\widehat{t})={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U:{\pi }_{kl}>0\}}{I}_k}}{I}_l\frac{\text{Cov}(I_k,I_l)}{{\pi }_{kl}}\frac{y_k}{{\pi }_k}\frac{y_l}{{\pi }_l}+{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}{\pi }_k}+I_l\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}{\pi }_l}\right)}},$$

where $a_{kl},b_{kl}$ are positive real numbers such that $1/a_{kl}+1/b_{kl}=1$ for all pairs $k,l$ with ${\pi }_{kl}=0.$ The estimator is guaranteed to produce an expected value greater than or equal to the true variance for all designs, and is thus conservative. We state this property formally:

Proposition 2. The expected value of ${\widehat{\text{Var}}}_C(\widehat{t}),$

$\text{E}\left[{\widehat{\text{Var}}}_C(\widehat{t})\right]\ge \text{Var}(\widehat{t}).$

Proof. By Young's inequality,

$$\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}}+\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}}\ge \left|y_k\right|\left|y_l\right|,$$

if $1/a_{kl}+1/b_{kl}=1.$ Define $A^{*}$ such that,

$A^{*}={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}}+\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}}}}\ge {\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left|y_k\right|\left|y_l\right|}}\ge {\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}{y}_k{y}_l}}=A$

and

$A^{*}\ge {\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left|y_k\right|\left|y_l\right|}}\ge {\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}-y_k{y}_l}}=-A.$

Therefore

$\text{Var}(\widehat{t})+A+A^{*}\ge \text{Var}(\widehat{t}).$

The associated Horvitz-Thompson estimator of $A^{*}$ would be

${\widehat{A}}^{*}={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}{\pi }_k}+I_l\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}{\pi }_l}\right)}},$

which is unbiased by $E(I_k)={\pi }_k$ and $E(I_l)={\pi }_l.$

Since $E({\widehat{A}}^{*})=A^{*},$ by Proposition 1,

$$\text{E}\left[{\displaystyle \sum _{k\in s^0}{\displaystyle \sum _{l\in s^0}\frac{\text{Cov}(I_k,I_l)}{{\pi }_{kl}}}}\frac{y_k}{{\pi }_k}\frac{y_l}{{\pi }_l}+{\widehat{A}}^{*}\right]=\text{Var}(\widehat{t})+A+A^{*}$$

$$\text{E}\left[{\widehat{\text{Var}}}_C(\widehat{t})\right]\ge \text{Var}(\widehat{t}).$$

Substituting terms,

$$\text{E}\left[{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U:{\pi }_{kl}>0\}}{I}_k}}{I}_l\frac{\text{Cov}(I_k,I_l)}{{\pi }_{kl}}\frac{y_k}{{\pi }_k}\frac{y_l}{{\pi }_l}+{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}{\pi }_k}+I_l\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}{\pi }_l}\right)}}\right]\ge \text{Var}(\widehat{t}).$$

${\widehat{\text{Var}}}_C(\widehat{t})$ is justified as a conservative estimator for the case when $A$ is not known to be positive. This estimator is unbiased under a special condition:

Corollary 1. If, for all pairs $k,l$ such that ${\pi }_{kl}=0,$ (i) ${\left|y_k\right|}^{a_{kl}}={\left|y_l\right|}^{b_{kl}}$ and (ii) $-y_k{y}_l=\left|y_k\right|\left|y_l\right|,$ 

$\text{E}\left[{\widehat{\text{Var}}}_C(\widehat{t})\right]=\text{Var}(\widehat{t}).$

Proof. By (i), (ii) and Young's inequality,

$$\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}}+\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}}=\left|y_k\right|\left|y_l\right|=-y_k{y}_l.$$

Therefore,

$A^{*}={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\frac{{\left|y_k\right|}^{a_{kl}}}{a_{kl}}+\frac{{\left|y_l\right|}^{b_{kl}}}{b_{kl}}}}={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left|y_k\right|\left|y_l\right|}}={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}-y_k{y}_l}}=-A.$

It follows that $\text{Var}(\widehat{t})+A+A^{*}=\text{Var}(\widehat{t})$ and $\text{E}\left[{\widehat{\text{Var}}}_C(\widehat{t})\right]=\text{Var}(\widehat{t}).$

If any units $k,l$ are in clusters (i.e., $\mathrm{Pr}(I_k\ne I_l)=0),$ these units should be totaled into one larger unit before estimation. Combining units will tend to reduce the bias of the variance estimator because only pairs of cluster-level totals will be included in $A^{*},$ as opposed to all constituent pairs.

3.2  Simplified special case

In general, it would be difficult to assign optimal values of $a_{kl}$ and $b_{kl}$ for all pairs $k,l$ such that ${\pi }_{kl}=0.$ Instead, we examine one intuitive case, assigning all $a_{kl}=b_{kl}=2:$

$${\widehat{\text{Var}}}_{C2}(\widehat{t})={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U:{\pi }_{kl}>0\}}{I}_k}}{I}_l\frac{\text{Cov}(I_k,I_l)}{{\pi }_{kl}}\frac{y_k}{{\pi }_k}\frac{y_l}{{\pi }_l}+{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{y_k^2}{2{\pi }_k}+I_l\frac{y_l^2}{2{\pi }_l}\right)}}.$$

As a special case of ${\widehat{\text{Var}}}_C(\widehat{t}),{\widehat{\text{Var}}}_{C2}(\widehat{t})$ is also conservative:

Corollary 2. The expected value of ${\widehat{\text{Var}}}_{C2}(\widehat{t}),$

$\text{E}\left[{\widehat{\text{Var}}}_{C2}(\widehat{t})\right]\ge \text{Var}(\widehat{t}).$

Proof. For all pairs $k,l$ such that ${\pi }_{kl}=0,1/a_{kl}+1/b_{kl}=1.$ Proposition 1 therefore holds.

The choice to set all $a_{kl}=b_{kl}=2$ is justified by the fact that it will yield the lowest value of the estimator ${\widehat{\text{Var}}}_C(\widehat{t})$ subject to the constraint that $a_{kl}$ and $b_{kl}$ are fixed as constants $a$ and $b$ over all $k,l.$

Corollary 3. Among the class of estimators ${\widehat{\text{Var}}}_{\text{Cab}}(\widehat{t}),$ defined as the set of estimators ${\widehat{\text{Var}}}_C(\widehat{t})$ such that all $a_{kl}=a$ and all $b_{kl}=b,{\widehat{\text{Var}}}_{C2}(\widehat{t})={\min }_{a,b}\left[{\widehat{\text{Var}}}_{\text{Cab}}(\widehat{t})\right].$

Proof. By simple algebra,

$$\begin{array}{cc}\hfill {\widehat{\text{Var}}}_{\text{Cab}}(\widehat{t})& ={\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U:{\pi }_{kl}>0\}}{I}_k}}{I}_l\frac{\text{Cov}(I_k,I_l)}{{\pi }_{kl}}\frac{y_k}{{\pi }_k}\frac{y_l}{{\pi }_l}+{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^a}{a{\pi }_k}+I_l\frac{{\left|y_l\right|}^b}{b{\pi }_l}\right)}}\\ & =\widehat{\text{Var}}(\widehat{t})+{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^a}{a{\pi }_k}+I_l\frac{{\left|y_l\right|}^b}{b{\pi }_l}\right)}}\hfill \\ & =\widehat{\text{Var}}(\widehat{t})+\frac{1}{2}{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(I_k\frac{{\left|y_k\right|}^a}{a{\pi }_k}+I_l\frac{{\left|y_l\right|}^b}{b{\pi }_l}+I_k\frac{{\left|y_k\right|}^b}{b{\pi }_k}+I_l\frac{{\left|y_l\right|}^a}{a{\pi }_l}\right)}}\hfill \\ & =\widehat{\text{Var}}(\widehat{t})+\frac{1}{2}{\displaystyle \sum _{k\in U}{\displaystyle \sum _{l\in \{U\backslash k:{\pi }_{kl}=0\}}\left(\frac{I_k}{{\pi }_k}\left(\frac{{\left|y_k\right|}^a}{a}+\frac{{\left|y_k\right|}^b}{b}\right)+\frac{I_l}{{\pi }_l}\left(\frac{{\left|y_l\right|}^a}{a}+\frac{{\left|y_l\right|}^b}{b}\right)\right)}}.\hfill \end{array}$$

By Young's inequality, given $1/a+1/b=1,{\left|y_k\right|}^a/a+{\left|y_k\right|}^b/b\ge y_k^2,$ but equality must hold if $a=b=2.$ Similarly, ${\left|y_l\right|}^a/a+{\left|y_l\right|}^b/b\ge y_l^2,$ but equality must hold if $a=b=2.$ Since $I_k/{\pi }_k\ge 0$ and $I_l/{\pi }_l\ge 0,$ any choice $a\ne b$ can only yield ${\widehat{\text{Var}}}_{\text{Cab}}(\widehat{t})\ge {\widehat{\text{Var}}}_{C2}(\widehat{t}).$

Given all values of $y_k$ and $y_l,$ it is possible to derive an optimal vector of $a_{kl}$ and $b_{kl}$ values that varies over $k,l,$ but such a derivation may not be of practical value.

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