# Upper bound for covariance of Hortvitz-Thompson Estimators

I need to bound on a covariance quantity that has come up in a sampling problem. $$\widehat{Y}$$ and $$\widehat{T}$$ are Horvitz-Thompson estimators of population totals, $$Y=\sum_{i=1}^N y_i$$ and $$T=\sum_{i=1}^N t_i$$, and take the following forms: $$\widehat{Y} = \sum_{i=1}^N \frac{Z_i y_i}{p_i}$$ $$\widehat{T} = \sum_{i=1}^N \frac{Z_i t_i}{p_i}$$

where $$Z_i \in \{0, 1\}$$ is a binary random variable indicating whether unit $$i$$ is included in the sample or not, $$p_i = P(Z_i = 1)$$ is the known probability of unit $$i$$ inclusion in the sample. $$Z_i$$'s are not independent and their joint distribution is known. In other words, we know the quantities $$p_{ij} = P(Z_i=1 \,\&\, Z_j=1)$$. Finally, we can assume $$y_i$$'s and $$t_i$$'s are finite and positive.

Can one find a minimal $$A$$ that satisfies the expression below? $$Var(\widehat{T})Var(\widehat{Y}) - 2E[\widehat{T}] Cov(\widehat{Y}, \widehat{Y}\widehat{T}) + A \ge 0$$ The quantity above shows up in expectation of a variance estimator and satisfying the inequality above implies the estimate is conservative (in expectation). Ideally I need $$A$$ to be in terms of simple quantities for which I have unbiased or conservative estimates (e.g. I have an unbiased estimate for $$E[\widehat{Y}^2]$$ or $$E[\widehat{T}]$$ but I don't have a good estimate for $$Cov(\widehat{Y}, \widehat{T})E[\widehat{Y}]$$).

This problem is equivalent to finding an upper-bound on the following covariance quantity:

$$Cov(\widehat{Y}, \widehat{Y}\widehat{T})$$

I tried to use Young inequality $$ab \le \frac{a^2}{2} + \frac{b^2}{2}$$ and Cauchy-Swartz, but both ended up with a complicated expression involving 3rd or 4th moments. Appreciate any insights.

• I think you might have overgeneralized your problem. For instance, in the general setting the covariance between $Y$ and $YT$ can be infinite. Also, the joint probabilities of the $Z_i$ are usually known in this setting.
– whuber
Commented Aug 1, 2023 at 20:34
• Thanks @whuber. Modified the question a bit with more context. Commented Aug 2, 2023 at 2:40

As pointed out in the comments, in general the covariance could be infinite, so the upper bound might be infinity. Nevertheless, it might benefit you to consider the following decomposition (which assumes that all relevant quantities are finite):

\begin{align} \mathbb{C}(Y,YT) &= \mathbb{E}(Y^2 T) - \mathbb{E}(Y) \mathbb{E}(YT) \\[6pt] &= \mathbb{E}(Y^2 T) - \mathbb{E}(Y) [ \mathbb{E}(Y) \mathbb{E}(T) + \mathbb{C}(Y,T) ] \\[6pt] &= \mathbb{E}(Y^2 T) - \mathbb{E}(Y)^2 \mathbb{E}(T) - \mathbb{C}(Y,T) \mathbb{E}(Y) \\[6pt] &= \mathbb{E}(Y^2 T) - \mathbb{E}(Y^2) \mathbb{E}(T) + [\mathbb{E}(Y^2) - \mathbb{E}(Y)^2] \mathbb{E}(T) - \mathbb{C}(Y,T) \mathbb{E}(Y) \\[6pt] &= \mathbb{C}(Y^2, T) + \mathbb{V}(Y) \mathbb{E}(T) - \mathbb{C}(Y,T) \mathbb{E}(Y) \\[6pt] &= \mathbb{C}(Y^2, T) - \mathbb{C}(Y,T) \mathbb{E}(Y) + \mathbb{V}(Y) \mathbb{E}(T). \\[6pt] \end{align}

This decomposition allows you to express $$\mathbb{C}(Y,YT)$$ in terms of the underlying moments of $$Y$$, $$Y^2$$ and $$T$$. It also "separates" $$Y$$ and $$T$$ in the sense that you no longer have a term $$YT$$ (though you do still look at covariances between these parts). If you are willing to specify relevant moment parameters for the $$Y_i$$ and $$T_i$$ values (and specify if they are independent) then you might be able to obtain the relevant moments in the decomposition, which would let you use this decomposition to compute the covariance at issue.

• Thank you @Ben. I added more context to my question while you were responding. This decomposition works great, but the issue is I don't have an unbiased estimate for $Cov(Y, T)E(Y)$ in your solution. Ideally, the upper bound should be a quantity for which I have an unbiased estimate, so when I subtract the estimate, the my whole expression becomes upper-bounded by zero in expectation. Commented Aug 2, 2023 at 4:16
• At line (4) you inserted a term that is equivalent to $\mathbb V(Y)\mathbb E(T),$ thereby altering the value and ruining the equality. It believe you meant not to include the "$-\mathbb E(Y^2)\mathbb E(T)$" term when you did that. It's unclear what these manipulations accomplish, though: they only re-express one third multivariate moment of $(Y,T)$ in terms of other moments through third order.
– whuber
Commented Aug 2, 2023 at 13:37
• @whuber: I'm not seeing the error in line (4) --- I think I just add and subtract the same term (equivalent to adding zero). Re the purpose, I suppose I see the value here as being that it separates the $Y$ and $T$ in each part of the decomposition (which might potentially be useful if there are independence assumptions that simplify things).
– Ben
Commented Aug 2, 2023 at 14:10
• Thanks (+1) -- I wasn't reading the parentheses correctly. I suppose if one considers a covariance between a function of $Y$ and a function of $T$ as "separating" them, then you have accomplished that.
– whuber
Commented Aug 2, 2023 at 17:48
• I confess that this answer is really only a partial step towards potentially solving the issue --- it might or might not be useful depending on the underlying independence assumptions for the pieces making up $Y$ and $T$. The "separation" shown here could potentially be useful in my opinion.
– Ben
Commented Aug 3, 2023 at 0:34