Estimating cell probabilities in contingency tables with constraints on marginals/conditionals by geometric programming with applications

Abstract

Contingency tables are often used to display the multivariate frequency distribution of variables of interest. Under the common multinomial assumption, the first step of contingency table analysis is to estimate cell probabilities. It is well known that the unconstrained maximum likelihood estimator (MLE) is given by cell counts divided by the total number of observations. However, in the presence of (complex) constraints on the unknown cell probabilities or their functions, the MLE or other types of estimators may often have no closed form and have to be obtained numerically. In this paper, we focus on finding the MLE of cell probabilities in contingency tables under two common types of constraints: known marginals and ordered marginals/conditionals, and propose a novel approach based on geometric programming. We present two important applications that illustrate the usefulness of our approach via comparison with existing methods. Further, we show that our GP-based approach is flexible, readily implementable, effort-saving and can provide a unified framework for various types of constrained estimation of cell probabilities in contingency tables.

Appendices

Appendix A: Proof of Theorem 1

Theorem 1 deals with three situations in three-way contingency tables, as described in Sect. 2.1.

Proof for situation (i):

The relaxed problem is a GP because (i) the objective is a monomial function that is a special case of posynomial functions; and (ii) all “$\le$” constraints are posynomial functions of $p_{ijk}$. Let $\{\bar{p}_{ijk}\}$ denote one optimal solution to the GP.

• 1. If $\{\bar{p}_{ijk}\}$ satisfies all the equality constraints in (2.2) , then $\{\bar{p}_{ijk}\}$ provides an optimal solution to (2.2) , including both empty and nonempty cells. The proof is done in this case.

• 2. Suppose there exists at least one constraint with “$<$” held at $\{\bar{p}_{ijk}\}$ so that $\{\bar{p}_{ijk}\}$ cannot provide a solution to the original optimization problem in (2.2). Without loss of generality, assume $\sum _{j,k}\bar{p}_{i^{1}jk}<p_{i^{1}++}$. Then $\sum _{i}\sum _{j,k}\bar{p}_{ijk}<\sum _{i}p_{i++}=1$, which gives $\sum _{j}\sum _{i,k}\bar{p}_{ijk}<\sum _{j}p_{+j+}$ and $\sum _{k}\sum _{i,j}\bar{p}_{ijk}<\sum _{k}p_{++k}$. So there exist $j^{1}$ and $k^{1}$ s.t. $\sum _{i,k}\bar{p}_{ij^{1}k}<p_{+j^{1}+}$ and $\sum _{i,j}\bar{p}_{ijk^{1}}<p_{++k^{1}}$. Then the cell $(i^{1},j^{1},k^{1})$ must be empty, i.e. $n_{i^{1}j^{1}k^{1}}=0$. Otherwise, we can construct $\{\tilde{p}_{ijk}\}$ by increasing $\bar{p}_{i^{1}j^{1}k^{1}}$ by a very small amount while keeping all the other $\bar{p}_{ijk}$ unchanged, so that the related three “$\le$” constraints still hold. Doing so would increase the value of $\mathcal {L}$ if the cell $(i^{1},j^{1},k^{1})$ is nonempty, which contradicts that $\{\bar{p}_{ijk}\}$ is an optimal solution to the GP.

• 3. Since the cell $(i^{1},j^{1},k^{1})$ is empty, we can increase $\bar{p}_{i^{1}j^{1}k^{1}}$, while keeping the probabilities of all the other cells unchanged (without increasing the value of $\mathcal {L}$), until at least one of the three equality constraints $\sum _{j,k}p_{i^{1}jk}=p_{i^{1}++}$, $\sum _{i,k}p_{ij^{1}k}=p_{+j^{1}+}$ and $\sum _{i,j}p_{ijk^{1}}=p_{++k^{1}}$ holds (i.e., the one with the least slack). Denote the increased value by $\bar{p}_{i^{1}j^{1}k^{1}}^{*}$. Now update $\{\bar{p}_{ijk}\}$ by letting $\bar{p}_{i^{1}j^{1}k^{1}}=\bar{p}_{i^{1}j^{1}k^{1}}^{*}$. Note that the updated $\{\bar{p}_{ijk}\}$ not only provides an alternative optimal solution to the GP, but also changes at least one inequality constraint from the strict “$<$” sign to the “$=$” sign.

• 4. Repeat steps 1 through 3 until all the “$<$” constraints are changed to the corresponding “$=$” constraints.

After the above steps, the adjusted $\{\bar{p}_{ijk}\}$ provides an optimal solution to both the GP and the original optimization problem in (2.2). Note that all the adjustments are done to empty cells. Thus, the GP provides an optimal solution to all non-empty cells in (2.2).

Proof for situation (ii):

This proof is omitted for brevity because it is similar to the proof in situation (i) with very slight modifications.

Proof for situation (iii):

Let $\{\bar{p}_{ijk}\}$ denote one optimal solution to the GP.

• 1. If $\{\bar{p}_{ijk}\}$ satisfies all the equality constraints in (2.4) , then $\{\bar{p}_{ijk}\}$ provides an optimal solution to (2.4), including both empty and nonempty cells. This concludes the proof in this case.

• 2. Suppose there exists at least one constraint in which “$<$” holds at $\{\bar{p}_{ijk}\}$ so that $\{\bar{p}_{ijk}\}$ cannot provide a solution to the original optimization problem in (2.4). Without loss of generality, assume $\sum _{k}\bar{p}_{i^{*}j^{*}k}<p_{i^{*}j^{*}+}$. Then $\sum _{j}\sum _{k}\bar{p}_{i^{*}jk}<\sum _{j}p_{i^{*}j+}=\sum _{k}p_{i^{*}+k}$, which gives $\sum _{k}\sum _{j}\bar{p}_{i^{*}jk} <\sum _{k}p_{i^{*}+k}$. So there exists $k^{*}$ s.t. $\sum _{j}\bar{p}_{i^{*}jk^{*}}<p_{i^{*}+k^{*}}$. Then the cell $(i^{*},j^{*},k^{*})$ must be empty, i.e. $n_{i^{*}j^{*}k^{*}}=0$. Otherwise, we can construct $\{\tilde{p}_{ijk}\}$ by increasing $\bar{p}_{i^{*}j^{*}k^{*}}$ by a very small amount while keeping all the other $\bar{p}_{ijk}$ unchanged, so that the related three “$\le$” constraints still hold. Doing so would increase the value of $\mathcal {L}$ if the cell $(i^{*},j^{*},k^{*})$ is nonempty, which contradicts that $\{\bar{p}_{ijk}\}$ is an optimal solution to the GP.

• 3. Since the cell $(i^{*},j^{*},k^{*})$ is empty, we can increase $\bar{p}_{i^{*}j^{*}k^{*}}$, while keeping the probabilities of all the other cells unchanged (without increasing the value of $\mathcal {L}$), until at least one of the three equality constraints $\sum _{j,k}p_{i^{*}jk}=p_{i^{*}++}$, $\sum _{i,k}p_{ij^{*}k}=p_{+j^{*}+}$ and $\sum _{i,j}p_{ijk^{*}}=p_{++k^{*}}$ holds (i.e., the one with the least slack). Denote the increased value as $\bar{p}_{i^{*}j^{*}k^{*}}^{*}$. Now update $\{\bar{p}_{ijk}\}$ by letting $\bar{p}_{i^{*}j^{*}k^{*}}=\bar{p}_{i^{*}j^{*}k^{*}}^{*}$. Note that the updated $\{\bar{p}_{ijk}\}$ not only provides an alternative optimal solution to the GP, but also changes at least one inequality constraint from the strict “$<$” sign to the “$=$” sign.

• 4. Repeat steps 1 through 3 until all the “$<$” constraints are changed to the corresponding “$=$” constraints.

After the above steps, the adjusted $\{\bar{p}_{ijk}\}$ provides an optimal solution to both the GP and the original optimization problem in (2.4). Note that all the adjustments are done to empty cells. Thus, the GP provides an optimal solution to all non-empty cells in (2.4).

Appendix B: Proof of Theorem 4

We need to show that the GP (3.5) achieves the optimal value only when (3.4) is satisfied. Let $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$ denote the optimal solution to the GP. Suppose there exists at least one of the terms among $\sum _{jk}p_{jk|i}$ and $\sum _{i}p_{i++}$ with “$<$1” held at $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$.

If for some $i$, say $i^{*},\, \sum _{jk}\bar{p}_{jk|i^{*}}<1$, then we can obtain $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ by setting $\tilde{p}_{j^{*}k^{*}|i^{*}}=\bar{p}_{j^{*}k^{*}|i^{*}}+1-\sum _{jk}\bar{p}_{jk|i^{*}}$, $\tilde{p}_{jk|i}=\bar{p}_{jk|i}$ for $i\ne i^{*}$ or $j\ne j^{*}$ or $k\ne k^{*}$, and $\tilde{p}_{i++}=\bar{p}_{i++}$ for all $i$, where $(j^{*},k^{*})$ satisfies $n_{i^{*}j^{*}k^{*}}>0$. Note that $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ satisfies all the inequality constraints in (3.5) with $\sum _{jk}\tilde{p}_{jk|i^{*}}=1$. Since $\tilde{p}_{j^{*}k^{*}|i^{*}}>\bar{p}_{j^{*}k^{*}|i^{*}}$, the objective function is smaller at $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$, indicating $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$ is not optimal. Hence, $\sum _{jk}\bar{p}_{jk|i}=1$ must hold for all $i$.

If $\sum _{i}\bar{p}_{i++}<1$, then we can obtain $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ by setting $\tilde{p}_{r++}=\bar{p}_{r++}+1-\sum _{i}\bar{p}_{i++}$, $\tilde{p}_{i++}=\bar{p}_{i++}$ for $i\ne r$, and $\tilde{p}_{jk|i}=\bar{p}_{jk|i}$ for all $i,j,k$. Note that

$$\begin{aligned} \bar{p}_{1++}=\tilde{p}_{1++}\le \bar{p}_{2++}=\tilde{p}_{2++}\le \cdots \le \bar{p}_{r++}<\tilde{p}_{r++} \end{aligned}$$

and $\sum _{i}\tilde{p}_{i++}=1$ so that $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ satisfies all the inequality constraints in (3.6). Since $\tilde{p}_{r++}>\bar{p}_{r++}$, the objective function is smaller at $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$, indicating $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$ is not optimal. Hence, $\sum _{i}\bar{p}_{i++}=1$ must hold.

Appendix C: Proof of Theorem 5

Again, we need to show that the GP (3.6) achieves the optimal value only when (3.4) is satisfied. Since $\mathcal {C}\subset \mathcal {J}\times \mathcal {K}$, the complement $\bar{\mathcal {C}}$ is nonempty. Let $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$ denote the optimal solution to the GP. Suppose there exists at least one of the terms among $\sum _{jk}p_{jk|i}$ and $\sum _{i}p_{i++}$ with “$<$1” held at $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$.

If for some $i$, say $i^{*},\, \sum _{jk}\bar{p}_{jk|i^{*}}<1$, then we can obtain $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ by setting $\tilde{p}_{j^{*}k^{*}|i^{*}}=\bar{p}_{j^{*}k^{*}|i^{*}}+1-\sum _{jk}\bar{p}_{jk|i^{*}}$, $\tilde{p}_{jk|i}=\bar{p}_{jk|i}$ for $i\ne i^{*}$ or $j\ne j^{*}$ or $k\ne k^{*}$, where $(j^{*},k^{*})\in \bar{\mathcal {C}}$ and $n_{i^{*}j^{*}k^{*}}>0$, and $\tilde{p}_{i++}=\bar{p}_{i++}$ for all $i$. Note that $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ satisfies all the inequality constraints in (3.6) with $\sum _{jk}\tilde{p}_{jk|i^{*}}=1$. Since $\tilde{p}_{j^{*}k^{*}|i^{*}}>\bar{p}_{j^{*}k^{*}|i^{*}}$, the objective function is smaller at $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$, indicating $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$ is not optimal. Hence, $\sum _{jk}\bar{p}_{jk|i}=1$ must hold for all $i$.

If $\sum _{i}\bar{p}_{i++}<1$, then a similar argument can be made as in the proof of Theorem 4 to show that $\sum _{i}\bar{p}_{i++}=1$ must hold; that is, we can obtain $\{\tilde{p}_{jk|i},\tilde{p}_{i++}\}$ by setting $\tilde{p}_{1++}=\bar{p}_{1++}+1-\sum _{i}\bar{p}_{i++}$, $\tilde{p}_{i++}=\bar{p}_{i++}$ for $i\ne 1$, and $\tilde{p}_{jk|i}=\bar{p}_{jk|i}$ for all $i,j,k$, which gives a smaller value of the objective function than $\{\bar{p}_{jk|i},\bar{p}_{i++}\}$.