Conditional Allocation Given a Pooled Count

Why Condition on a Pooled Count?

Suppose a study records outcomes in several categories, but an analysis focuses only on a selected group of them. The total number of observations in that group may be known, while the separate counts within it are not. The natural question is then: given the pooled total, how is that total allocated among the selected categories?

The answer is another multinomial distribution. Its sample size is the observed pooled total, and its category probabilities are the original probabilities renormalized to sum to one within the selected group. This conditional law describes the allocation without needing to model how many observations fell outside the group.

The \(q_i\)'s describe the relative shares of the selected categories. They sum to one because \(\sum_{i\in S}q_i=\sum_{i\in S}p_i/p_S=1\). The key fact is that, once \(M=m\) is given, the count vector within the selected set has the same form as counts from \(m\) trials with probabilities \(q_i\).

The Conditional Allocation Theorem

Here \(m\) must be a possible pooled count. If \(0<p_S<1\), every integer \(m\) from 0 to \(n\) is possible. If \(p_S=1\), then \(M=n\) with probability one, so the only possible conditioning value is \(m=n\). A category with \(p_i=0\) has \(q_i=0\) and must receive count zero with conditional probability one.

To prove the result when \(0<p_S<1\), combine all categories outside \(S\) into one group. The joint probability of the selected counts \(x_i\) and the pooled total \(m\) is \[ \frac{n!}{(n-m)!\prod_{i\in S}x_i!}(1-p_S)^{n-m}\prod_{i\in S}p_i^{x_i}. \] The pooled count \(M\) is binomial with parameters \(n\) and \(p_S\), so \[ P(M=m)=\frac{n!}{m!(n-m)!}p_S^m(1-p_S)^{n-m}. \] Dividing the joint probability by this positive marginal probability gives \[ \begin{aligned} P\bigl(X_i=x_i,\ i\in S\mid M=m\bigr) &=\frac{n!(1-p_S)^{n-m}\prod_{i\in S}p_i^{x_i}} {(n-m)!\prod_{i\in S}x_i!} \cdot \frac{m!(n-m)!}{n!p_S^m(1-p_S)^{n-m}}\\ &=\frac{m!}{\prod_{i\in S}x_i!} \prod_{i\in S}\left(\frac{p_i}{p_S}\right)^{x_i}, \end{aligned} \] where the last equality uses \(\sum_{i\in S}x_i=m\). If \(p_S=1\), all trials fall in \(S\), so \(M=n\); the original multinomial probability already has the asserted form with \(m=n\) and \(q_i=p_i\). This proves the theorem. \(\square\)

The formula has a useful interpretation: conditional on being in the selected group, a trial has probability \(q_i\) of belonging to category \(i\). Conditioning on the group total fixes how many trials are allocated within the group, not the allocation itself.

Worked Allocations

Conditional Means and Dependence

Once the conditional multinomial law is established, its moments describe the expected allocation and the way its components vary together. For each selected category \(i\), the conditional mean is proportional to its share \(q_i\). Counts in distinct selected categories are negatively associated: if one category receives more of a fixed total, fewer observations remain for the others.

For a derivation, represent the conditional allocation as the counts from \(m\) independent categorical trials with probabilities \(q_i\), which have exactly the conditional multinomial probability just derived. Let \(I_{\ell i}\) equal 1 when trial \(\ell\) is in category \(i\), and 0 otherwise. Then \(X_i=\sum_{\ell=1}^m I_{\ell i}\), with \(E[I_{\ell i}]=q_i\) and \(\operatorname{Var}(I_{\ell i})=q_i(1-q_i)\). Independence across trials gives \[ E[X_i]=\sum_{\ell=1}^m q_i=mq_i,\qquad \operatorname{Var}(X_i)=\sum_{\ell=1}^m q_i(1-q_i)=mq_i(1-q_i). \] For distinct categories \(i,j\), a single trial cannot be in both, so \(I_{\ell i}I_{\ell j}=0\). Thus \[ \operatorname{Cov}(I_{\ell i},I_{\ell j}) =E[I_{\ell i}I_{\ell j}]-E[I_{\ell i}]E[I_{\ell j}] =0-q_iq_j=-q_iq_j. \] Indicators from different trials are independent and have covariance zero. Summing the covariances over the \(m\) trials gives \(\operatorname{Cov}(X_i,X_j)=-mq_iq_j\). These calculations apply to the conditional law, proving all three formulas. \(\square\)

The negative covariance is a consequence of the fixed total: \( \sum_{i\in S}X_i=m\). In particular, once all but one selected count are known, the final count is determined. The counts are not conditionally independent, except in degenerate cases.

Common Mistake: Forgetting to Renormalize

A frequent error is to use the original \(p_i\)'s as the conditional allocation probabilities. In the first example, using 0.2 and 0.3 directly would give probabilities that sum to only 0.5, so they cannot describe the allocation of all 6 observations in the selected group. The correct probabilities are \(0.2/0.5=0.4\) and \(0.3/0.5=0.6\), which sum to one.

Another error is to confuse the pooled event with a specified allocation. Knowing \(M=m\) says only that the selected counts add to \(m\); it does not say how the total is split. The conditional multinomial law supplies the probabilities of the possible splits. To calculate an event involving several splits, add the probabilities of the disjoint allocations it contains.