Composing Thinning and Superposition

Thinning and Superposition in Either Order

Several independent arrival sources may feed one system, while each arrival is classified, retained, or discarded according to its source. The resulting rates can be calculated by thinning each source and then combining the category streams. Alternatively, one can combine the sources first and classify arrivals in the pooled stream. The rates agree when the pooled classification probabilities account for the relative sizes of the source rates.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $N_r(t)$, for $r=1,\ldots,m$, be independent Poisson processes with rates $\lambda_r\ge 0$. Thus, for $0\le s\le t$, the count $N_r(t)-N_r(s)$ has a Poisson distribution with mean $\lambda_r(t-s)$, and counts on disjoint time intervals are independent. Each arrival from source $r$ receives exactly one category $c\in\{1,\ldots,k\}$, independently of the other arrivals and of the source processes, with probability $p_{rc}$. Assume $p_{rc}\ge0$ and $\sum_{c=1}^k p_{rc}=1$ for each source $r$. Probabilities equal to zero or one are allowed.

Definition: The category-$c$ process $N_c(t)$ counts all arrivals from all sources assigned category $c$ by time $t$. The rate contributed by source $r$ to category $c$ is $\lambda_r p_{rc}$. The total rate of category $c$ is therefore $R_c=\sum_{r=1}^m\lambda_rp_{rc}$.

The formula for $R_c$ adds source-specific contributions, but the formula alone does not establish that category counts are Poisson or that different categories are independent. The next result proves both facts and gives the precise sense in which thinning and superposition can be composed.

Theorem (Composition of thinning and superposition): Under the assumptions above, the processes $N_1,\ldots,N_k$ are mutually independent Poisson processes with respective rates $R_1,\ldots,R_k$. If $\Lambda=\sum_{r=1}^m\lambda_r>0$, then the same category-specific rates result by first superposing the source processes into a rate-$\Lambda$ process and assigning each pooled arrival category $c$ with probability $$ q_c=\frac{R_c}{\Lambda} =\sum_{r=1}^m\frac{\lambda_r}{\Lambda}p_{rc}. $$

Proof. Fix an interval of length $h\ge0$, and let $Z_c$ be the number of source-$r$ arrivals in that interval assigned category $c$. Conditional on there being $n$ source-$r$ arrivals, their independent category assignments give the probability generating function $(\sum_{c=1}^k p_{rc}z_c)^n$. Since the source count is Poisson with mean $\lambda_rh$, its unconditional joint probability generating function is

\begin{aligned} \mathbb{E}\left[\prod_{c=1}^k z_c^{Z_c}\right] &=\sum_{n=0}^{\infty}e^{-\lambda_rh}\frac{(\lambda_rh)^n}{n!} \left(\sum_{c=1}^k p_{rc}z_c\right)^n\\ &=\exp\left(\lambda_rh\left(\sum_{c=1}^k p_{rc}z_c-1\right)\right)\\ &=\prod_{c=1}^k\exp\left(\lambda_rhp_{rc}(z_c-1)\right). \end{aligned}

The last expression is a product of the probability generating functions of independent Poisson variables with means $\lambda_rhp_{rc}$. Thus, within this interval, the source-$r$ category counts are independent Poisson variables with those means. Different source processes and their assignments are independent, so multiplying the generating functions over $r$ gives

\prod_{r=1}^m\prod_{c=1}^k \exp\left(\lambda_rhp_{rc}(z_c-1)\right) = \prod_{c=1}^k\exp\left(hR_c(z_c-1)\right).

This is the joint probability generating function of independent Poisson counts, one for each category, with means $hR_c$. The argument applies separately to any time interval. Counts on disjoint intervals are independent because the original source processes have independent increments and the category assignments are independent across arrivals. Hence each $N_c(t)$ is a Poisson process of rate $R_c$, and the category processes are mutually independent.

For the pooled construction, the superposition of the source processes is a Poisson process with rate $\Lambda$, by the superposition result. A pooled arrival has source $r$ with probability $\lambda_r/\Lambda$, as in the source-label distribution for superposed streams. Its probability of category $c$ is consequently $\sum_r(\lambda_r/\Lambda)p_{rc}=R_c/\Lambda=q_c$. These probabilities sum to one because $$ \sum_{c=1}^k q_c =\frac{1}{\Lambda}\sum_{r=1}^m\lambda_r\sum_{c=1}^k p_{rc} =\frac{1}{\Lambda}\sum_{r=1}^m\lambda_r=1. $$ Thinning the pooled rate-$\Lambda$ process with these probabilities gives category-$c$ rate $\Lambda q_c=R_c$. The same generating-function argument proves that the resulting category processes are independent Poisson processes. If $\Lambda=0$, every source rate is zero, so all category processes are identically zero; the ratio defining $q_c$ is unnecessary. $\square$

The theorem compares category-specific processes, not necessarily every detail retained in the data. If source labels are kept, source and category jointly identify the contribution $\lambda_rp_{rc}$. If source labels are discarded, choosing the pooled probabilities $q_c$ preserves the category processes but does not preserve information about which source produced a particular arrival.

Worked Example: Two Sources and Three Categories

Source 1 has rate $6$, and source 2 has rate $3$. Each arrival is assigned to category $A$, category $B$, or category $C$, with probabilities $(\tfrac12,\tfrac13,\tfrac16)$ for source 1 and $(\tfrac13,\tfrac12,\tfrac16)$ for source 2. Find the category rates, then find the probabilities to use if the sources are superposed first.

For source 1, the three category rates are $6\cdot\tfrac12=3$, $6\cdot\tfrac13=2$, and $6\cdot\tfrac16=1$. These probabilities sum to $\tfrac12+\tfrac13+\tfrac16=1$. For source 2, they are $3\cdot\tfrac13=1$, $3\cdot\tfrac12=\tfrac32$, and $3\cdot\tfrac16=\tfrac12$; its probabilities also sum to $\tfrac13+\tfrac12+\tfrac16=1$. Adding the contributions gives

R_A=3+1=4,\qquad R_B=2+\tfrac32=\tfrac72,\qquad R_C=1+\tfrac12=\tfrac32.

The pooled rate is $\Lambda=6+3=9$. The corresponding pooled probabilities are $q_A=4/9$, $q_B=(7/2)/9=7/18$, and $q_C=(3/2)/9=1/6$. They sum to $4/9+7/18+1/6=8/18+7/18+3/18=1$. Multiplication by the pooled rate checks the rates: $9(4/9)=4$, $9(7/18)=7/2$, and $9(1/6)=3/2$. The three category processes are mutually independent Poisson processes with these rates.

Combining Successive Thinning Stages

Thinning may occur in stages. For example, an arrival may first be retained and then, if retained, assigned to one of several categories. When the stage decisions are independent across arrivals, the probability of a final category is the product of the probabilities along its path through the stages. This converts a multistage procedure into the single categorical assignment used in the theorem.

Proposition: Suppose source $r$ has rate $\lambda_r$, and an arrival is first retained with probability $a_r$. Conditional on retention, it is assigned category $c$ with probability $b_{rc}$, where $\sum_c b_{rc}=1$. If the two decisions are made independently across arrivals, the final category-$c$ rate from source $r$ is $\lambda_ra_rb_{rc}$, and the final category processes from all sources are independent Poisson processes.

Proof. For one source-$r$ arrival, the probability of ending in category $c$ is the probability of retention times the conditional probability of category $c$, namely $a_rb_{rc}$. The probability of being discarded at the first stage is $1-a_r$. Thus the full procedure is a single categorical assignment with probabilities $a_rb_{rc}$ for category $c$ and $1-a_r$ for first-stage discard; these probabilities sum to $a_r\sum_c b_{rc}+1-a_r=1$. Apply the generating-function calculation in the theorem to this enlarged set of categories. It gives independent Poisson category processes with rates $\lambda_ra_rb_{rc}$, as claimed. $\square$

Worked Example: Retain, Then Classify

Two independent sources have rates $8$ and $4$. Source 1 is retained with probability $\tfrac34$, then classified as $A$ with conditional probability $\tfrac23$ and $B$ with probability $\tfrac13$. Source 2 is retained with probability $\tfrac12$, then classified as $A$ with conditional probability $\tfrac14$ and $B$ with probability $\tfrac34$. Find the final rates of $A$, $B$, and all discarded arrivals.

For source 1, the probability of final category $A$ is $\tfrac34\cdot\tfrac23=\tfrac12$, so its $A$ rate is $8\cdot\tfrac12=4$. Its probability of $B$ is $\tfrac34\cdot\tfrac13=\tfrac14$, giving $B$ rate $8\cdot\tfrac14=2$. The probability discarded at the first stage is $1-\tfrac34=\tfrac14$, giving discarded rate $8\cdot\tfrac14=2$.

For source 2, the probability of $A$ is $\tfrac12\cdot\tfrac14=\tfrac18$, giving rate $4\cdot\tfrac18=\tfrac12$. The probability of $B$ is $\tfrac12\cdot\tfrac34=\tfrac38$, giving rate $4\cdot\tfrac38=\tfrac32$. The first-stage discard probability is $1-\tfrac12=\tfrac12$, giving rate $4\cdot\tfrac12=2$. Combining sources yields

R_A=4+\tfrac12=\tfrac92,\qquad R_B=2+\tfrac32=\tfrac72,\qquad R_{\text{discard}}=2+2=4.

Check: The output rates sum to $\tfrac92+\tfrac72+4=8+4=12$, the total input rate. The per-source probabilities also sum to one: for source 1, $\tfrac12+\tfrac14+\tfrac14=1$; for source 2, $\tfrac18+\tfrac38+\tfrac12=1$. Therefore the rates account for every arrival exactly once.

Rates in a Combined Model

Once the category rates are known, the theorem also gives count distributions on any fixed interval. Over an interval of length $t$, category $c$ has a Poisson count with mean $tR_c$, and the category counts are independent. For example, if $A$ and $B$ are two distinct categories, then the probability of exactly one $A$ arrival and no $B$ arrivals is the product of the corresponding Poisson probabilities. This use of independence depends on assigning each arrival to exactly one category with independent marks; it would not follow merely from knowing the expected rates.

Worked Example: A Count Event After Pooling

Two independent sources have rates $4$ and $2$. Source 1 assigns arrivals to categories $A,B,C$ with probabilities $(\tfrac12,\tfrac14,\tfrac14)$; source 2 uses $(\tfrac13,\tfrac12,\tfrac16)$. Find the probability that during an interval of length $3$ there is exactly one $A$ arrival and no $B$ arrivals, with no restriction on $C$.

The category probabilities for each source sum to one: $\tfrac12+\tfrac14+\tfrac14=1$, and $\tfrac13+\tfrac12+\tfrac16=1$. The category rates are $$ R_A=4\cdot\tfrac12+2\cdot\tfrac13=2+\tfrac23=\tfrac83,\qquad R_B=4\cdot\tfrac14+2\cdot\tfrac12=1+1=2, $$ and $$ R_C=4\cdot\tfrac14+2\cdot\tfrac16=1+\tfrac13=\tfrac43. $$ Their sum is $\tfrac83+2+\tfrac43=4+2=6$, equal to the pooled input rate.

Over length $3$, the $A$ count is Poisson with mean $3\cdot\tfrac83=8$, and the $B$ count is an independent Poisson variable with mean $3\cdot2=6$. Thus the requested probability is $$ \mathbb{P}(N_A(3)=1,\ N_B(3)=0) =\left(e^{-8}\frac{8^1}{1!}\right)\left(e^{-6}\frac{6^0}{0!}\right) =8e^{-14}. $$ The unrestricted $C$ count contributes no extra factor: summing its probabilities over all nonnegative counts gives $1$. This verifies that pooling first and using the category probabilities $q_A=(8/3)/6=4/9$, $q_B=2/6=1/3$, and $q_C=(4/3)/6=2/9$ gives the same answer.

A Common Rate-Weighting Mistake

A frequent error is to average source-specific category probabilities equally when the source rates differ. The pooled probability is a rate-weighted average, not an unweighted average. For instance, suppose rates are $6$ and $3$, and the probability of category $A$ is respectively $\tfrac12$ and $\tfrac13$. The correct category rate is $6\cdot\tfrac12+3\cdot\tfrac13=3+1=4$. The pooled probability is therefore $4/(6+3)=4/9$.

Wrong calculation: The unweighted average probability is $(\tfrac12+\tfrac13)/2=\tfrac5{12}$, which would give pooled rate $9\cdot\tfrac5{12}=\tfrac{15}{4}$, not $4$. The mistake treats a source with rate $6$ as though it contributed the same number of arrivals as a source with rate $3$. The correction is to weight each probability by its source rate: $q_A=\sum_r(\lambda_r/\Lambda)p_{rA}$.

For a reliable calculation, first check that the category probabilities for each source sum to one. Then multiply each source rate by its probability of entering the category, add those contributions, and verify that the category rates sum to the total input rate when every arrival receives exactly one category. If there is a retained-versus-rejected split, include rejection as a category in that final check.

Check Your Understanding

Use the rate-weighting formulas and verify each result by checking the category probabilities or rates.

Two independent sources have rates $5$ and $1$. Their probabilities of category $A$ are $\tfrac25$ and $1$. Find the category-$A$ rate and the pooled probability of $A$.
Explain why mutually exclusive categorical assignments give independent Poisson category processes in the theorem, despite each arrival going to only one category.
A source of rate $10$ is retained with probability $\tfrac35$, then assigned category $B$ with conditional probability $\tfrac14$. Find the $B$ rate and the first-stage discard rate.
For the combined model in the last worked example, check that the pooled category probabilities sum to one and recover each category rate by multiplying by the total rate.
Give a concrete reason why an unweighted average of source-specific category probabilities can give an incorrect pooled rate.

Composing Thinning and Superposition

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