Splitting Poisson Counts Into Categories

Why Split a Poisson Count?

A Poisson model describes a total count over a specified observation window. Often, each counted event also belongs to a category: incoming requests may be assigned a type, or inspected items may be classified by defect. If the total count follows a Poisson distribution and the category labels are assigned independently according to fixed probabilities, the count in each category also has a Poisson distribution. Moreover, the category counts are independent.

The result is useful because it lets us answer questions about particular categories without first listing every possible total count. The key is to distinguish the total event rate from the category probabilities. A category's rate is the total rate multiplied by the probability that an event belongs to that category.

Definition: Suppose the total number of events in a fixed observation window is $N\sim\operatorname{Poisson}(\lambda)$. Each event is independently assigned to exactly one of $k$ categories, with category $j$ having probability $p_j$, where $p_j\geq 0$ and $\sum_{j=1}^{k}p_j=1$. Let $X_j$ be the number of events assigned to category $j$. This is a Poisson splitting model. The category-specific count $X_j$ has a Poisson distribution with mean $\lambda p_j$, and the counts $X_1,\ldots,X_k$ are mutually independent.

The assumptions are substantive. The events must be counted over the same window for the total and all categories; every event must receive one category; and the category probabilities must be fixed and the assignments independent from event to event. The labels must also be assigned independently of the total count. Under these conditions, category $j$ has expected count $\lambda p_j$ in the window.

Category Rates and Probabilities

If the total mean count is $\lambda$ per observation window, then the mean count in category $j$ is

\lambda_j=\lambda p_j.

The category count therefore has probability mass function

P(X_j=x)=e^{-\lambda p_j}\frac{(\lambda p_j)^x}{x!}, \qquad x=0,1,2,\ldots.

If a total rate is given per unit of exposure, the same multiplication applies to rates. For example, if the total rate is $r$ events per hour and category $j$ has probability $p_j$, its rate is $rp_j$ events per hour. Over $t$ hours, its mean count is $rtp_j$. The unit of time or other exposure must be kept consistent.

Worked Example: Finding Category Means

Suppose a hypothetical service desk receives a Poisson-distributed mean of 16 requests per shift. Each request is independently labeled as account help with probability $0.25$, software help with probability $0.50$, or another type with probability $0.25$. Find the mean count for each category and the probability of exactly three account-help requests in a shift.

The probabilities sum to $0.25+0.50+0.25=1$, so the categories are exhaustive. The splitting assumptions say that each request receives one label and that the labels are assigned independently with the stated probabilities. The category means are

\lambda_{\text{account}}=16(0.25)=4,\qquad \lambda_{\text{software}}=16(0.50)=8,\qquad \lambda_{\text{other}}=16(0.25)=4.

Thus the account-help count $X_{\text{account}}$ follows a Poisson distribution with mean 4 requests per shift. The probability of exactly three such requests is

P(X_{\text{account}}=3) =e^{-4}\frac{4^3}{3!} =e^{-4}\frac{64}{6} \approx 0.1954.

The model assigns about a 19.54% probability to exactly three account-help requests in a shift. The result concerns that category's count; it does not require the total number of requests to equal 16, since 16 is the mean, not a fixed total.

Why the Category Counts Are Independent

The independence result can seem surprising. If the total were fixed, an increase in one category's count would leave fewer events available for the others. But here the total is random and Poisson-distributed. The random variation in the total changes the relationship: after the total is allowed to vary according to its Poisson model, the category counts are independent.

A direct derivation shows this. Consider any nonnegative counts $x_1,\ldots,x_k$, and write $m=\sum_{j=1}^{k}x_j$. For the category counts to have these values, the total must be $m$. Given a total of $m$, the probability of these category counts is the multinomial probability: there are $m!/\prod_j x_j!$ ways to arrange the labels, and each arrangement has probability $\prod_j p_j^{x_j}$. Combining that with the Poisson probability of a total of $m$ gives

\begin{aligned} P(X_1=x_1,\ldots,X_k=x_k) &=P(N=m)P(X_1=x_1,\ldots,X_k=x_k\mid N=m)\\ &=e^{-\lambda}\frac{\lambda^m}{m!} \left(\frac{m!}{\prod_{j=1}^{k}x_j!} \prod_{j=1}^{k}p_j^{x_j}\right)\\ &=e^{-\lambda}\prod_{j=1}^{k} \frac{(\lambda p_j)^{x_j}}{x_j!}\\ &=\prod_{j=1}^{k} \left(e^{-\lambda p_j}\frac{(\lambda p_j)^{x_j}}{x_j!}\right). \end{aligned}

In the third line, $m=\sum_j x_j$ makes $\lambda^m\prod_j p_j^{x_j}=\prod_j(\lambda p_j)^{x_j}$. Also, because the category probabilities sum to 1, $\prod_j e^{-\lambda p_j}=e^{-\lambda\sum_j p_j}=e^{-\lambda}$. The final expression is a product of the Poisson probability mass functions for the separate category counts. A joint probability that factors this way for every possible set of counts establishes that the counts are mutually independent, with $X_j\sim\operatorname{Poisson}(\lambda p_j)$.

This derivation also explains why the result is not the same as fixing a total and then distributing it. Conditional on a particular fixed total, category counts compete for that total. The independence claim applies to the counts when the total itself has the Poisson model and is not held fixed.

Theorem: Under the Poisson splitting assumptions, $X_1,\ldots,X_k$ are mutually independent and $X_j\sim\operatorname{Poisson}(\lambda p_j)$. Therefore, for specified nonnegative counts $x_1,\ldots,x_k$, $$ P(X_1=x_1,\ldots,X_k=x_k) =\prod_{j=1}^{k}e^{-\lambda p_j}\frac{(\lambda p_j)^{x_j}}{x_j!}. $$

Worked Probabilities for a Category and for Several Categories

Worked Example: Probability of at Least Two Rare Events

Suppose a hypothetical support line receives a Poisson-distributed mean of 8 calls per hour. Independently, each call has probability $0.10$ of being classified as urgent. Find the probability of at least two urgent calls in one hour.

The category probability is fixed and the calls are classified independently, so the urgent-call count $U$ is Poisson with mean

\lambda_U=8(0.10)=0.8\text{ urgent calls per hour}.

“At least two” is the complement of zero or one. Using the Poisson probability mass function,

\begin{aligned} P(U\geq 2) &=1-P(U=0)-P(U=1)\\ &=1-e^{-0.8}\frac{0.8^0}{0!} -e^{-0.8}\frac{0.8^1}{1!}\\ &=1-e^{-0.8}(1+0.8)\\ &\approx 0.1912. \end{aligned}

The model gives about a 19.12% probability of at least two urgent calls in an hour. The rate $0.8$ is a mean count per hour, not the probability of receiving any urgent call.

Worked Example: Joint Counts in Two Categories

In a hypothetical inspection process, the total number of flagged items per shift follows a Poisson distribution with mean 10. Each flagged item is independently classified as type A with probability $0.20$, type B with probability $0.30$, or type C with probability $0.50$. Find the probability that a shift has exactly one type A item and exactly four type B items, with no restriction on the type C count.

The category probabilities sum to 1 and the labels are independent, so the category counts are independent Poisson variables. Their means per shift are

\lambda_A=10(0.20)=2,\qquad \lambda_B=10(0.30)=3,\qquad \lambda_C=10(0.50)=5.

The type C count is unrestricted, so its probabilities sum to 1 and it need not appear in the calculation. Independence gives

\begin{aligned} P(X_A=1,\ X_B=4) &=P(X_A=1)P(X_B=4)\\ &=\left(e^{-2}\frac{2^1}{1!}\right) \left(e^{-3}\frac{3^4}{4!}\right)\\ &=e^{-5}\left(2\right)\left(\frac{81}{24}\right)\\ &\approx 0.04548. \end{aligned}

Thus the probability is about 4.548% per shift. This event does not require the total to be five: any type C count is allowed, and the total is consequently at least five.

Using the Model Carefully

Poisson splitting is useful when the scientific or operational question concerns particular types of events rather than only the overall count. It converts a total rate into category-specific rates and supports calculations for category counts or combinations of counts. If only some categories are of interest, their counts still have independent Poisson distributions with means $\lambda p_j$, provided the remaining events can be grouped into an “other” category. The probability of an event belonging to that remaining group is the leftover probability.

The method depends on the classification mechanism, not just on the fact that the total count is Poisson. If category probabilities change from event to event, or if labels are dependent, the simple independent Poisson conclusion need not hold. Likewise, if the total is fixed rather than Poisson, one should not use this theorem to claim that the resulting category counts are independent.

Common Mistake

A common error is to treat the total mean as though it were the number of events guaranteed to occur, then reason that category counts must add to that mean. For example, a total mean of 10 with category probabilities $0.20$ and $0.30$ does not mean there will be exactly two events in the first category and three in the second. Those are the category means, found as $10(0.20)=2$ and $10(0.30)=3$; the realized counts vary from window to window.

Another error is to multiply a category probability by a count probability rather than by the total Poisson mean. The category-specific Poisson parameter is $\lambda p_j$, not $p_j$ and not $\lambda$ alone. First calculate that parameter, with its time or exposure unit, and then use the Poisson probability formula.

Key takeaway: A Poisson total with mean $\lambda$, independently classified into categories with probabilities $p_j$, produces independent category counts with means $\lambda p_j$. Check the classification assumptions, compute each category mean, and then calculate probabilities using the appropriate Poisson distribution.

Check Your Understanding

Answer each question using the splitting assumptions where they apply.

A total count has a Poisson mean of 12 per day. Each event is independently assigned to category A with probability $0.25$. What is the mean count in category A per day?
A Poisson total has mean 5 per hour, and each event independently receives one of two labels with probabilities $0.40$ and $0.60$. What are the two category means per hour?
Why are the category counts from a Poisson splitting model independent even though counts assigned to categories would compete if the total were fixed?
A category has mean count $0.6$ per observation window. Write an expression for the probability of at least one event in that category.
Name one assumption that could fail and prevent the independent-Poisson conclusion from being justified.