Exact Counts from Unequal-Rate Sources
When several independent Poisson streams are observed over the same interval, an exact-count question can ask for the total number of arrivals, the number from each source, or both. Unequal rates affect both the total rate and the relative likelihood of different source compositions. The key calculation is to keep these two contributions distinct: the total count has a Poisson probability, while the allocation of that total among sources has a multinomial probability.
Let \(N_r(t)\), for \(r=1,\ldots,m\), be independent Poisson processes on a common probability space \((\Omega,\mathcal{F},\mathbb{P})\), with rates \(\lambda_r\ge 0\). For an interval \((a,a+h]\), where \(h\ge 0\), write \(C_r=N_r(a+h)-N_r(a)\) for the count from source \(r\). Set \(\theta_r=\lambda_r h\) and \(\Lambda=\sum_{r=1}^m\lambda_r\). By the Poisson-process count law, \(C_r\) has a Poisson distribution with mean \(\theta_r\); independence of the processes makes these counts independent.
For a Poisson random variable with mean zero, the count is zero with probability one. Thus, if \(\theta_r=0\), any composition with \(c_r>0\) has probability zero. If \(h=0\), every source count is zero. The conditional calculations below require \(\Lambda h>0\) and a conditioning event of positive probability; the all-zero-rate case is handled separately.
Joint Count and Conditional Composition Formulas
Proof. Independence of the source processes implies independence of their counts on the interval. Multiplying their Poisson probabilities gives the displayed joint formula when all means are positive. If a mean is zero, the corresponding count is zero almost surely, giving exactly the stated convention.
To find the total-count distribution, sum the joint probabilities over all nonnegative integer vectors whose entries sum to \(n\). Since \(\theta_r=h\lambda_r\), this gives $$ \mathbb{P}(C=n)=e^{-\Lambda h} \sum_{\sum_r c_r=n}\prod_{r=1}^m\frac{(h\lambda_r)^{c_r}}{c_r!} =e^{-\Lambda h}\frac{(h\Lambda)^n}{n!}. $$ The last equality follows by expanding \((\sum_r h\lambda_r)^n\) with the multinomial theorem: the term for \((c_1,\ldots,c_m)\) has coefficient \(n!/\prod_r c_r!\). This is the Poisson probability with mean \(\Lambda h\).
Now take a composition with total \(n\). Its conditional probability is its joint probability divided by \(\mathbb{P}(C=n)\). For \(\Lambda h>0\), cancellation gives $$ \frac{e^{-\Lambda h}\prod_r(h\lambda_r)^{c_r}/c_r!} {e^{-\Lambda h}(h\Lambda)^n/n!} =\frac{n!}{\prod_r c_r!}\prod_r\left(\frac{\lambda_r}{\Lambda}\right)^{c_r}, $$ because \(\sum_r c_r=n\). This proves the conditional formula. If some \(\lambda_r=0\), compositions with \(c_r>0\) have probability zero in both formulas, and the same result holds with the zero-probability convention. \(\square\)
The joint formula is usually the shortest route when the source counts are all specified. The conditional formula is convenient when the total is specified and several source compositions satisfy the event. In either route, the factorials matter: the factor \(n!/\prod_r c_r!\) counts the different ways the total can be allocated to the specified source counts.
Worked Example: One Arrival from the First Source and Two from the Second
Two independent sources have rates \(\lambda_1=2\) and \(\lambda_2=3\) arrivals per unit time. Find the probability that in an interval of length \(1\), there are exactly three arrivals in total, with one from source 1 and two from source 2.
Here \(\theta_1=2\cdot1=2\), \(\theta_2=3\cdot1=3\), and \(\Lambda=2+3=5\). The specified source counts already sum to three, so the joint formula gives $$ \mathbb{P}(C_1=1,C_2=2) =e^{-2}\frac{2^1}{1!}\,e^{-3}\frac{3^2}{2!} =e^{-5}\cdot 2\cdot\frac{9}{2} =9e^{-5}. $$
Check using the total and composition. The total has mean \(5\), so $$ \mathbb{P}(C=3)=e^{-5}\frac{5^3}{3!} =e^{-5}\frac{125}{6}. $$ Given a total of three, the source probabilities are \(2/5\) and \(3/5\). The conditional probability of one from source 1 and two from source 2 is $$ \frac{3!}{1!2!}\left(\frac25\right)\left(\frac35\right)^2 =3\cdot\frac25\cdot\frac9{25} =\frac{54}{125}. $$ Multiplying gives \(e^{-5}(125/6)(54/125)=9e^{-5}\), agreeing with the joint calculation.
Worked Example: A Three-Source Composition
Three independent sources have rates \(1\), \(2\), and \(3\) per unit time. In an interval of length \(1\), find the probability of exactly four arrivals with source counts \((0,1,3)\).
The total rate is \(\Lambda=1+2+3=6\), and the three means over the interval are \(1\), \(2\), and \(3\). The joint formula gives $$ \begin{aligned} \mathbb{P}(C_1=0,C_2=1,C_3=3) &=e^{-1}\frac{1^0}{0!}\, e^{-2}\frac{2^1}{1!}\, e^{-3}\frac{3^3}{3!}\\ &=e^{-6}\cdot1\cdot2\cdot\frac{27}{6} =9e^{-6}. \end{aligned} $$
Check using the conditional composition. The total-count probability is $$ \mathbb{P}(C=4)=e^{-6}\frac{6^4}{4!} =e^{-6}\frac{1296}{24} =54e^{-6}. $$ Given four arrivals, the composition probability is $$ \frac{4!}{0!1!3!} \left(\frac16\right)^0 \left(\frac26\right)^1 \left(\frac36\right)^3 =4\cdot\frac13\cdot\frac18 =\frac16. $$ The product is \(54e^{-6}\cdot\frac16=9e^{-6}\), which verifies the answer. The zero count from source 1 contributes a factor of \(1\), not an extra arrival probability.
An event may specify a total and a range of possible source counts rather than one composition. In that case, add the probabilities of the disjoint compositions that satisfy the event. Equivalently, sum their conditional probabilities given the total and multiply by the probability of that total. These are finite sums for a fixed total, so no approximation is needed.
Worked Example: At Least Two Arrivals from One Source
Two independent sources have rates \(\lambda_1=1\) and \(\lambda_2=2\) per unit time. Find the probability that there are exactly three arrivals in an interval of length \(1\), with at least two from source 1.
The total rate is \(1+2=3\). The allowed compositions are \((2,1)\) and \((3,0)\); the inequality “at least two” includes both counts \(2\) and \(3\). From the joint formula, $$ \begin{aligned} \mathbb{P}(C=3,\ C_1\ge2) &=\mathbb{P}(C_1=2,C_2=1)+\mathbb{P}(C_1=3,C_2=0)\\ &=e^{-3}\frac{1^2}{2!}\frac{2^1}{1!} +e^{-3}\frac{1^3}{3!}\frac{2^0}{0!}\\ &=e^{-3}\left(1+\frac16\right) =\frac76e^{-3}. \end{aligned} $$
Check using the conditional distribution. The total probability is $$ \mathbb{P}(C=3)=e^{-3}\frac{3^3}{3!} =\frac92e^{-3}. $$ Given \(C=3\), the probability that source 1 contributes at least two is the sum for counts \(2\) and \(3\): $$ \begin{aligned} \mathbb{P}(C_1\ge2\mid C=3) &=\binom32\left(\frac13\right)^2\left(\frac23\right) +\binom33\left(\frac13\right)^3\\ &=\frac{2}{9}+\frac{1}{27} =\frac{7}{27}. \end{aligned} $$ Multiplying gives \((9/2)e^{-3}\cdot(7/27)=(7/6)e^{-3}\), as required.
Checks That Prevent Common Errors
A frequent mistake is to multiply the total-count probability by the probability of one particular ordering of source labels, without accounting for all orderings that yield the requested composition. For example, in the first worked example, using only \((2/5)(3/5)^2\) as the conditional probability gives $$ e^{-5}\frac{125}{6}\left(\frac25\right)\left(\frac35\right)^2 =3e^{-5}, $$ which is not the requested probability. The composition \((1,2)\) can occur in \(\binom31=3\) different source-label orders. Including that factor gives \(3\cdot3e^{-5}=9e^{-5}\), matching the joint calculation. More generally, use the multinomial coefficient \(n!/\prod_r c_r!\) for a specified composition.
Also distinguish “exactly this composition” from “any composition satisfying a condition.” If the event allows multiple count vectors, add the probabilities of all allowed vectors; do not select only one. Conversely, if the event specifies the counts from every source, the joint product already gives the exact probability, and no additional ordering factor should be added to that product.
Finally, conditional source proportions are \(\lambda_r/\Lambda\), not \(\lambda_r\) themselves. They are defined when \(\Lambda>0\). If all rates are zero, the total is zero with probability one and the only possible composition is \((0,\ldots,0)\); conditioning on a positive total is then undefined because that event has probability zero. This boundary case is different from conditioning on a total under positive combined rate.
These formulas are useful whenever a question involves an exact count from several independent sources: compute the total probability and the source allocation separately, or use the independent Poisson probabilities directly. Calculating both ways when practical is a strong check on factorials, rate proportions, and event boundaries.
Check Your Understanding
For each question, identify whether a joint count formula or a conditional composition formula is more direct, and verify the arithmetic.
- Two independent sources have rates \(1\) and \(3\) per unit time. In an interval of length \(1\), find the probability of exactly two arrivals, both from source 2.
- Three sources have rates \(1\), \(1\), and \(2\). Given that the total count in a unit interval is \(2\), find the probability of composition \((1,0,1)\).
- Explain why the event “total \(3\), with at least two from source 1” requires adding more than one composition probability.
- For rates \(0\) and \(2\) over an interval of length \(1\), which source-count compositions have positive probability when the total is \(2\)?
- For two rates \(2\) and \(3\) over a unit interval, identify the missing factor in the incorrect conditional composition probability \((2/5)(3/5)^2\) for counts \((1,2)\).