Forward Equations for Poisson Counts

From Short Intervals to Forward Equations

A forward equation describes how the probability of occupying a state changes with time. For a Poisson counting process, a count can increase only through arrivals: over a short interval, the probability of one arrival is approximately the rate times the interval length, while the probability of two or more arrivals is of smaller order. Combining these local probabilities with independent increments gives a differential equation for each possible count.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$. Let $(N(t))_{t\geq 0}$ be a counting process with state space $S=\{0,1,2,\ldots\}$, starting at $N(0)=0$. Use its natural filtration $\mathcal{F}_t=\sigma(N(s):0\leq s\leq t)$. Assume the process has independent, stationary increments, and that for a rate $\lambda\geq 0$, as $h\downarrow 0$,

\mathbb{P}(N(h)=0)=1-\lambda h+o(h),\qquad \mathbb{P}(N(h)=1)=\lambda h+o(h),\qquad \mathbb{P}(N(h)\geq 2)=o(h).

Here $o(h)$ denotes a quantity whose ratio to $h$ tends to zero as $h\downarrow0$. These short-interval conditions are used for intervals of length $h$; stationarity of increments gives them for every interval of that length. Define $p_n(t)=\mathbb{P}(N(t)=n)$ for $n\geq0$. The initial count is known exactly, so $p_0(0)=1$ and $p_n(0)=0$ for every $n\geq1$.

Theorem (Forward equations for Poisson counts): Under the assumptions above, for $t\geq0$ the count probabilities satisfy the right-derivative equations at $t=0$ and the ordinary differential equations for $t>0$: $$ p_0'(t)=-\lambda p_0(t),\qquad p_n'(t)=\lambda p_{n-1}(t)-\lambda p_n(t)\quad(n\geq1), $$ with initial conditions $p_0(0)=1$ and $p_n(0)=0$ for every $n\geq1$. The derivative at $t=0$ is understood from the right.

The equation at state $0$ has no inflow: once an arrival occurs, the count is no longer zero. For $n\geq1$, the probability at state $n$ gains mass when the process moves from $n-1$ to $n$, and loses mass when it moves from $n$ to $n+1$. The proof makes this flow argument precise and controls the possibility of multiple arrivals in a short interval.

Proof. Fix $t\geq0$ and $h>0$, and write $Y=N(t+h)-N(t)$ for the increment over the next interval. Independent, stationary increments imply that $Y$ is independent of $N(t)$ and has the same distribution as $N(h)$. Since counts cannot decrease, for $n\geq0$, $$ p_n(t+h)=\sum_{k=0}^{n}p_{n-k}(t)\mathbb{P}(Y=k). $$ Only $k\leq n$ can contribute because $N(t)$ cannot be negative. For $n=0$, this gives $p_0(t+h)=p_0(t)\mathbb{P}(Y=0)$, and therefore $$ p_0(t+h)=p_0(t)\bigl(1-\lambda h+o(h)\bigr). $$ For $n\geq1$, separate the terms $k=0$ and $k=1$. The sum of terms with $2\leq k\leq n$ is nonnegative and at most $\mathbb{P}(Y\geq2)=o(h)$, since the probabilities $p_{n-k}(t)$ are at most $1$ and their sum over this subset is at most $1$. Thus $$ p_n(t+h)=p_n(t)(1-\lambda h+o(h)) +p_{n-1}(t)(\lambda h+o(h))+o(h). $$ Subtract $p_n(t)$, divide by $h$, and let $h\downarrow0$. This gives the stated right derivatives and equations.

To verify that the derivative is also the ordinary derivative when $t>0$, first note that $$ |p_n(t+h)-p_n(t)|\leq\mathbb{P}(N(t+h)-N(t)>0)=\lambda h+o(h). $$ Indeed, the two indicators of the event $N(\cdot)=n$ can differ only if the count changes. The same bound applies to an interval ending at $t$, so each $p_n$ is continuous. Now apply the already derived forward-increment formula over $[t-h,t]$, divide by $h$, and let $h\downarrow0$. Continuity gives the same limit as for the right derivative. This proves the equations for $t>0$ as well as the right-derivative statement at $0$. Finally, $N(0)=0$ gives the initial conditions directly. $\square$

Reading the Equations State by State

The count states form a one-way sequence. In a time interval of length $h$, the probability of a single arrival is $\lambda h+o(h)$. From state $n-1$, that arrival contributes probability to state $n$; from state $n$, it removes probability from state $n$. Events with two or more arrivals have probability $o(h)$, so they do not contribute to the first-order derivative. This also explains why no probability from state $n+1$ appears in the forward equation for $p_n$: the process cannot move downward into $n$.

In the usual generator notation, the transition rates from state $n$ are $q_{n,n+1}=\lambda$ and $q_{n,n}=-\lambda$, with all other entries in that row zero. Each row sums to zero. For $n=0$, the same rule gives the equation $p_0'=-\lambda p_0$. The coordinate equations above are the forward equations written without requiring operations on an infinite matrix.

Worked Example: The Zero-Count Boundary at Rate 2

Suppose $\lambda=2$. The probability of no arrivals by time $t$ is $p_0(t)$. Over a further interval of length $h$, the count remains zero only if it was zero at time $t$ and there are no arrivals in the interval. Independent increments therefore give $$ p_0(t+h)=p_0(t)\bigl(1-2h+o(h)\bigr). $$ Subtracting $p_0(t)$, dividing by $h$, and taking the limit gives $p_0'(t)=-2p_0(t)$, with $p_0(0)=1$.

The first positive state receives probability from state $0$ and loses probability to state $2$. Its equation and initial condition are $$ p_1'(t)=2p_0(t)-2p_1(t),\qquad p_1(0)=0. $$ At $t=0$, the equations give $p_0'(0)=-2$ and $p_1'(0)=2(1)-2(0)=2$. The signs are consistent: initially, probability leaves state $0$ and enters state $1$. Substituting the initial values into the two equations gives those same derivatives, checking the boundary and the first inflow equation.

Worked Example: The Equation for Count 3 at Rate 3/2

Let $\lambda=\tfrac32$, and consider the probability $p_3(t)$. To be in state $3$ at time $t+h$, the process must either be in state $3$ and have no arrival, or be in state $2$ and have one arrival, up to events with two or more arrivals. Thus $$ p_3(t+h)=p_3(t)\bigl(1-\tfrac32h+o(h)\bigr) +p_2(t)\bigl(\tfrac32h+o(h)\bigr)+o(h). $$ The probability of all the omitted multiple-arrival contributions is at most $\mathbb{P}(N(h)\geq2)=o(h)$. It follows that $$ p_3'(t)=\tfrac32p_2(t)-\tfrac32p_3(t),\qquad p_3(0)=0. $$ The neighboring equation is $p_2'(t)=\tfrac32p_1(t)-\tfrac32p_2(t)$, with $p_2(0)=0$. At time $0$, the equations yield $p_2'(0)=\tfrac32(0)-\tfrac32(0)=0$ and $p_3'(0)=\tfrac32(0)-\tfrac32(0)=0$. The initial derivative of $p_1$, by contrast, is $\tfrac32p_0(0)-\tfrac32p_1(0)=\tfrac32$. Thus the first-order change begins in state $1$; states $2$ and $3$ require successive arrivals. Substitution verifies each stated initial derivative.

Probability Flow Across a Cutoff

The individual equations also give a useful result for the probability of being at or below a fixed count. For an integer $m\geq0$, define $F_m(t)=\mathbb{P}(N(t)\leq m)=\sum_{n=0}^{m}p_n(t)$. Only arrivals from state $m$ to state $m+1$ carry probability out of this collection of states. The exact identity is as follows.

Proposition (Forward equation for a count cutoff): For every fixed integer $m\geq0$, $$ F_m'(t)=-\lambda p_m(t). $$

Proof. Since $m$ is finite, $F_m$ is a finite sum of differentiable probabilities, so its derivative is the sum of their derivatives. The equation for state $0$ contributes $-\lambda p_0(t)$. For each $n$ from $1$ through $m$, the contribution is $\lambda p_{n-1}(t)-\lambda p_n(t)$. Therefore $$ F_m'(t)=-\lambda p_0(t)+\sum_{n=1}^{m}\bigl(\lambda p_{n-1}(t)-\lambda p_n(t)\bigr) =-\lambda p_m(t), $$ because every term involving $p_0,\ldots,p_{m-1}$ appears once with a positive sign and once with a negative sign. When $m=0$, the sum from $1$ to $m$ is empty and the same conclusion follows directly from $p_0'=-\lambda p_0$. This covers the boundary case and proves the identity. $\square$

Worked Example: Probability of a Count at Most 2

Take $\lambda=1$ and $F_2(t)=p_0(t)+p_1(t)+p_2(t)$. The first three forward equations are $$ p_0'=-p_0,\qquad p_1'=p_0-p_1,\qquad p_2'=p_1-p_2. $$ Adding them gives $$ F_2'(t)=(-p_0)+(p_0-p_1)+(p_1-p_2)=-p_2(t). $$ The terms for states $0$ and $1$ cancel: probability moves within the set $\{0,1,2\}$ when the count rises from $0$ to $1$, or from $1$ to $2$. Only a rise from $2$ to $3$ leaves the set. Initially, $F_2(0)=1+0+0=1$, and $F_2'(0)=-p_2(0)=0$. This matches the initial conditions and the cutoff identity.

Common Mistake: Reversing the Flow

A frequent error is to write $p_n'(t)=\lambda p_{n+1}(t)-\lambda p_n(t)$, treating a transition from $n+1$ as if it could enter state $n$. That is not the forward equation for this counting process: counts never decrease. For example, with $\lambda=2$ and $n=1$, the incorrect expression is $2p_2(t)-2p_1(t)$. The correct expression is $2p_0(t)-2p_1(t)$, because the only single-arrival inflow to state $1$ comes from state $0$. At time $0$, the incorrect expression gives $2p_2(0)-2p_1(0)=0$, while the correct equation gives $2p_0(0)-2p_1(0)=2$. This check against the initial state exposes the error immediately.

The equations specify how all count probabilities evolve and where their probability mass flows. Their initial conditions are essential: every positive-count state starts with probability zero, while state $0$ starts with probability one. The equations can then be solved to obtain explicit count probabilities; the derivation here establishes the system they must satisfy.

Check Your Understanding

Use the short-interval assumptions and the state-to-state flow direction to answer each question.

For a rate $\lambda$, write the forward equation and initial condition for $p_0(t)$.
For a fixed $n\geq1$, identify the inflow and outflow terms in the equation for $p_n(t)$.
Why can the contribution from two or more arrivals be omitted when deriving a first derivative?
For $m=1$, use the equations for $p_0$ and $p_1$ to verify $F_1'(t)=-\lambda p_1(t)$.
What incorrect term appears if one treats a transition from $n+1$ to $n$ as an inflow? Why is that transition impossible here?

Forward Equations for Poisson Counts

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