Solving the Poisson Forward Equations

From Forward Equations to Count Probabilities

The forward equations describe how the probability of each count changes over time. Solving them one count at a time turns that local-in-time description into an explicit distribution. The first equation determines the probability of no arrivals; each subsequent equation uses the probability at the preceding count level. This recursive structure is what makes the Poisson formula emerge.

Work on a probability space $(\Omega,\mathcal{F},\mathbb{P})$, and let $(N_t)_{t\ge0}$ be a Poisson counting process with rate $\lambda>0$, state space $S=\{0,1,2,\ldots\}$, and time index $t\ge0$. Thus $N_0=0$ almost surely. For $n\ge0$, write $p_n(t)=\mathbb{P}(N_t=n)$. The forward equations, obtained from the infinitesimal transition probabilities, are

p_0'(t)=-\lambda p_0(t), \qquad p_n'(t)=-\lambda p_n(t)+\lambda p_{n-1}(t)\quad(n\ge1),

with initial conditions $p_0(0)=1$ and $p_n(0)=0$ for $n\ge1$. It is convenient to set $p_{-1}(t)=0$, so that the same equation $p_n'=-\lambda p_n+\lambda p_{n-1}$ applies for every $n\ge0$. The term $-\lambda p_n(t)$ accounts for probability leaving count $n$, while $\lambda p_{n-1}(t)$ accounts for probability entering count $n$ from count $n-1$.

Proposition (Recursive solution): For each $n\ge1$, if $p_{n-1}$ is known, the forward equation and initial condition determine $p_n$ by $$ p_n(t)=\lambda e^{-\lambda t}\int_0^t e^{\lambda s}p_{n-1}(s)\,ds. $$ For $n=0$, the solution is $p_0(t)=e^{-\lambda t}$. Each equation has a unique differentiable solution on $t\ge0$.

Proof. For $n=0$, multiply $p_0'(t)=-\lambda p_0(t)$ by $e^{\lambda t}$. The product rule gives $\frac{d}{dt}(e^{\lambda t}p_0(t))=0$, so $e^{\lambda t}p_0(t)$ is constant. Its value at $t=0$ is $1$, and therefore $p_0(t)=e^{-\lambda t}$. For $n\ge1$, multiply $p_n'(t)+\lambda p_n(t)=\lambda p_{n-1}(t)$ by $e^{\lambda t}$. The product rule yields $\frac{d}{dt}(e^{\lambda t}p_n(t))=\lambda e^{\lambda t}p_{n-1}(t)$. Integrating from $0$ to $t$ and using $p_n(0)=0$ gives $e^{\lambda t}p_n(t)=\lambda\int_0^t e^{\lambda s}p_{n-1}(s)\,ds$, which is the stated recursion. For uniqueness, the difference $u$ of any two solutions with the same $p_{n-1}$ satisfies $u'+\lambda u=0$ and $u(0)=0$. Hence $\frac{d}{dt}(e^{\lambda t}u(t))=0$, and $e^{\lambda t}u(t)=0$ for every $t$. Thus $u(t)=0$. Applying this argument successively from $n=0$ proves uniqueness at every count level. $\square$

The Induction That Gives the Poisson Formula

The recursion produces a simple pattern: each increase in the count level adds one power of $\lambda t$ and divides by the new count. The next theorem proves that pattern for all nonnegative integers, rather than relying on a few calculated cases.

Theorem (Solution of the Poisson forward equations): For every $t\ge0$ and every integer $n\ge0$, the solution of the forward equations with the stated initial conditions is $$ p_n(t)=e^{-\lambda t}\frac{(\lambda t)^n}{n!}. $$

Proof. We use induction on $n$. At $n=0$, the base equation has the solution $p_0(t)=e^{-\lambda t}$, which agrees with the formula because $(\lambda t)^0/0!=1$, including at $t=0$. Now let $n\ge1$, and suppose the formula holds for $n-1$ at every $s\ge0$. The recursive solution gives

\begin{aligned} p_n(t) &=\lambda e^{-\lambda t}\int_0^t e^{\lambda s}p_{n-1}(s)\,ds\\ &=\lambda e^{-\lambda t}\int_0^t e^{\lambda s} \left(e^{-\lambda s}\frac{(\lambda s)^{n-1}}{(n-1)!}\right)\,ds\\ &=\lambda e^{-\lambda t}\frac{\lambda^{n-1}}{(n-1)!} \int_0^t s^{n-1}\,ds\\ &=\lambda e^{-\lambda t}\frac{\lambda^{n-1}}{(n-1)!} \frac{t^n}{n}\\ &=e^{-\lambda t}\frac{(\lambda t)^n}{n!}. \end{aligned}

The integral evaluation holds also at $t=0$, since both sides are then zero for $n\ge1$. This completes the induction. The recursive-solution proposition establishes that each level has only one solution once the preceding level is fixed, so the formula is the unique solution of the forward equations. $\square$

A formula for each count should also describe a probability distribution: its terms must add to one. Here that check follows directly from the exponential series.

Corollary (Normalization): For each $t\ge0$, the probabilities given by the theorem sum to one.

Proof. Set $x=\lambda t$, which is nonnegative. The exponential series gives $\sum_{n=0}^{\infty}x^n/n!=e^x$. Therefore $$ \sum_{n=0}^{\infty}p_n(t) =e^{-\lambda t}\sum_{n=0}^{\infty}\frac{(\lambda t)^n}{n!} =e^{-x}e^x=1. $$ Every term is nonnegative, so these probabilities account for the full mass on the countable state space $S$. $\square$

Worked Solutions

Worked Example: The First Two Count Levels

Take rate $\lambda=3$. Solve the equations for $p_0(t)$ and $p_1(t)$, and evaluate them at $t=\tfrac12$.

The base equation gives $p_0(t)=e^{-3t}$. For the next level, use the recursion with $p_0(s)=e^{-3s}$:

\begin{aligned} p_1(t) &=3e^{-3t}\int_0^t e^{3s}e^{-3s}\,ds\\ &=3e^{-3t}\int_0^t 1\,ds\\ &=3t e^{-3t}. \end{aligned}

At $t=\tfrac12$, the exponent is $-3(\tfrac12)=-\tfrac32$. Thus $p_0(\tfrac12)=e^{-3/2}$, and $p_1(\tfrac12)=3(\tfrac12)e^{-3/2}=\tfrac32e^{-3/2}$.

Check: Differentiate the functions before evaluating. We have $p_0'(t)=-3e^{-3t}$, equal to $-3p_0(t)$. Also $p_1'(t)=3e^{-3t}-9t e^{-3t}$, while the forward equation gives $-3p_1(t)+3p_0(t)=-9t e^{-3t}+3e^{-3t}$. These expressions agree; at $t=\tfrac12$, both equal $-\tfrac32e^{-3/2}$.

Worked Example: Counts Through Three Arrivals

Take rate $\lambda=2$ and time $t=1$. Compute the probabilities of zero, one, two, and three arrivals, and check their forward equations at that time.

Here $\lambda t=2$ and $e^{-\lambda t}=e^{-2}$. Applying the formula separately at each count gives $p_0(1)=e^{-2}$, $p_1(1)=e^{-2}(2/1!)=2e^{-2}$, $p_2(1)=e^{-2}(2^2/2!)=2e^{-2}$, and $p_3(1)=e^{-2}(2^3/3!)=\tfrac43e^{-2}$.

Check by substitution: For general $t$, these four functions are $e^{-2t}$, $2t e^{-2t}$, $2t^2e^{-2t}$, and $\tfrac43t^3e^{-2t}$. At $t=1$, the derivative of $p_0$ is $-2e^{-2}$, matching $-2p_0(1)$. The derivative of $p_1$ is $2e^{-2}-4e^{-2}=-2e^{-2}$, matching $-2p_1(1)+2p_0(1)=-4e^{-2}+2e^{-2}$. The derivative of $p_2$ is $4e^{-2}-4e^{-2}=0$, matching $-2p_2(1)+2p_1(1)=0$. Finally, the derivative of $p_3$ is $4e^{-2}-\tfrac83e^{-2}=\tfrac43e^{-2}$, matching $-2p_3(1)+2p_2(1)=-\tfrac83e^{-2}+4e^{-2}=\tfrac43e^{-2}$. Each computed value satisfies its forward equation.

Worked Example: Exactly Two Arrivals at a Later Time

Take rate $\lambda=\tfrac12$ and time $t=4$. Find the probability of exactly two arrivals, and verify it using the recursion rather than just substituting into the closed form.

First, $\lambda t=(\tfrac12)(4)=2$, so the formula gives $p_2(4)=e^{-2}2^2/2!=2e^{-2}$. To verify it recursively, the preceding level is $p_1(s)=e^{-s/2}(\tfrac12s)$. Hence

\begin{aligned} p_2(4) &=\tfrac12 e^{-2}\int_0^4 e^{s/2} \left(e^{-s/2}\tfrac{s}{2}\right)\,ds\\ &=\tfrac14e^{-2}\int_0^4 s\,ds\\ &=\tfrac14e^{-2}\left[\frac{s^2}{2}\right]_0^4\\ &=\tfrac14e^{-2}(8-0)=2e^{-2}. \end{aligned}

The recursive computation and the closed form agree. The cancellation of $e^{s/2}$ with $e^{-s/2}$ in the integrand is the same integrating-factor mechanism used at every count level.

A Common Error and Its Correction

A common mistake is to keep the incoming term in the equation for $p_1$ but omit the outgoing term. For rate $\lambda=1$, that mistaken equation would be $p_1'(t)=p_0(t)=e^{-t}$, with $p_1(0)=0$. Integrating gives the incorrect answer $p_1(t)=1-e^{-t}$, so at $t=1$ it predicts $1-e^{-1}$.

The correct forward equation is $p_1'(t)=-p_1(t)+p_0(t)$. The negative term is essential: probability leaves state $1$ when another arrival occurs and the count becomes $2$. The correct solution is $p_1(t)=te^{-t}$, so at $t=1$ it gives $e^{-1}$. Substitution exposes the error: the derivative of the mistaken function is $e^{-t}$, whereas the right-hand side of the correct equation at that function is $-(1-e^{-t})+e^{-t}=2e^{-t}-1$. At $t=1$, these are $e^{-1}$ and $2e^{-1}-1$, which are not equal. In solving each equation, retain both the probability flowing out of count $n$ and the probability flowing in from count $n-1$.

The recursive method is useful whenever the forward equations have a one-way structure in the state index: first solve the equation at the lowest state, then use that solution as the forcing term for the next equation. For Poisson counts, the initial condition at zero and the repeated integrating-factor step determine every count probability. The induction proves the formula for all $n$, and normalization confirms that the resulting probabilities form a complete distribution.

Check Your Understanding

Use the forward equations and initial conditions to answer the following. When checking a proposed expression, substitute it into the equation for the relevant count.

For $\lambda=4$, solve the equation for $p_0(t)$ and state its value at $t=\tfrac12$.
Starting from $p_0(t)=e^{-\lambda t}$, use the recursive integral formula to derive $p_1(t)$.
In the induction step for count $n\ge1$, evaluate $\int_0^t s^{n-1}\,ds$ and show how the factor $n!$ appears.
For $\lambda=1$ and $t=2$, calculate $p_2(t)$ exactly and verify it with the recursion from $p_1(s)$.
Why does the exponential series prove that the probabilities over all counts sum to one?

Solving the Poisson Forward Equations

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