This equation shows up when considering a derivation for the Poison process when considering a small timestep, \(\Delta t\).

\[ q_k (t + \Delta t) - q_k(t) = \lambda \Delta t [q_{k-1}(t) - q_k(t)] \]

Which becomes:

\[ \frac{dq_k(t)}{dt} = \lambda [q_{k-1}(t) - q_k(t)] \]

We also have as boundary conditions, \(q_0(0) = 1\), and \(q_k(0) = 0\) for $k > 0$. (This comes from the construction, which I won’t get into here, but makes sense because \(q\) is a probability, namely \(q_k(t) = Pr(X_t = k)\), and the process \(X_t\) starts at \(0\)).

First, we have:

\[\frac{dq_0(t)}{dt} = -\lambda q_0(t) \] \[ q_0(t) = C_0e^{-\lambda t} = 1\] \[ q_0(t) = e^{-\lambda t} \]

Next, let us just solve for \(q_1\) to get a feel for what is going on (also, taking up the dot notation to indicate time derivatives, and omitting the function argument \(t\) for \(q_k\)’s): \[ \dot q_1 = \lambda[e^{-\lambda t} - q_1] \] \[ \dot q_1 + \lambda q_1 = \lambda e^{-\lambda t} \]

We find solutions to the homogenous equation: \[ \dot q_1 + \lambda q_1 = 0 \]

Which are of the form: \[ q_1(t) = C_1e^{-\lambda t} \]

Here, however, \(C_1\) is not a constant but instead \(C_1(t)\), which solves the inhomogeneous equation, so we have:

\[ \dot C_1(t) e^{-\lambda t} -\lambda C_1(t) e^{-\lambda t} + \lambda C_1(t) e^{-\lambda t} = \lambda e^{-\lambda t} \] \[ \dot C_1(t) = \lambda \] \[ C_1(t) = \lambda t\] \[ \dot q_1(t) = \lambda t e^{-\lambda t} \]

In general, we have: \[ \dot q_k + \lambda q_k = \lambda q_{k-1} \] \[q_k = C_k(t) e^{-\lambda t} \] \[\dot C_k(t) e^{-\lambda t} - \lambda C_k(t) e^{-\lambda t} + \lambda C_k(t) e^{-\lambda t} = \lambda q_{k-1}\] \[\dot C_k(t) e^{-\lambda t} = \lambda q_{k-1} = \lambda C_{k-1}(t) e^{-\lambda t} \] \[\dot C_k(t) = \lambda C_{k-1}(t)\]

Finally, we have from the last line: \[C_k(t) = \int C_{k-1}(t)dt\] \[C_0 = 1\] \[C_1 = \lambda t\] \[C_2 = \lambda^2 \frac{t^2}{2!} \] \[…\] \[C_k = \frac{(\lambda t)^k}{k!}\] \[q_k(t) = e^{-\lambda t} \frac{(\lambda t)^k}{k!}\]