Poisson process

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Poisson process

Consider the random process \((X_t)\). The time \(t\) is continuous, \(t\in[0;\infty)\). The random variable \(X_t\) counts the number of “arrivals” on \([0;t]\).

We assume that

1. \(X_0 = 0\).

2. “Stationary increments”. The number of arrival during any time interval \([t;t+h]\) depends only on the length \(h\) of the interval and not on starting time \(t\).

3. “Independent increments”.

4. For small time interval length \(h\) the probability of exactly one arrival is approximately proportional to \(h\). \[ \P(X_{t+h} - X_t = 1) = \lambda h + o(h). \]

5. For small time interval length \(h\) the probability of two or more arrivals is negligible compared to \(h\).

\[ \P(X_{t+h} - X_t \geq 2) = o(h). \]

Let’s recap that \(o(h)\) is any function of \(h\) such that \[ \lim_{n\to\infty} \frac{o(h)}{h} = 0. \]

From the last two assumptions we deduce that \(\P(X_{t+h} - X_t \geq 2) = 1 - \lambda h + o(h)\).