The Poisson distribution counts random events in a window; the exponential distribution measures the wait between them. They are two views of the same model, the Poisson process: arrivals scattered in time at a steady rate, with no memory.
Idea clave
In a Poisson process with rate $\lambda$: the number of arrivals in any window of length $t$ is Poisson with mean $\lambda t$; counts in non-overlapping windows are independent; and the gaps between arrivals are independent exponential times with mean $\frac{1}{\lambda}$. The $k$th arrival comes after an expected $\frac{k}{\lambda}$.
Por qué funciona
Start the clock at any moment. The first arrival comes after time $t$ exactly when the window $[0, t]$ holds no arrivals, so $$P(\text{first wait} > t) = P(N(t) = 0) = e^{-\lambda t}:$$ the wait is exponential with rate $\lambda$. Because separate windows are independent, the same argument restarts after every arrival, so every gap is exponential and the gaps are independent. The time to the $k$th arrival is the sum of $k$ gaps, with mean $\frac{k}{\lambda}$ by linearity.
Worked examples
Ejemplo resuelto: Counts in a window
Customers arrive as a Poisson process at 6 per hour. Find the probability of exactly 2 arrivals in 20 minutes.
- 20 minutes is $\frac{1}{3}$ hour, so the mean count is $\lambda t = 6 \cdot \frac{1}{3} = 2$.
- $$P(N = 2) = e^{-2}\frac{2^{2}}{2!} = 2e^{-2} \approx 0.271$$
Respuesta: about $0.271$
Ejemplo resuelto: Waiting
With the same process, find the probability that nobody arrives in the next 15 minutes, and the expected wait for the 3rd customer.
- No arrivals in $\frac{1}{4}$ hour: $e^{-6/4} = e^{-1.5} \approx 0.223$.
- Each gap has mean $\frac{1}{6}$ hour = 10 minutes, so the 3rd arrival is expected after 30 minutes.
Respuesta: about $0.223$; 30 minutes.
Ejemplo resuelto: Given the total
Exactly 4 customers arrived during one hour. Find the probability that exactly 1 of them came in the first 15 minutes.
- Given the total, the 4 arrival times behave like 4 independent uniform points in the hour.
- Each lands in the first quarter-hour with probability $\frac{1}{4}$, so the count there is binomial.
- $\binom{4}{1}\left(\frac{1}{4}\right)\left(\frac{3}{4}\right)^{3} = \frac{27}{64} \approx 0.422$. The rate 6 dropped out.
Respuesta: $\frac{27}{64}$
Error común
Giving the count in a window of length $t$ the mean $\lambda$ instead of $\lambda t$, or giving the gaps the mean $\lambda$ instead of $\frac{1}{\lambda}$. Check: a higher rate means more arrivals and shorter gaps, so the count's mean grows with $\lambda$ and the gap's mean shrinks.