Derive Poisson distribution from probability per time of event Suppose we have a probability per time $\lambda$ that something (e.g. nuclear decay, random walk takes a step, etc.) happens.
It is a known result that the probability that $n$ events happen in a time interval of length $T$, is given by the Poisson distribution
$$P(n) = \frac{e^{-\lambda T} (\lambda T)^n}{n!} \, .$$ 
How do we prove this?
 A: The Poisson distribution describes the probability of a certain number ($n$) of unlikely events ($p\ll 1$) happening given $N$ opportunities.
This is like doing a very unfair coin toss $N$ times, with the probability $p$ of the coin turning up heads. The number of heads would follow the binomial distribution:
$$P(n|p,N) = ~^{N}C_n~p^n (1-p)^{N-n} =\frac{N!}{(N-n)!~n!} p^n (1-p)^{N-n}$$
Now it remains to prove that when $N\rightarrow \infty$ and $p\rightarrow 0$ while $Np\rightarrow \lambda T$, that the above converges to the known result. In essence, I argue that when you make the number of opportunities tend to infinity, you go from a discrete to a continuous approach; but as long as you are careful with your infinities, the result should still be valid.
First, we find an approximation for $(1-p)^{N-n}$. Taking the log, we get
$$\log\left((1-p)^{N-n}\right) = (N-n)\log(1-p)\approx (N-n)\cdot (-p)$$
Since $N\gg n$, we obtain $(1-p)^{N-n}\approx e^{-Np}$
Next, we approximate the $~^N C_n$ term using Stirling's approximation that $\log N! \approx N\log N - N$ and noting that $n\ll N$
Then
$$\begin{align} \log\left(\frac{N!}{(N-n)!}\right) &= N\log N - N - (N-n)\log(N-n) + (N-n) \\
&=N\log N - (N-n)\log(N-n) - n\\
&= N \log N -(N-n)\left(\log(N)+\log\left(1-\frac{n}{N}\right)\right) -
 n\\
&\approx N\log N -(N-n)\left(\log(N)-\frac{n}{N}\right) - n\\
&\approx n\log N + n -n \log n - n\\
&=n\log(N-n)\end{align}$$
It follows that $\frac{N!}{(N-n)! n!} \approx \frac{N^n}{n!}$
Finally, we note that $pN = \lambda T$, and we get
$$P(n|N,p) = \frac{N^n p^n e^{-Np}}{n!}$$
this easily rearranges to
$$P(n) = \frac{(\lambda T)^n e^{-\lambda T}}{n!}$$
Which is the result we set out to prove.
I made use of this article to remind me of some of the steps in this.
A: Let $A^n_t$ be the event: exactly $n$ point events happened over a time interval $t$. Then, for small $\Delta t$ 
$$\begin{align}P(A^n_{t+\Delta t}) &= P( A^n_t \cap A^0_{\Delta t }) + P(A^{n-1}_t \cap A^1_{\Delta t }) \\
&= P(A^n_t) P (A^0_{\Delta t }) + P(A^{n-1}_t )P( A^1_{\Delta t })\\
\end{align}
$$
where we've used independence of occurrence.
Now, defining $p_n(t) \equiv P( A^n_t)$ and 
 $\lambda = \lim_{\Delta t\to 0} p_1(\Delta t)/\Delta t$, and taking the limit for $\Delta t \to 0$ we get
$$ p_n(t+\delta t)=p_n(t)(1 - \lambda \, \delta t) +  p_{n-1}(t) \lambda \, \delta t $$
which leads to the differential equations:
$$p'_n(t)= \begin{cases}
-\lambda ( p_n(t) - p_{n-1}(t) ) & n>0\\
-\lambda  p_n(t) & n=0
\end{cases}$$
with the initial conditions
$$p_n(0)=\begin{cases}0 & n>0\\1 & n=0 \, .\end{cases}$$
The solution is given by the Poisson distribution:
$$p_n(t)=\frac{(\lambda t)^n \exp(-\lambda t)}{n!} \, .$$
A: Probability distribution of time until next event
First we calculate the probability density that a time $t$ passes without any event happening.
Divide $t$ into $N$ small intervals each of length $dt = t/N$.
Defining $\lambda$ as the probability per time that the event occurs, the probability that no event occurs within any one short time interval is approximately $(1 - \lambda dt)$.
Therefore, the probability that no event happens in any of the intervals, but then does happen within an interval of length $dt$ right at the end of all the intervals is
$$\left( \prod_{i=1}^N \left( 1 - \lambda dt \right) \right) \lambda dt
= \left( 1 - \frac{\lambda t}{N} \right)^N \lambda dt
\stackrel{N\rightarrow \infty}{=} \lambda dt \exp \left( - \lambda t \right) \, .$$
In other words, given a starting time $0$, the probability density that no event has happened after a time $t$, but then happens right at $t$ is $\lambda \exp(-\lambda t)$.
Probability of multiple events
Now we ask the probability that we get $n$ events in the time interval $T$.
Suppose the first event happens at $t_1$, the second even happens at $t_2$, etc.
We therefore have a series of intervals
$$\{[0, t_1], [t_1, t_2], \ldots [t_n, T] \} $$
with events happening at the end of each interval.
The probability that our events occur in this way is
$$P(t_1, t_2, \ldots t_n) = \lambda \exp(-\lambda t_1) \lambda \exp(-\lambda (t_2 - t_1)) \cdots \lambda \exp(-\lambda(T - t_n)) = \lambda^n \exp(-\lambda T) \, .$$
Of course, any arrangement of $\{t_1, t_2, \ldots t_n \}$ such that $t_1 < t_2 < \ldots < t_n$ counts as an $n$-event arrangement, so we have to add up the probabilities of all these possible arrangements, i.e. the probability of $n$ events is
\begin{align}
P(n \text{ events})
&= \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-1}}^T dt_n P(t_1, t_2, \ldots t_n) \\
&= \lambda^n \exp(-\lambda T) \int_0^T dt_1 \int_{t_1}^T dt_2 \cdots \int_{t_{n-1}}^T dt_n
\, .
\end{align}
The multiple integral is the volume of a right simplex and has value $T^n/n!$, so the final result is
$$P(n\text{ events}) = \frac{(\lambda T)^n \exp(-\lambda T)}{n!} $$
which is the Poisson distribution with mean $\lambda T$.
Related

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*Direct evaluation of the simplex volume integral

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