I've seen the formula most commonly derived as a continuum generalization of a binomial random variable with large $n$, small $p$ and finite $\lambda = np$ yielding

$$ \lim_{n \to \infty} \binom{n}{x} p^x(1-p)^{n-x} = e^{-\lambda}\frac{\lambda ^ x}{x!}$$

It follows, from this derivation, that $$ \lim_{n \to \infty } = (1-p)^{n-x} = e^{-\lambda}$$ yields the probability of failing infinitely many times when the success rate is $\lambda$.

However, from this approach, I could not grok the remaining term

$$\frac { \lambda ^ x } {x!} $$


What insightful derivations (perhaps, from generalizations) of the Poisson random variable exist which leaves an intuition for each of the terms?

My Answer:

My answer, https://math.stackexchange.com/a/2727388/338817 comes from geometric approach to Gamma function intuition (https://math.stackexchange.com/a/1651961/338817) which I quote:

Note that $\frac{t^n}{n!}$ is the volume of the set $S_t=\{(t_1,t_2,\dots,t_n)\in\mathbb R^{n}\mid t_i\geq 0\text{ and } t_1+t_2+\cdots+t_n\leq t\}$

  • $\begingroup$ So you specifically want a sort of "combinatorial" interpretation of the expression $e^{-\lambda}\frac{\lambda^n}{n!}$, rather than just any intuitive explanation of the Poisson distribution? $\endgroup$ – Jack M Mar 28 '18 at 23:19
  • $\begingroup$ @JackM, any precise intuition will help, not necessarily combinatorial $\endgroup$ – jaslibra Mar 29 '18 at 3:02
  • $\begingroup$ You may find my explanation of the exponential distribution inspiring. The connection to the Poisson is that both the ED and PD are connected to the Poisson process - in the language of the linked answer, the Poisson is the distribution of the number of births in a given interval of time. This is a very different approach to the "limit of binomials" approach, however. $\endgroup$ – Jack M Mar 29 '18 at 11:28
  • $\begingroup$ You say you want "precise intuition". That seems a bit contradictory; if an explanation is completely precise, then it's a proof (which may or may not be intuitive, depending on the reader). $\endgroup$ – tparker Apr 6 '18 at 12:33
  • $\begingroup$ @tparker terrytao.wordpress.com/career-advice/…, I think seek a post-rigorous understanding. $\endgroup$ – jaslibra Apr 6 '18 at 20:27

Since you asked for an intuition, and there are many online derivations of the pdf of the Poisson distribution (e.g. here or here), which already follow a mathematically strict sequence, I'm giving it a shot at looking at it almost as a mnemonic construction.

So the pdf is

$$f_X(x=k)=\frac{\lambda^k\mathrm e^{-\lambda}}{k!}$$

What about thinking of the Poisson parameter $\lambda$ as somewhat reflecting the odds of an event happening in any time period. After all, it is a rate (events/time period), and hence, the higher the rate, the more likely it will be that a certain number of events takes place in a given time period. Further, you already mention how the pdf of the Poisson is derived from the binomial, allowing $n$ to go to infinity; and in the binomial distribution, the expectation is $np,$ equal to $\lambda$ in the Poisson: $p=\frac{\lambda}{n}.$

Notice, for instance, that in the derivation of the pdf of the Poisson, $\left(\frac{\lambda}{n}\right)^k$ is precisely introduced as the $p^k$ (the probability of $k$ successes) in the binomial pmf, $\binom{n}{k}p^k(1-p)^{n-k}.$ The denominator $n^k$ is later eliminated as we calculate the limit $n\to\infty,$ and indeed, $\lambda^k$ is "left over" from this initial probability formula.

Now, in the pdf you have the term raised to the $k$ power, i.e. $\lambda^k$, and it makes intuitive sense, because each occurrence is independent from the preceding and subsequent. So if we are calculating the probability of $k$ iid events happening in a time period, we shouldn't be surprised to end up with $\underbrace{\lambda\cdot\lambda \cdots\lambda}_k=\lambda^k$.

Since these events are indistinguishable from each other, it is not surprising either that we have to prevent over-counting by dividing by the number of permutations of these events, $k!.$ This, in fact is the exact role of the term in the combinations formula of $\binom{n}{k}=\frac{n!}{(n-k)!\color{blue}{k!}}.$

And for the term $e^{-\lambda}$ we could bring into play the inter-arrival time following an exponential distribution: as the rate $\lambda$ increases, the inter-arrival time decreases. We can think of this factor as decreasing the probability of a low $k$ number of events when the rate $\lambda$ factor is high.

  • $\begingroup$ @Antoni_Parellada, it doesn't make sense to think of $\lambda$ as a probability, because $\lambda \in \mathbb{R}$ and also if $\lambda$ has units events/time, then what units would you through onto $p$, the probability of the event? $\endgroup$ – jaslibra Mar 29 '18 at 0:58
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    $\begingroup$ @jaslibra I’m not necessarily defending the analogy (which is great by the way, +1) but here’s something to consider: there’s a difference between $\lambda$ being a literal probability and being a measure/indication of a probability. $\endgroup$ – gen-ℤ ready to perish Mar 29 '18 at 1:00
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    $\begingroup$ @jaslibra To consider $\lambda$ as a probability measure is an atrocity (committed by me after much hedging). I believe that was clearly and explicitly stated, not just in the opening remarks, but also in the careful introduction of the idea. I was shooting for an intuition, but if it doesn't help you, I'll delete the answer. $\endgroup$ – Antoni Parellada Mar 29 '18 at 1:03
  • $\begingroup$ @ChaseRyanTaylor, thanks for pointing that out. I'm looking for intuitive, but my intuition should still be precise in understanding exactly what $\lambda ^ k$ should represent. $\endgroup$ – jaslibra Mar 29 '18 at 1:12
  • $\begingroup$ @AntoniParellada You don't need to delete your answer. And an up-vote would be helpful if you think this deserves more attention $\endgroup$ – jaslibra Mar 29 '18 at 1:16

Suppose $k$ successes occur in an interval $[0, t)$ and let their times be given by the $k$-tuple $(t_1, \dots, t_k), t_i \leq t$.

The set of events where exactly $n$ successes occur can be measured as $$ \int_0^{t} \int_0^{t - x_1} \cdots \int_0^{ t - \sum_{i = 1}^{n-1} x_i } \int_0^{ t - \sum_{i = 1}^{n} x_i } dx_n dx_{n-1} dx_{n-2} \dots dx_2 dx_1 = \frac{ t^n } { n! }$$

Importantly, the size of the sample space of all events is measured by considering the size of all possible $k$-tuples, $\forall n \geq 0$:

$$ \sum_{k = 0}^{\infty} \frac{ t^k }{ k! } = e^t$$

Taking the ratio of these size of these sets yields the probability that $n$ events occur in the interval $[0, t)$.

$$\boxed{ P \{ X = n \} = e^{-t} \frac{ t^n }{ n! } }$$


More generally, the event rate can be made non-homogeneous with a scalar function $\lambda(t)$. When the rate is constant for all time, i.e, $\lambda(t) = \lambda$, we write

$$P(X = n) = e^{-\lambda t}\frac{ (\lambda t)^n } { n! }$$

Letting $t = 1$ gives the process on a unit time interval, scaled by $\lambda$. Although we're really interested in $[0, 1)$, it's really as if we're looking at the interval $[0, \lambda)$.


You're basically there with your limit. The ratio of two factorials is the product of $x$ numbers from $n-x+1$ to $n$, so for $n\gg x$ the result is approximately $n^x$.

  • 1
    $\begingroup$ How does this make intuitive sense in the context of the probability? $\endgroup$ – jaslibra Mar 28 '18 at 21:49

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