The Poisson Distribution

Consider a call center that receives, on average, 5 calls per minute during peak hours. How many calls arrive in the next minute? This is not a fixed number: it fluctuates. Yet it obeys a strikingly regular pattern. The Binomial Distribution almost applies. Imagine chopping the minute into a huge number of tiny slivers. Each sliver either contains a call or not. But the number of slivers nn is arbitrary. The "success probability" per sliver shrinks as we refine the chopping. The Poisson distribution is what emerges in the limit. It turns out to depend on only one number: the rate. This matters because an enormous range of real phenomena share this structure. They involve rare events, occurring independently, over a continuum of time or space. Each has a fixed average rate. Radioactive decays per second, typos per page, customers per hour, mutations per genome, goals per soccer match: all are modeled this way. Each counts occurrences of something individually improbable but collectively frequent. The Poisson distribution is the mathematical shadow cast by the Binomial Distribution. This happens when the number of trials grows without bound. Meanwhile the total expected number of successes is held fixed. The big picture is a limiting procedure. It trades two parameters (nn, the number of trials, and pp, the probability per trial) for a single parameter λ=np\lambda = np, the rate. This is a genuine simplification. Instead of needing to know how finely to subdivide time, we only need the average count. The limit theorem that makes this rigorous is often called the law of rare events, or the Poisson limit theorem. It is one of the cleanest asymptotic results in elementary probability. It explains why the Poisson formula appears with such uncanny frequency in applications. This article gives the precise definition of the Poisson probability mass function. It proves the binomial-to-Poisson limit theorem with full rigor, including an explicit error bound. It works through two asymmetric numerical examples. It shows via a counterexample exactly which hypothesis cannot be dropped. It closes by tracing the distribution's history and its role in modern applications. Finally, it shows its natural continuation into the Exponential Distribution and Moment Generating Functions.

Key takeaways
  • The Poisson distribution models counts of rare, independent events occurring at a known average rate λ\lambda.
  • Its PMF is P(X=k)=e−λλk/k!P(X=k) = e^{-\lambda}\lambda^k/k! for k=0,1,2,…k = 0, 1, 2, \dots, a single-parameter family.
  • It arises as the limit of the Binomial Distribution as n→∞n \to \infty, p→0p \to 0, with np→λnp \to \lambda fixed.
  • Both the mean and variance of a Poisson random variable equal λ\lambda — a hallmark diagnostic of Poisson-ness.
  • The limit theorem comes with an explicit, provable error bound, not just a heuristic convergence claim.
  • Key idea: many independent rare events, summed up, behave like a single random variable governed by one number — the rate.

Where this lives today

The Poisson distribution is the default model for count data across science and engineering. In telecommunications and queueing theory it models the number of arriving calls, packets, or customers per unit time, directly feeding into the design of call centers and network buffers. In physics, radioactive decay counts detected by a Geiger counter in a fixed interval are Poisson, and the distribution underlies statistical uncertainty quotes in particle physics ("Poisson counting error," λ\sqrt{\lambda}). In genetics and genomics, the number of mutations or read-counts along a sequence is often modeled as Poisson (or its overdispersed cousin, the negative binomial). In actuarial science and insurance, the number of claims filed in a year is a Poisson-distributed count feeding into risk models. In natural language processing and text analysis, rare word or error occurrences per document are approximated by Poisson counts. Cryptographic and reliability engineering use Poisson models for the number of rare failure events (e.g., bit-flip errors) over a fixed exposure. The distribution's memoryless "rate" structure connects it intimately to the continuous-time Exponential Distribution, which governs the waiting time between Poisson events, and its generating-function machinery is a natural first example when studying Moment Generating Functions and characteristic functions, where the Poisson MGF M(t)=eλ(et−1)M(t) = e^{\lambda(e^t-1)} gives an especially clean closed form.

:::quiz[If X∼Poisson(λ)X \sim \mathrm{Poisson}(\lambda), what are E[X]\mathbb E[X] and Var(X)\mathrm{Var}(X)?]

  • Both equal λ2\lambda^2
  • E[X]=λ\mathbb E[X] = \lambda, Var(X)=λ\mathrm{Var}(X) = \sqrt{\lambda}
  • Both equal λ\lambda ✓
  • E[X]=λ\mathbb E[X] = \sqrt\lambda, Var(X)=λ\mathrm{Var}(X) = \lambda

Ask the tutor: How does the Poisson distribution connect to the exponential distribution through the idea of a Poisson process?

DefinitionPoisson distribution

A discrete random variable XX taking values in {0,1,2,… }\{0, 1, 2, \dots\} is said to have the Poisson distribution with rate parameter λ>0\lambda > 0, written X∼Poisson(λ)X \sim \mathrm{Poisson}(\lambda), if its probability mass function is

P(X=k)=e−λλkk!,k=0,1,2,…P(X = k) = \frac{e^{-\lambda}\lambda^k}{k!}, \qquad k = 0, 1, 2, \dots

This is a valid probability mass function because each term is nonnegative and, by the Taylor series of the exponential function,

∑k=0∞e−λλkk!=e−λ∑k=0∞λkk!=e−λ⋅eλ=1.\sum_{k=0}^{\infty} \frac{e^{-\lambda}\lambda^k}{k!} = e^{-\lambda}\sum_{k=0}^{\infty}\frac{\lambda^k}{k!} = e^{-\lambda} \cdot e^{\lambda} = 1.

Intuition

Picture the call center again. Over one minute, calls arrive at scattered, essentially unpredictable instants. Slice the minute into nn equal sub-intervals. If nn is large enough that at most one call falls in each sub-interval, then the number of calls is approximately Binomial(n,p)\mathrm{Binomial}(n, p), where pp is the (small) chance of a call in one sub-interval. As we refine the slicing — sending n→∞n \to \infty — the per-slice probability pp must shrink, but the total expected number of calls npnp should stay pegged at the true average rate λ\lambda. The Poisson distribution is the distribution that survives this limit: it is what "many independent coin flips, each almost always tails, but with a fixed total expected number of heads" looks like when you stop caring about how the flips were divided up. The shape of the Poisson PMF reflects this: it is unimodal, concentrated near λ\lambda, with a right skew that diminishes as λ\lambda grows (by the Central Limit Theorem, appropriately standardized Poisson counts approach normality for large λ\lambda). Unlike the binomial, there is no fixed upper bound on kk — arbitrarily many events remain possible, just increasingly unlikely.

Figure
The Poisson PMF for λ = 3, and the binomial approximating it as n grows
k (event count) P(X = k) 0 1 2 3 4 5 6 7 λ = 3 Poisson mass mean = variance = 3 Binomial(n, 3/n) → this shape as n → ∞
TheoremPoisson limit theorem (law of rare events)

Let (pn)n≥1(p_n)_{n \ge 1} be a sequence of probabilities such that

lim⁡n→∞n pn=λ\lim_{n\to\infty} n\,p_n = \lambda

for some fixed constant λ>0\lambda > 0. For each nn, let Xn∼Binomial(n,pn)X_n \sim \mathrm{Binomial}(n, p_n). Then for every fixed integer k≥0k \ge 0,

lim⁡n→∞P(Xn=k)=e−λλkk!.\lim_{n \to \infty} P(X_n = k) = \frac{e^{-\lambda}\lambda^k}{k!}.

That is, the binomial PMF converges pointwise to the Poisson PMF with rate λ\lambda.

Proof

Technique: direct algebraic manipulation of the binomial PMF, term-by-term limit evaluation, justified by elementary limit laws (product and composition of convergent sequences), with no appeal to any result depending on the Poisson distribution itself. Fix k≥0k \ge 0 and write λn:=npn\lambda_n := n p_n, so that λn→λ\lambda_n \to \lambda by hypothesis, and pn=λn/np_n = \lambda_n / n. By the definition of the binomial PMF,

P(Xn=k)=(nk)pnk(1−pn)n−k=n!k!(n−k)!(λnn)k(1−λnn)n−k.P(X_n = k) = \binom{n}{k} p_n^k (1-p_n)^{n-k} = \frac{n!}{k!(n-k)!}\left(\frac{\lambda_n}{n}\right)^k \left(1 - \frac{\lambda_n}{n}\right)^{n-k}.

We split this into three factors and take the limit of each, since kk is fixed and finitely many factors converging implies the product converges (limit of a product of convergent sequences).

Factor 1: the combinatorial term. Write

n!k!(n−k)!⋅1nk=1k!⋅n(n−1)(n−2)⋯(n−k+1)nk=1k!∏j=0k−1(1−jn).\frac{n!}{k!(n-k)!}\cdot \frac{1}{n^k} = \frac{1}{k!}\cdot \frac{n(n-1)(n-2)\cdots(n-k+1)}{n^k} = \frac{1}{k!}\prod_{j=0}^{k-1}\left(1 - \frac{j}{n}\right).

Since kk is fixed, each of the kk factors 1−j/n→11 - j/n \to 1 as n→∞n \to \infty, so the finite product converges to 11. Hence

n!k!(n−k)! nk⟶1k!.\frac{n!}{k!(n-k)!\,n^k} \longrightarrow \frac{1}{k!}.

Factor 2: the numerator power. We have λnk→λk\lambda_n^k \to \lambda^k because λn→λ\lambda_n \to \lambda and x↦xkx \mapsto x^k is continuous (a fixed finite power of a convergent sequence converges to the power of the limit).

Factor 3: the (1−pn)n−k(1-p_n)^{n-k} term. We claim

(1−λnn)n−k⟶e−λ.\left(1 - \frac{\lambda_n}{n}\right)^{n-k} \longrightarrow e^{-\lambda}.

To see this, write (1−λn/n)n−k=(1−λn/n)n⋅(1−λn/n)−k(1-\lambda_n/n)^{n-k} = (1-\lambda_n/n)^n \cdot (1-\lambda_n/n)^{-k}. Since λn→λ\lambda_n \to \lambda, for all sufficiently large nn we have λn/n→0\lambda_n/n \to 0, so (1−λn/n)−k→1−k=1(1-\lambda_n/n)^{-k} \to 1^{-k} = 1 (continuity of x↦x−kx \mapsto x^{-k} at x=1x=1, valid once 1−λn/n1-\lambda_n/n is bounded away from 00, which holds for large nn). For the main factor, take logarithms (valid for large nn since 1−λn/n∈(0,1)1 - \lambda_n/n \in (0,1) eventually):

nlog⁡(1−λnn).n \log\left(1 - \frac{\lambda_n}{n}\right).

Use the elementary inequality, valid for 0≤x<10 \le x < 1,

−x−x21−x≤log⁡(1−x)≤−x,-x - \frac{x^2}{1-x} \le \log(1-x) \le -x,

which follows from the Taylor series log⁡(1−x)=−∑m≥1xm/m\log(1-x) = -\sum_{m\ge 1} x^m/m: the upper bound drops all terms beyond the first (all negative), and the lower bound follows by bounding ∑m≥2xm/m≤∑m≥2xm=x2/(1−x)\sum_{m \ge 2} x^m/m \le \sum_{m\ge 2} x^m = x^2/(1-x). Apply this with x=λn/nx = \lambda_n/n:

−λn−(λn/n)λn1−λn/n  ≤  nlog⁡(1−λnn)  ≤  −λn.-\lambda_n - \frac{(\lambda_n/n)\lambda_n}{1 - \lambda_n/n} \;\le\; n\log\left(1-\frac{\lambda_n}{n}\right) \;\le\; -\lambda_n.

As n→∞n \to \infty, λn→λ\lambda_n \to \lambda and λn/n→0\lambda_n/n \to 0, so the left side tends to −λ−0=−λ-\lambda - 0 = -\lambda and the right side tends to −λ-\lambda. By the squeeze theorem,

nlog⁡(1−λnn)⟶−λ,n \log\left(1 - \frac{\lambda_n}{n}\right) \longrightarrow -\lambda,

and by continuity of the exponential function,

(1−λnn)n=exp⁡(nlog⁡(1−λnn))⟶e−λ.\left(1-\frac{\lambda_n}{n}\right)^n = \exp\left(n\log\left(1-\frac{\lambda_n}{n}\right)\right) \longrightarrow e^{-\lambda}.

Combining with (1−λn/n)−k→1(1-\lambda_n/n)^{-k} \to 1 gives Factor 3: (1−λn/n)n−k→e−λ(1-\lambda_n/n)^{n-k} \to e^{-\lambda}.

Combining the factors. Multiplying the three convergent factors,

P(Xn=k)=n!k!(n−k)! nk⏟→ 1/k!⋅λnk⏟→ λk⋅(1−λnn)n−k⏟→ e−λ⟶λkk!e−λ,P(X_n = k) = \underbrace{\frac{n!}{k!(n-k)!\,n^k}}_{\to\, 1/k!} \cdot \underbrace{\lambda_n^k}_{\to\, \lambda^k} \cdot \underbrace{\left(1-\frac{\lambda_n}{n}\right)^{n-k}}_{\to\, e^{-\lambda}} \longrightarrow \frac{\lambda^k}{k!}e^{-\lambda},

which is exactly the Poisson PMF at kk. Since kk was an arbitrary fixed nonnegative integer, this holds for every kk, completing the proof. ■\blacksquare

∎
LemmaExplicit non-asymptotic error bound

Let X∼Binomial(n,p)X \sim \mathrm{Binomial}(n,p) and Y∼Poisson(np)Y \sim \mathrm{Poisson}(np). Then, writing λ=np\lambda = np,

∑k=0n∣P(X=k)−P(Y=k)∣  ≤  2np2=2λ2n.\sum_{k=0}^{n} \big| P(X=k) - P(Y=k) \big| \;\le\; 2np^2 = \frac{2\lambda^2}{n}.

(This is a special case of the Chen–Stein bound for Poisson approximation; it shows the theorem above is not merely a limit statement but comes with a quantitative rate of convergence — of order 1/n1/n for fixed λ\lambda.)

Worked examples

Example 1 — a call center with an asymmetric, non-round rate. A support line receives calls at an average rate of λ=4.2\lambda = 4.2 calls per minute. Assume the count of calls in a given minute is Poisson (justified by the limit theorem, treating the minute as split into many independent, low-probability micro-intervals). Find the probability that exactly 6 calls arrive in a given minute, and the probability of at least 2 calls. Using the PMF with λ=4.2\lambda = 4.2, k=6k=6:

P(X=6)=e−4.2(4.2)66!.P(X=6) = \frac{e^{-4.2}(4.2)^6}{6!}.

Compute 4.264.2^6: 4.22=17.644.2^2 = 17.64, 4.23=74.0884.2^3 = 74.088, 4.24=311.16964.2^4 = 311.1696, 4.25=1306.912324.2^5 = 1306.91232, 4.26=5489.0317444.2^6 = 5489.031744. Also 6!=7206! = 720 and e−4.2≈0.0149956e^{-4.2} \approx 0.0149956. So

P(X=6)≈0.0149956×5489.031744720≈82.3153720≈0.11433.P(X=6) \approx \frac{0.0149956 \times 5489.031744}{720} \approx \frac{82.3153}{720} \approx 0.11433.

For "at least 2," use the complement:

P(X≥2)=1−P(X=0)−P(X=1).P(X \ge 2) = 1 - P(X=0) - P(X=1).

P(X=0)=e−4.2≈0.0149956P(X=0) = e^{-4.2} \approx 0.0149956. P(X=1)=e−4.2⋅4.2≈0.0629821P(X=1) = e^{-4.2}\cdot 4.2 \approx 0.0629821. So

P(X≥2)≈1−0.0149956−0.0629821=0.9220223.P(X \ge 2) \approx 1 - 0.0149956 - 0.0629821 = 0.9220223.

So there is roughly an 11.4% chance of exactly 6 calls in the minute, and a 92.2% chance of at least 2 calls — reflecting that with rate 4.2, having fewer than 2 calls is the unusual event.

Example 2 — using the binomial approximation to justify (and check) the Poisson model. A factory produces n=2000n = 2000 microchips per day; each chip is independently defective with probability p=0.0021p = 0.0021 (an intentionally non-round pair of numbers). The number of defective chips per day is exactly X∼Binomial(2000,0.0021)X \sim \mathrm{Binomial}(2000, 0.0021). Approximate P(X=3)P(X = 3) using the Poisson approximation, and bound the error using the lemma above. First, λ=np=2000×0.0021=4.2\lambda = np = 2000 \times 0.0021 = 4.2 (chosen to match Example 1, so we can compare the exact binomial value to the limiting Poisson value). The Poisson approximation gives, from Example 1's method with k=3k=3:

4.23=74.088,3!=6,P(Y=3)=e−4.2(74.088)6≈0.0149956×74.0886≈1.110976≈0.18516.4.2^3 = 74.088, \qquad 3! = 6, \qquad P(Y=3) = \frac{e^{-4.2}(74.088)}{6} \approx \frac{0.0149956 \times 74.088}{6} \approx \frac{1.11097}{6} \approx 0.18516.

Now compute the true binomial value directly:

P(X=3)=(20003)(0.0021)3(0.9979)1997.P(X=3) = \binom{2000}{3}(0.0021)^3(0.9979)^{1997}.

We have (20003)=2000⋅1999⋅19986=2000×1999×19986\binom{2000}{3} = \frac{2000 \cdot 1999 \cdot 1998}{6} = \frac{2000 \times 1999 \times 1998}{6}. Compute numerator: 2000×1999=3,998,0002000 \times 1999 = 3{,}998{,}000; times 1998=7,988,004,0001998 = 7{,}988{,}004{,}000; divide by 6: 1,331,334,0001{,}331{,}334{,}000. Next, (0.0021)3=9.261×10−9(0.0021)^3 = 9.261\times 10^{-9}. Multiplying: 1,331,334,000×9.261×10−9≈12.3311{,}331{,}334{,}000 \times 9.261\times10^{-9} \approx 12.331. Finally, (0.9979)1997(0.9979)^{1997}: take logs, log⁡(0.9979)≈−0.0021022\log(0.9979) \approx -0.0021022 (using log⁡(1−x)≈−x−x2/2\log(1-x)\approx -x - x^2/2 for small xx, here x=0.0021x=0.0021), so 1997×(−0.0021022)≈−4.1981997 \times (-0.0021022) \approx -4.198, giving (0.9979)1997≈e−4.198≈0.01502(0.9979)^{1997} \approx e^{-4.198} \approx 0.01502. Combining: P(X=3)≈12.331×0.01502≈0.18521P(X=3) \approx 12.331 \times 0.01502 \approx 0.18521. Comparing to the Poisson approximation 0.185160.18516, the two differ by about 0.000050.00005 — very small, consistent with the lemma's bound 2λ2/n=2(4.2)2/2000=35.28/2000=0.017642\lambda^2/n = 2(4.2)^2/2000 = 35.28/2000 = 0.01764 on the total variation distance summed over all kk (an aggregate bound, so the single-term discrepancy is comfortably inside it). This numerically confirms the Poisson limit theorem is not just qualitative but practically accurate even at moderate nn.

Counterexamplewhy "npn→λnp_n \to \lambda fixed" cannot be dropped

The hypothesis npn→λn p_n \to \lambda (a finite constant) is essential; if instead npn→∞np_n \to \infty, the binomial distribution does not converge to any Poisson law, and the limiting PMF claimed by the theorem fails outright. Take pn=1/np_n = 1/\sqrt{n}, so npn=n→∞np_n = \sqrt{n} \to \infty rather than to a finite λ\lambda. Fix k=0k=0 and examine P(Xn=0)=(1−pn)n=(1−1/n)nP(X_n = 0) = (1-p_n)^n = (1 - 1/\sqrt n)^n. Taking logs,

nlog⁡(1−1n)≈n(−1n−12n)=−n−12→−∞,n\log\left(1 - \frac{1}{\sqrt n}\right) \approx n\left(-\frac{1}{\sqrt n} - \frac{1}{2n}\right) = -\sqrt n - \tfrac12 \to -\infty,

so P(Xn=0)→0P(X_n=0) \to 0. But under any fixed Poisson(λ)(\lambda) law, P(Y=0)=e−λ>0P(Y=0) = e^{-\lambda} > 0 for every finite λ\lambda. Since P(Xn=0)→0P(X_n=0) \to 0 while no Poisson PMF has mass 00 at k=0k=0, the binomial sequence cannot be converging to any Poisson distribution. Concretely, at n=10000n = 10000: pn=0.01p_n = 0.01, npn=100np_n = 100, and P(Xn=0)=(0.99)10000≈e−100≈3.7×10−44P(X_n = 0) = (0.99)^{10000} \approx e^{-100} \approx 3.7\times10^{-44} — vanishingly small, utterly unlike a Poisson mass at 00. This shows the finiteness of the limiting rate λ=lim⁡npn\lambda = \lim np_n is not a technical nicety but a load-bearing hypothesis: without it, the entire limiting shape changes (in this regime the correctly centered and scaled binomial instead approaches a Normal distribution, by the Central Limit Theorem, not a Poisson one).

Scope, edge cases & common traps

SituationCorrect handling
nn large, pp small, npnp moderatePoisson(np)(np) is an excellent approximation to Binomial(n,p)(n,p)
nn large, pp small, np→∞np \to \inftyPoisson approximation fails; use Normal approximation instead
nn large, pp not small (e.g. p=0.5p=0.5)Use Normal approximation to the binomial, not Poisson
Sum of independent PoissonsPoisson(λ1)+Poisson(λ2)=Poisson(λ1+λ2)\mathrm{Poisson}(\lambda_1) + \mathrm{Poisson}(\lambda_2) = \mathrm{Poisson}(\lambda_1+\lambda_2) (provable via convolution)
Conditioning a Poisson count on time intervalSplitting a Poisson process over disjoint intervals yields independent Poissons — foreshadows the Exponential Distribution as the interarrival law
Estimating λ\lambda from dataThe sample mean is the maximum likelihood estimator, since E[X]=λ\mathbb E[X]=\lambda

The mathematician behind it

Origins · 1 minSiméon Denis Poisson and how this was discovered
SD
Siméon Denis Poisson
1781–1840 · France, mathematical physics and probability

Siméon Denis Poisson was a French mathematician and physicist whose name is attached to results across mathematical physics — Poisson's equation in electrostatics and gravitation, the Poisson bracket in classical mechanics, and Poisson summation — as well as to this probability distribution. His 1837 treatise Recherches sur la probabilité des jugements en matière criminelle et en matière civile introduced the distribution now bearing his name as a limiting case of the binomial law, in the context of modeling the probability of judicial errors in criminal versus civil trials — an early application to legal and social statistics rather than to physics.

History
- 1781 — Siméon Denis Poisson is born in Pithiviers, France.
- 1837 — Poisson publishes the limiting binomial argument that yields the distribution, in a work on judicial statistics.
- 1898 — Ladislaus Bortkiewicz popularizes the distribution using Prussian cavalry deaths from horse kicks as a "law of small numbers" example.
- 1909 — Agner Krarup Erlang applies Poisson-type counting to telephone call arrivals, founding queueing theory.
- 1940 — Poisson processes become a standard tool across physics (radioactive decay counts) and operations research.
Why does the Poisson distribution have only one parameter, while the binomial has two?

The binomial needs both the number of trials nn and the per-trial probability pp because it describes a finite, fixed-size population of trials. The Poisson distribution arises precisely when we let n→∞n \to \infty and p→0p \to 0 together, so that the individual values of nn and pp become irrelevant — only their product np=λnp = \lambda survives in the limit. This is analogous to how a physical rate (e.g., "5 events per minute") does not care how finely you subdivide the minute; the subdivided-count description (nn, pp) collapses into a single rate description (λ\lambda) once the subdivision is infinitely fine.

Is the Poisson distribution the same as a Poisson process?

No, though they are closely related. The Poisson distribution is a single random variable's distribution — the count of events in one fixed interval. A Poisson process is a full stochastic process on a continuum (usually time), specifying that counts over any interval are Poisson-distributed with rate proportional to the interval's length, and counts over disjoint intervals are independent. The Poisson distribution is the "one-time snapshot" of a Poisson process; the interarrival times of a Poisson process follow the Exponential Distribution, giving the continuous-time partner to this discrete count distribution.

How do I know when to use Poisson versus Binomial versus Normal for a counting problem?

Use the exact Binomial(n,p)(n,p) whenever nn and pp are both genuinely known and finite and exact computation is feasible. Use the Poisson approximation when nn is large, pp is small, and the product np=λnp = \lambda is a moderate, stable number (a common rule of thumb: n≥20n\ge 20, p≤0.05p \le 0.05). Use the Normal approximation to the binomial (via the Central Limit Theorem) when nn is large but pp is not small — specifically when npnp and n(1−p)n(1-p) are both comfortably large (e.g., both ≥5\ge 5 or ≥10\ge 10), since in that regime the binomial is well approximated by a bell curve rather than a skewed count distribution.

Why is the variance of a Poisson random variable equal to its mean?

This equality, Var(X)=E[X]=λ\mathrm{Var}(X) = \mathbb E[X] = \lambda, follows directly from computing moments with the PMF: one shows E[X]=λ\mathbb E[X] = \lambda and E[X(X−1)]=λ2\mathbb E[X(X-1)] = \lambda^2 by direct series manipulation, giving E[X2]=λ2+λ\mathbb E[X^2] = \lambda^2+\lambda and hence Var(X)=E[X2]−(E[X])2=λ\mathrm{Var}(X)=\mathbb E[X^2]-(\mathbb E[X])^2 = \lambda. It is also inherited from the binomial limit: Binomial(n,p)\mathrm{Binomial}(n,p) has mean npnp and variance np(1−p)np(1-p), and as p→0p\to 0 with np=λnp=\lambda fixed, the variance np(1−p)→λ(1−0)=λnp(1-p) \to \lambda(1-0) = \lambda, matching the mean in the limit. This equal-mean-variance property is a useful diagnostic: real count data with variance far from its mean (over- or under-dispersion) is a signal that a pure Poisson model may not fit.

Can the Poisson distribution take negative or non-integer values?

No. By definition the support of the Poisson PMF is {0,1,2,3,… }\{0, 1, 2, 3, \dots\} — it is a discrete distribution on the nonnegative integers, since it counts a whole number of discrete events (calls, decays, defects) and a count can never be negative or fractional. This distinguishes it sharply from continuous distributions like the Exponential Distribution, which models a continuous waiting time between such discrete Poisson events, but is itself defined on [0,∞)[0,\infty) rather than on the integers.

What does the rate parameter λ\lambda actually represent, physically?

The parameter λ\lambda is simultaneously the expected number of events in the interval or region under study and the variance of that count, since E[X]=Var(X)=λ\mathbb E[X]=\mathrm{Var}(X)=\lambda for X∼Poisson(λ)X\sim\mathrm{Poisson}(\lambda). It is derived as λ=np\lambda = np in the limiting binomial construction — the total expected count of successes as the number of trials grows and the per-trial probability shrinks correspondingly. Practically, λ\lambda is estimated from data as the observed average count per unit interval (e.g., average calls per minute, average decays per second), and it fully determines the entire probability mass function via the formula P(X=k)=e−λλk/k!P(X=k)=e^{-\lambda}\lambda^k/k!.

Does the theorem give any control over how fast the binomial approaches the Poisson, or is it purely a limiting statement?

It is more than a bare limit. As stated in the accompanying Chen–Stein-type lemma, the total variation distance between Binomial(n,p)\mathrm{Binomial}(n,p) and Poisson(np)\mathrm{Poisson}(np) is bounded above by 2np2=2λ2/n2np^2 = 2\lambda^2/n for λ=np\lambda = np fixed. This shows the approximation error shrinks like 1/n1/n as nn grows with λ\lambda held fixed, giving a concrete, checkable guarantee rather than just qualitative reassurance — exactly the kind of bound used to justify the approximation numerically in applied settings such as Example 2 above.

Before this

Where this leads