Poisson Distribution

The Poisson distribution is the universal law of counting rare, independent things. Point a Geiger counter at a weak source, watch photons trickle onto a detector, tally galaxies in a patch of sky, or log defects in a wafer — whenever many independent opportunities each carry a tiny chance of an event, the number $k$ you count obeys $P(k)=e^{-\lambda}\lambda^{k}/k!$ . A single parameter, the expected count $\lambda$ , fixes the whole distribution. That economy is no accident: the microscopic bookkeeping of how many opportunities there were and how small each chance was washes out, leaving only their product $\lambda$ — which is why one law describes phenomena that share no physics beyond independence and a constant rate. Two roads lead here, and this article walks both. The first is the binomial limit: slice the window into $n$ tiny cells, each independently holding an event with probability $p=\lambda/n$ , and let $n\to\infty$ at fixed $\lambda=np$ . The second is the Poisson process: assume one event in an instant $\mathrm{d}t$ with probability $\nu\,\mathrm{d}t$ , two-at-once negligible, disjoint intervals independent — and solve the resulting master equation. Both yield the identical mass function with $\lambda=\nu t$ . The distribution's signature is stark: its mean and variance are equal, $\langle k\rangle=\operatorname{Var}(k)=\lambda$ . That single equality is the origin of the $\sqrt{N}$ error bar stamped on every counting experiment, and — through the clean generating function $G(s)=e^{\lambda(e^{s}-1)}$ — of the fact that independent Poisson streams simply add their rates. In the large- $\lambda$ limit the distribution relaxes into a Gaussian, placing Poisson at the hinge between the discrete world of Bernoulli trials and the continuous world of the normal law.

Key takeaways

The Poisson distribution gives the probability of $k$ independent events in a fixed window at constant average rate: $P(k)=e^{-\lambda}\lambda^{k}/k!$ , with $k=0,1,2,\dots$ and the single parameter $\lambda>0$ .
It is the rare-event limit of the binomial — $n\to\infty$ , $p\to0$ , product $np=\lambda$ fixed — so only the expected total survives, not the microscopic count of opportunities.
Its defining signature is $\langle k\rangle=\operatorname{Var}(k)=\lambda$ ; the relative fluctuation is $\sigma/\langle k\rangle=1/\sqrt{\lambda}$ , the universal $\sqrt{N}$ law of detectors.
The generating function $G(s)=e^{\lambda(e^{s}-1)}$ makes every cumulant equal $\lambda$ and makes independent streams add: Pois $(\lambda_1)+$ Pois $(\lambda_2)=$ Pois $(\lambda_1+\lambda_2)$ .
For large $\lambda$ it tends to a Gaussian of mean and variance $\lambda$ (central limit theorem); for small $\lambda$ it is sharply peaked at $k=0$ — the law of rare events.
Key idea: Poisson counts arise whenever many independent chances for a rare event combine at fixed expected total λ; the resulting distribution has the unique property mean = variance = λ, and adds like a conserved current under independence.

The core insight

Poisson counts arise whenever many independent chances for a rare event combine at a fixed expected total $\lambda$ ; the resulting distribution has the unique property mean = variance = $\lambda$ , and it adds like a conserved current under independence.

Where this lives today

The Poisson law is the quiet grammar of modern measurement. Every low-count experiment at the frontier — dark-matter searches, neutrino counting, faint-source X-ray and gamma-ray astronomy — abandons Gaussian $\chi^{2}$ fitting for the Poisson likelihood, because when you expect three events you cannot pretend the count is normally distributed. Single-photon detectors report arrivals that are Poisson for coherent (laser) light, and the deviations are physics: quantum optics engineers deliberately produce sub-Poissonian, antibunched light ( $F<1$ ) as a signature of nonclassical states and a resource for metrology, while thermal light is super-Poissonian. In positron-emission tomography, images are reconstructed from Poisson-distributed coincidence counts, and the noise floor of every PET and CT scan is the same $\sqrt{N}$ law that limits a Geiger counter. Schottky's shot noise, $\langle\Delta I^{2}\rangle=2eI\,\Delta f$ , is nothing but variance-equals-mean applied to the arrival of discrete charges, setting the fundamental sensitivity limit of photodiodes, transistors, and gravitational-wave interferometers alike — and the same mathematics governs network packet arrivals, insurance claims, and the spatial statistics of galaxies.

The physical picture

Hold a Geiger counter near a weak radioactive source. The sample contains an astronomical number of atoms, each with a minuscule, independent probability of decaying in the next second. No single atom matters; what matters is that their collective expected count over the second is some finite number $\lambda$ . The reading fluctuates around $\lambda$ from second to second, and the precise law of that scatter is the Poisson distribution. The crucial move is that the binomial needs two numbers — the count of opportunities $n$ and the chance each $p$ — but the Poisson keeps only their product $\lambda=np$ . The details of how the rare total was assembled evaporate; any superposition of independent rare contributions looks Poisson, which is why the law is everywhere.

Figure

The Poisson mass function for λ = 4 counts per window: the distribution peaks jointly at k = 3 and k = 4, the mean sits at ⟨k⟩ = λ = 4 with spread σ = √λ = 2, and the single-bar value P(2) = 8e⁻⁴ ≈ 0.1465

Because variance equals mean, counting noise has a fixed character: the absolute spread grows like $\sqrt{\lambda}$ but the relative spread $1/\sqrt{\lambda}$ shrinks, so reaching one-percent precision demands about $10^{4}$ counts — the reason every counting experiment trades integration time for accuracy. And because two independent Poisson streams merge into one of the summed rate, the distribution is the atom of point processes the way the Gaussian is the atom of continuous noise. The second figure shows the other end of the story: as $\lambda$ grows the ragged, right-skewed staircase smooths into a symmetric bell.

Figure

As λ grows the Poisson distribution becomes symmetric and approaches a Gaussian of the same mean and variance: for λ = 10 the bars (Poisson) track the smooth curve (Gaussian, mean = variance = 10, σ = √10 ≈ 3.16), with the peak P(10) ≈ 0.1251 against the Gaussian value 0.1262

The core equations

The whole subject hangs on one mass function. It assigns to every non-negative integer $k$ a probability built from just $\lambda$ , and it is the object every later relation differentiates, sums, or takes limits of.

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

where: $k$ — number of events observed (a non-negative integer); $\lambda$ — expected number of events in the window, with $\lambda>0$ ; $P(k)$ — probability of observing exactly $k$ events. Reading it: the factor $\lambda^{k}/k!$ rises then falls, peaking near $k\approx\lambda$ , while $e^{-\lambda}$ is the normalising constant that makes the total probability one — because $\sum_{k\ge0}\lambda^{k}/k!=e^{\lambda}$ is precisely the exponential series. The mean and the variance are the two numbers you most often need, and for Poisson they collapse into one. This equality — not the shape of the curve — is what makes a distribution "Poisson".

\langle k\rangle=\lambda,\qquad \operatorname{Var}(k)=\langle k^{2}\rangle-\langle k\rangle^{2}=\lambda \qquad (2)

where: $\langle k\rangle$ — first moment (the average count); $\operatorname{Var}(k)$ — variance; both equal the single parameter $\lambda$ . Reading it: the standard deviation is $\sigma=\sqrt{\lambda}$ , so the fractional uncertainty $\sigma/\langle k\rangle=1/\sqrt{\lambda}$ falls as counts accumulate — the mathematical seed of the $\sqrt{N}$ error bar. Nearly every property of the distribution flows most easily from its moment-generating function, which for Poisson is remarkably compact. Derivatives at the origin give the moments; its logarithm gives the cumulants.

G(s)=\langle e^{sk}\rangle=\exp\!\left[\lambda\left(e^{s}-1\right)\right],\qquad \ln G(s)=\lambda\sum_{n=1}^{\infty}\frac{s^{n}}{n!}\;\Rightarrow\;\kappa_{n}=\lambda\ \ \forall n \qquad (3)

where: $s$ — auxiliary variable (set $s\to it$ for the characteristic function); $G(s)$ — moment-generating function; $\kappa_{n}$ — the $n$ -th cumulant. Reading it: because $\ln G$ is linear in $\lambda$ with every Taylor coefficient equal to $\lambda$ , all cumulants are $\lambda$ — mean, variance, and beyond — a compact restatement of equation (2). The generating function also settles how Poisson variables combine. For independent variables the generating function of a sum is the product of the individual ones, and the Poisson form is closed under that product.

X\sim\mathrm{Pois}(\lambda_1),\ Y\sim\mathrm{Pois}(\lambda_2)\ \text{independent}\ \Rightarrow\ X+Y\sim\mathrm{Pois}(\lambda_1+\lambda_2) \qquad (4)

where: $X,Y$ — independent Poisson counts with rates $\lambda_1,\lambda_2$ ; the sum is Poisson with the added rate. Reading it: merging two independent streams gives one Poisson stream — the "superposition" property that makes Poisson the building block of point processes. Finally, the shape coefficients read straight off the cumulants and quantify how the distribution loses its skew as $\lambda$ grows: skewness $=\kappa_3/\kappa_2^{3/2}=1/\sqrt{\lambda}$ and excess kurtosis $=\kappa_4/\kappa_2^{2}=1/\lambda$ . Both vanish as $\lambda\to\infty$ , so a large- $\lambda$ Poisson is symmetric and Gaussian-tailed — the content of the second figure.

Derivation APoisson as the limit of the binomial

The goal is to show, cleanly and with every limit justified, that the binomial distribution for $n$ independent trials of success probability $p$ becomes the Poisson distribution when the trials grow numerous and each success grows rare, with the expected number of successes $\lambda=np$ held fixed. This is the "law of rare events": it explains why the Poisson keeps only the product $np$ and forgets $n$ and $p$ separately, and it is the reason the same one-parameter law fits so many unrelated counting problems. We start from the exact binomial probability and take $n\to\infty$ with $p=\lambda/n$ , tracking the three pieces — the binomial coefficient, the success factor, and the survival factor — to their limits.

Step 1. Write the exact binomial probability of $k$ successes, $P_n(k)=\binom{n}{k}p^{k}(1-p)^{n-k}$ , and expand the coefficient as $\binom{n}{k}=\dfrac{n(n-1)\cdots(n-k+1)}{k!}$ , a ratio with exactly $k$ descending factors on top. Why: isolating the $k$ falling factors lets us see precisely which pieces survive the limit and which tend to one.

Step 2. Substitute $p=\lambda/n$ into the success factor and group powers of $n$ : $\binom{n}{k}p^{k}=\dfrac{\lambda^{k}}{k!}\displaystyle\prod_{j=0}^{k-1}\dfrac{n-j}{n}$ . Each factor $(n-j)/n\to1$ as $n\to\infty$ for fixed $k$ , so this whole block tends to $\lambda^{k}/k!$ . Why: with $k$ fixed and $n$ diverging, the finitely many corrections $j/n$ all vanish, leaving the Poisson numerator.

Step 3. Handle the survival factor by splitting the exponent: $\left(1-\dfrac{\lambda}{n}\right)^{n-k}=\left(1-\dfrac{\lambda}{n}\right)^{n}\left(1-\dfrac{\lambda}{n}\right)^{-k}$ . The second bracket $\to1$ , and the first is the classic definition of the exponential, $\left(1-\lambda/n\right)^{n}\to e^{-\lambda}$ . Why: separating the $n$ and the $-k$ powers isolates the one piece that carries the $e^{-\lambda}$ and confirms the remainder is harmless.

Step 4. Multiply the surviving pieces: $\dfrac{\lambda^{k}}{k!}\cdot 1\cdot e^{-\lambda}\cdot 1=e^{-\lambda}\dfrac{\lambda^{k}}{k!}=P(k)$ . Why: collecting the three limits reproduces equation (1) exactly, so the Poisson is literally the $n\to\infty$ , $p\to0$ , $np=\lambda$ face of the binomial. *Why the limit is tight: at $\lambda=4$ the binomial with $n=1000$ , $p=0.004$ already matches $P(k)$ to better than $10^{-3}$ , confirming that the approximation is excellent well before $n\to\infty$ .*

Go deeper · Derivation B · 1 minMean, variance, and the generating function

Derivation BMean, variance, and the generating function

The goal is to derive the Poisson signature — that the mean and the variance both equal $\lambda$ — directly and without ad-hoc index shifts, by building the moment-generating function $G(s)=\langle e^{sk}\rangle$ and reading the first two moments off its derivatives at the origin. Differentiated once more or expanded logarithmically, the same object delivers every higher cumulant, so it encapsulates the entire moment structure of the distribution. We first sum the defining series to obtain $G(s)$ in closed form, then differentiate twice and evaluate at $s=0$ .

Step 1. Insert the mass function into $G(s)=\sum_{k\ge0}e^{sk}e^{-\lambda}\lambda^{k}/k!$ and group $e^{sk}\lambda^{k}=(\lambda e^{s})^{k}$ , giving $G(s)=e^{-\lambda}\sum_{k\ge0}(\lambda e^{s})^{k}/k!=e^{-\lambda}e^{\lambda e^{s}}=\exp[\lambda(e^{s}-1)]$ . Why: the inner sum is again an exponential series, now with argument $\lambda e^{s}$ , which is what makes the Poisson generating function so clean.

Step 2. Differentiate once: $G'(s)=\lambda e^{s}\exp[\lambda(e^{s}-1)]$ , so $G'(0)=\lambda$ . Since $G'(0)=\langle k\rangle$ , the mean is $\langle k\rangle=\lambda$ . Why: the first derivative of the generating function at the origin is by construction the first moment, and the chain rule brings down exactly one factor of $\lambda e^{s}$ .

Step 3. Differentiate again with the product rule: $G''(s)=(\lambda e^{s}+\lambda^{2}e^{2s})\exp[\lambda(e^{s}-1)]$ , so $G''(0)=\lambda+\lambda^{2}=\langle k^{2}\rangle$ . Why: the second derivative at the origin is the second moment, and here it carries both a linear and a quadratic piece in $\lambda$ .

Step 4. Form the variance: $\operatorname{Var}(k)=\langle k^{2}\rangle-\langle k\rangle^{2}=(\lambda^{2}+\lambda)-\lambda^{2}=\lambda$ . The $\lambda^{2}$ terms cancel, leaving variance = mean = $\lambda$ . Why: this cancellation is the whole point — it is the algebraic origin of the counting-statistics signature and, equivalently, of $\kappa_{2}=\lambda$ from equation (3). *Why this is self-consistent: $\ln G(s)=\lambda(e^{s}-1)$ gives $\kappa_{n}=\lambda$ for all $n$ , so in particular $\kappa_1=\langle k\rangle=\lambda$ and $\kappa_2=\operatorname{Var}(k)=\lambda$ , in full agreement with the direct calculation above.*

Go deeper · Derivation C · 1 minPoisson from the process axioms

Derivation CPoisson from the process axioms

The goal is to reach the same mass function from a completely different starting point — not a limit of trials, but the differential description of a continuous-time process — so that the two derivations reinforce each other and expose the meaning of $\lambda=\nu t$ . We postulate a homogeneous Poisson process: in any short interval $[t,t+\mathrm{d}t]$ exactly one event arrives with probability $\nu\,\mathrm{d}t$ , none with probability $1-\nu\,\mathrm{d}t$ , two or more with probability of order $(\mathrm{d}t)^{2}$ (negligible), and non-overlapping intervals are independent. From these axioms we write a balance equation for $P_k(t)$ , pass to a differential-difference equation, and solve it.

Step 1. Balance the probability of $k$ events at time $t+\mathrm{d}t$ : either $k$ events had already arrived and none came in $\mathrm{d}t$ , or $k-1$ had arrived and exactly one came. This gives $P_k(t+\mathrm{d}t)=P_k(t)(1-\nu\,\mathrm{d}t)+P_{k-1}(t)\,\nu\,\mathrm{d}t$ . Why: the two-or-more-events term is higher order in $\mathrm{d}t$ and drops out, so only these two mutually exclusive histories contribute at first order.

Step 2. Rearrange and let $\mathrm{d}t\to0$ to obtain the master equations $\dfrac{\mathrm{d}P_k}{\mathrm{d}t}=-\nu P_k(t)+\nu P_{k-1}(t)$ for $k\ge1$ , together with $\dfrac{\mathrm{d}P_0}{\mathrm{d}t}=-\nu P_0(t)$ . With the initial condition $P_0(0)=1$ the zeroth equation integrates immediately to $P_0(t)=e^{-\nu t}$ . Why: the $k=0$ state can only be left, never entered, so it decays as a pure exponential — the survival probability of "no event yet".

Step 3. Solve the hierarchy by induction with the ansatz $P_k(t)=e^{-\nu t}(\nu t)^{k}/k!$ . Substituting into $\dot P_k=-\nu P_k+\nu P_{k-1}$ , the terms match order by order, and the base case $P_0=e^{-\nu t}$ holds; setting $\lambda=\nu t$ gives $P_k=e^{-\lambda}\lambda^{k}/k!$ . Why: identifying $\lambda=\nu t$ shows the parameter is literally the rate times the elapsed window, unifying the process picture with the binomial-limit result (1). *Why the solution is complete: the family $P_k(t)$ normalises to one at every $t$ since $\sum_k(\nu t)^k/k!=e^{\nu t}$ , and the continuum-rate result agrees exactly with the binomial limit once $\lambda=\nu t$ is identified.*

Limiting cases

The single parameter $\lambda$ sweeps the distribution through three qualitatively different regimes, and the same parameter fixes the counting noise everywhere in between.

Case	Result	Meaning
Small $\lambda$ ( $\lambda\ll1$ )	$P(0)\approx1-\lambda,\ \ P(1)\approx\lambda,\ \ P(k\ge2)=O(\lambda^{2})$	Sharply peaked at $k=0$ ; seeing two or more events is doubly unlikely — the pure "law of rare events" regime.
Binomial origin	$\displaystyle\lim_{n\to\infty}\binom{n}{k}p^{k}(1-p)^{n-k}=e^{-\lambda}\frac{\lambda^{k}}{k!},\ np=\lambda$	The rare-event limit of the binomial; an excellent approximation already for $n\gtrsim20$ , $p\lesssim0.05$ .
Large $\lambda$ ( $\lambda\gg1$ )	$\displaystyle P(k)\approx\frac{1}{\sqrt{2\pi\lambda}}\exp\!\left[-\frac{(k-\lambda)^{2}}{2\lambda}\right]$	Tends to a Gaussian of mean and variance $\lambda$ (central limit theorem); skewness $1/\sqrt{\lambda}\to0$ , so the shape becomes symmetric.
Relative fluctuation	$\displaystyle\frac{\sigma}{\langle k\rangle}=\frac{\sqrt{\lambda}}{\lambda}=\frac{1}{\sqrt{\lambda}}$	Fractional counting noise falls as $1/\sqrt{\lambda}$ — the universal $\sqrt{N}$ statistics behind every detector's error bar.

Common trapEqual mean and variance is an assumption, not a guarantee

The identity $\langle k\rangle=\operatorname{Var}(k)=\lambda$ holds only when the rate is genuinely constant and the events genuinely independent. Real data are frequently over-dispersed: correlations, or a rate that itself fluctuates from window to window, inflate the variance above the mean (a mixture of Poissons with different $\lambda$ is always over-dispersed). Assume Poisson and slap $\sqrt{N}$ error bars on such data, and you systematically underestimate the true uncertainty. The diagnostic is the Fano factor $F=\operatorname{Var}(k)/\langle k\rangle$ : $F=1$ is Poisson, $F>1$ is over-dispersed (clustering, bursts), and $F<1$ is sub-Poissonian (regularity, anti-bunching) — itself a signature of nonclassical light or dead-time effects. Never assume $F=1$ ; measure it.

Worked examples

1 · Counting statistics of a weak source (fully worked)

A Geiger counter near a weak sample registers on average $\lambda=4$ counts per second. Find the probability of exactly 2 counts in one second, the probability of at least one count, the standard deviation of a one-second reading, and the integration time needed for 1 % relative precision.

Step 1. The probability of exactly $k$ counts is $P(k)=e^{-4}4^{k}/k!$ . For $k=2$ : $P(2)=e^{-4}\cdot16/2!=8e^{-4}$ . With $e^{-4}=0.018316$ this is $P(2)=\mathbf{0.1465}$ . The window mean $\lambda=4$ is the only input; everything follows from it.

Step 2. The probability of at least one count is the complement of zero counts: $P(k\ge1)=1-P(0)=1-e^{-4}=1-0.018316=\mathbf{0.9817}$ . So a "silent" second is only about a $1.8\,\%$ event at this rate.

Step 3. The variance equals the mean, so $\operatorname{Var}=4$ and the standard deviation of a single one-second reading is $\sigma=\sqrt{\lambda}=\sqrt{4}=\mathbf{2\ \text{counts}}$ . The relative fluctuation is $\sigma/\lambda=2/4=50\,\%$ — a single second is a very noisy measurement.

Step 4. To reach $1\,\%$ relative precision we need $1/\sqrt{N}=0.01$ , i.e. $N=10^{4}$ total counts. At $4$ counts per second the required time is $t=N/\lambda=10^{4}/4=\mathbf{2500\ \text{s}}\approx42$ minutes — ten times better precision costs a hundredfold longer counting.

2 · Merging two streams, and the Gaussian crossover

Two independent radioactive sources are counted together; alone they would give Poisson counts at rates $\lambda_1=3$ and $\lambda_2=5$ per second. What distribution governs the combined per-second count, what is its standard deviation, and how close is its peak to the Gaussian approximation?

Step 1. By the additivity property (equation 4), the combined count is Poisson with rate $\lambda=\lambda_1+\lambda_2=3+5=\mathbf{8}$ per second. Independence is essential; correlated sources would violate additivity and give an over-dispersed sum.

Step 2. Because variance equals mean for the combined Poisson, $\operatorname{Var}=8$ and the standard deviation is $\sigma=\sqrt{8}=\mathbf{2.83\ \text{counts}}$ .

Step 3. Evaluate the peak probability exactly: $P(8)=e^{-8}\,8^{8}/8!$ . With $8^{8}=16\,777\,216$ , $8!=40\,320$ , and $e^{-8}=3.3546\times10^{-4}$ , this gives $P(8)=\mathbf{0.1396}$ .

Step 4. Compare with the large- $\lambda$ Gaussian of the same mean and variance evaluated at the mean: $\dfrac{1}{\sqrt{2\pi\lambda}}=\dfrac{1}{\sqrt{16\pi}}=\mathbf{0.1410}$ . The two agree to about $1\,\%$ already at $\lambda=8$ — the crossover to Gaussian behaviour is well under way even for a handful of expected counts, exactly as the vanishing skewness $1/\sqrt{8}\approx0.35$ foreshadows.

Physics of the portrait

The distribution carries the name of a mathematician who wrote it down almost in passing, inside a treatise on jury verdicts, and then largely moved on — leaving others, decades later, to discover that this obscure limit of the binomial governed everything from horse-kicks to radioactive decay.

Origins · 1 minSiméon Denis Poisson and how this was discovered

Siméon Denis Poisson

1781–1840 · probability, potential theory, and mathematical physics

Siméon Denis Poisson was a towering figure of early nineteenth-century French mathematical physics, a student of Laplace and Lagrange whose name is stamped across the subject: Poisson's equation of potential theory, the Poisson bracket of analytical mechanics, Poisson's ratio in elasticity, and the bright Poisson–Arago spot that vindicated the wave theory of light. The distribution that bears his name appeared in 1837 in his Recherches sur la probabilité des jugements en matière criminelle et en matière civile, where he studied the number of guilty verdicts among many independent, individually-unlikely trials and derived $e^{-\lambda}\lambda^{k}/k!$ as the limiting form of the binomial. Characteristically, Poisson treated it as a technical device rather than a centrepiece, and it drew little notice for half a century. Its rise came from others: Ladislaus Bortkiewicz, who in 1898 coined the "law of small numbers" and famously fitted the distribution to the annual horse-kick deaths in the Prussian cavalry, and the experimenters Rutherford and Geiger, whose 1910 alpha-particle counts (analysed with Harry Bateman) gave it its first great physical confirmation. Poisson scattered foundational results across physics and would likely be surprised that a footnote in a book about courtrooms became his most quoted legacy.

History

1837

— Poisson derives the distribution as a limit of the binomial in his treatise on judgments

1860

— Newcomb uses the law to model the apparent clustering of stars into groups

1898

— Bortkiewicz's law of small numbers fits Prussian cavalry horse-kick deaths

1910

— Rutherford, Geiger, and Bateman confirm Poisson statistics for alpha-particle counts

1918

— Schottky derives electronic shot noise from Poisson arrival statistics

Going further

Texts. Stone & Goldbart, Mathematics for Physics, §11 (probability and the Poisson process) · Arfken, Weber & Harris, Mathematical Methods for Physicists, 7th ed., Ch. 23 (probability and statistics) · Hassani, Mathematical Physics, chapters on probability distributions · Riley, Hobson & Bence, Mathematical Methods for Physics and Engineering, 3rd ed., Ch. 30 · Byron & Fuller, Mathematics of Classical and Quantum Physics, on distributions and generating functions · Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, Ch. VI (the definitive account of the Poisson process). The key results to keep in mind: (1) the binomial limit $\binom{n}{k}(\lambda/n)^{k}(1-\lambda/n)^{n-k}\to e^{-\lambda}\lambda^{k}/k!$ ; (2) normalisation via the exponential series $\sum_k\lambda^k/k!=e^{\lambda}$ ; (3) $G(s)=e^{\lambda(e^{s}-1)}$ and its first two derivatives giving $\langle k\rangle=\lambda$ , $\operatorname{Var}(k)=\lambda$ ; (4) the master-equation solution $P_k(t)=e^{-\nu t}(\nu t)^{k}/k!$ with $\lambda=\nu t$ . Numerical anchors: $P(2)=8e^{-4}=0.1465$ at $\lambda=4$ ; $\sigma=\sqrt{4}=2$ ; combined rate $3+5=8$ with $P(8)=0.1396$ versus Gaussian $0.1410$ ; peak $P(10)=0.1251$ versus Gaussian $0.1262$ at $\lambda=10$ .

Prerequisites. Binomial and Bernoulli Distributions · Random Variables and Probability Distributions · Expectation, Variance and Moments · Axioms of Probability. Leads to. Gaussian (Normal) Distribution · Central Limit Theorem · Moment-Generating and Characteristic Functions · Laws of Large Numbers and Convergence of Random Variables.

Ask the tutor: Ask me to walk through why a Geiger counter's noise scales as √N, to fit a Fano factor to your own count data, or to show how the Poisson slides into a Gaussian as λ grows.

Why is there only one parameter when the binomial has two?

The binomial needs both the number of trials $n$ and the per-trial success probability $p$ . In the rare-event limit those two collapse into a single meaningful combination, the expected count $\lambda=np$ . Sending $n\to\infty$ with $p$ fixed would give infinitely many expected events, which is unphysical for a rare process, and fixing $n$ while $p\to0$ would leave a trivial point mass at zero. Only the product $np$ stays finite and informative, so the Poisson forgets the microscopic split between "many chances" and "small chance each" and remembers only their product. That erasure is exactly why one law fits so many different underlying processes.

What exactly does it mean that mean equals variance?

It means the two most basic descriptors of a Poisson count — its average $\langle k\rangle$ and its spread $\operatorname{Var}(k)$ — are not independent; both equal $\lambda$ . Physically, the fluctuations are locked to the signal: a source averaging $\lambda$ counts has a standard deviation of exactly $\sqrt{\lambda}$ . This is why a measured count $N$ is reported as $N\pm\sqrt{N}$ , and why the fractional error $1/\sqrt{N}$ improves only as the square root of counting time. It is a genuine constraint, not something you can impose on any data — a set with mean 50 and variance 200 is simply not Poisson.

How can two completely different derivations give the same distribution?

Because they describe the same underlying idealisation from different angles. The binomial-limit derivation chops a window into many tiny independent cells and asks how many contain an event; the process derivation lets events arrive continuously in time at a constant rate with no memory. Both encode the same three assumptions — independence, a constant rate, and negligible chance of two events coinciding — and any set of assumptions that pins those down forces the same answer. The identification $\lambda=\nu t$ (rate times elapsed time) is the bridge: the number of tiny cells times the per-cell probability equals the rate times the window length.

When is it wrong to use Poisson statistics?

Whenever independence or a constant rate fails. If events cluster (bursts, avalanches, correlated triggers) or if the rate itself drifts or fluctuates between measurement windows, the variance exceeds the mean — the data are over-dispersed and Poisson $\sqrt{N}$ error bars understate the true scatter. The clean test is the Fano factor $F=\operatorname{Var}/\langle k\rangle$ : it should be one. A mixture of Poisson processes with different rates is always over-dispersed ( $F>1$ ); regularising effects such as detector dead time or nonclassical light can make it sub-Poissonian ( $F<1$ ). Always estimate $F$ from repeated measurements before trusting Poisson uncertainties.

Why does the Poisson turn into a Gaussian for large λ?

A Poisson count at rate $\lambda$ can be viewed as a sum of many independent Poisson counts at smaller rates that add up to $\lambda$ (additivity, equation 4). The central limit theorem then says the sum of many independent contributions tends to a Gaussian, and matching the first two moments fixes that Gaussian to have mean $\lambda$ and variance $\lambda$ . Quantitatively, the skewness $1/\sqrt{\lambda}$ and excess kurtosis $1/\lambda$ both vanish as $\lambda$ grows, so the distribution loses its right-skew and becomes symmetric. The approximation is already good near $\lambda\approx10$ and excellent by $\lambda\approx20$ — though the true Poisson is always discrete and never negative, unlike its Gaussian stand-in.

Is the Poisson distribution the same thing as the Poisson process?

Closely related but not identical. The Poisson distribution is the discrete probability law $P(k)=e^{-\lambda}\lambda^{k}/k!$ for a single count in a fixed window. The Poisson process is the full stochastic model of events arriving in continuous time at rate $\nu$ ; the number of arrivals it produces in any window of length $t$ is Poisson-distributed with $\lambda=\nu t$ . So the distribution is a snapshot and the process is the movie. The process carries extra structure the distribution alone does not — for instance, the waiting time between consecutive events is exponentially distributed.