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Adding positive quadratic confinement $t\int h^2/2$ to the membrane energy changes its height spectrum to $k_{B}T/(\sigma q^2+t)$. The continuous height-shift symmetry is explicitly broken. Connected correlation function decay on the finite length $\xi$ and the height-difference variance saturates for every internal dimension, although its saturation value can depend on a microscopic cutoff. A negative confinement coefficient gives an unstable quadratic model rather than pinning.

For free planar turning with constant $D_r$, no imposed torque and negligible inertia, model an unwrapped angle by the stochastic differential equationHere $W$ is standard one-dimensional Brownian motion, with increment variance $dt$, and brackets denote expected values over repeated realizations. The angular variance is conventionally reported in $\mathrm{ra}{\mathrm{d}}^2$. “Unwrapped” means recording accumulated turns continuously, rather than resetting the angle at each revolution. Directly subtracting wrapped camera angles creates artificial jumps; for an unmarked ellipsoid the wrapping period is $\pi$, and resolving successive turns additionally requires sufficiently frequent images.

For a planar arrow $\mathbf{u}=(\cos \varphi ,\sin \varphi )$, the orientational autocorrelation isFor an unoriented planar axis, use an autocorrelation unchanged by replacing either endpoint angle by that angle plus $\pi$:The same $D_r$ thus gives arrow and axis correlation times $1/D_r$ and $1/(4D_r)$, respectively, where a correlation time here is the lag at which the relevant curve reaches $1/e$. A short explanation is enough: $\Delta \varphi$ has a Gaussian distribution of variance $2D_{r}t$, so its characteristic function gives $⟨\cos (n\Delta \varphi )⟩=e^{-n^2{D}_{r}t}$. The integer $n$ selects what the observable can distinguish; it does not change the underlying angular spreading.

For an original numerical example, choose $D_r=0.25{\mathrm{s}}^{-1}$. After $2\mathrm{s}$, the angular variance is $1\mathrm{ra}{\mathrm{d}}^2$ and its root mean square displacement is $1\mathrm{rad}$, although the mean displacement is zero. The planar arrow autocorrelation is $e^{-0.5}\simeq 0.607$, while the planar axis autocorrelation is $e^{-2}\simeq 0.135$. Calling both “the orientation correlation” without specifying the observable would hide a factor of four in decay rate.

At an interior target point in regular fixed-design nonparametric regression, mean functions with Lipschitz bound $L$ have pointwise minimax risk of order $(L{\sigma }^2/n{)}^{2/3}$ for fixed positive $L,\sigma$ as $n\to \infty$. A smoothing bandwidth of order $({\sigma }^2/(n{L}^2){)}^{1/3}$ balances squared bias of an estimator and variance; the two-point lower bound with a triangular bump gives the matching lower rate.

The time of the $n$-th event of a Poisson process. At positive rate $\lambda$, the Poisson interarrival times $T_j-T_{j-1}$, with $T_0=0$, are independent and identically distributed random variables with the rate-$\lambda$ exponential distribution. In particular, $T_n$ has expected value $n/\lambda$ and variance $n/{\lambda }^2$. The inverse-count identity is $\{T_n\le t\}=\{N_t\ge n\}$.

The sum of independent Bernoulli random variables with possibly unequal success probabilities $p_i$ has a Poisson binomial distribution. Its probability generating function is $G_K(z)={\prod }_i(1-p_i+p_{i}z)$: expand the product to read off the probability mass function. Linearity of expectation and additivity of variance for independent random variables give $\mathbb{E}K={\sum }_i{p}_i$ and $\mathrm{Var}(K)={\sum }_i{p}_i(1-p_i)$. Equal $p_i=p$ recover the binomial distribution.

For three trials with success probabilities $1/10,3/10,4/5$, expanding $(9/10+z/10)(7/10+3z/10)(1/5+4z/5)$ gives masses $0.126,0.572,0.278,0.024$ at $0,1,2,3$. Their expected value is $1.2$ and variance is $0.46$. Replacing all three success probabilities by their average $0.4$ preserves the expected value but gives binomial distribution variance $0.72$, so it does not preserve the law.

For a rate-one Poisson process with Poisson arrival times $T_n$, suppose $t_n\to \infty$ and $t_n/n\to \lambda \in [0,\infty ]$. Thenwhere $1/\sqrt{\infty }=0$ and $\mathrm{BN}(\rho )$ denotes the bivariate normal distribution with standard normal marginal distributions, including the singular cases. In particular, equal scales give $\rho =-1$ and separated scales give $\rho =0$.

To prove this, fix real $x,y$ and put $s_n=n+y\sqrt{n}>0$ for large $n$. The event $(T_n-n)/\sqrt{n}\le y$ is exactly $N_{s_n}\ge n$. Counts at the two deterministic times $t_n,s_n$ are sums of a shared count on $[0,\min (t_n,s_n)]$ and an independent count on the remaining interval. The central limit theorem for a centered Poisson random variable of diverging parameter follows by writing a count at a real time $u$ as its count at $\lfloor u\rfloor$, a sum of independent unit increments, plus an independent remainder with expected value at most one. Markov's inequality makes that remainder negligible after division by $\sqrt{u}$.

Apply this central limit theorem to the two independent count blocks whenever their lengths diverge. A block with bounded length, or length negligible compared with the scale of its coordinate, contributes a term tending to zero in probability. The standardized pair of counts therefore has a joint normal limit whose correlation coefficient isThis includes $\lambda =1$: the difference block then has negligible variance after normalization. In the arrival event the standardized count at $s_n$ must be at least $(n-s_n)/\sqrt{s_n}\to -y$. Reversing this inequality changes the sign of the second limiting normal random variable, giving the asserted negative correlation coefficient. Continuity of the marginal distributions permits the moving thresholds and the weak inequalities, including at the singular limits.

Every $\rho \in [-1,0)$ is realized by $t_n=\lambda n$ with $\lambda ={\rho }^{-2}\ge 1$, and zero by $t_n=n^4$. Conversely, even when $t_n/n$ has no limit, it has a subsequence converging in the extended interval $[0,\infty ]$. If the original pair has a joint bivariate normal distribution limit, that subsequence has the same limit, forcing its correlation coefficient into $[-1,0]$.

The normalized coherent state has number amplitude $e^{-|\alpha {|}^2/2}{\alpha }^n/\sqrt{n!}$, hence the displayed Poisson law with mean and variance $|\alpha {|}^2$. A product of coherent modes has factorized number probabilities.

If $X_{\lambda }$ has a Poisson distribution, then $\mathbb{P}(X_{\lambda }\le \lfloor \lambda \rfloor )\to 1/2$ as $\lambda \to \infty$. The Poisson central limit theorem standardizes by mean $\lambda$ and variance $\lambda$; the cutoff $(\lfloor \lambda \rfloor -\lambda )/\sqrt{\lambda }$ tends to zero. Squeezing between fixed nearby cutoffs and using continuity of the standard normal distribution function gives the limit.

For $\lambda >0$, a Poisson random variable $N$ has probability mass functionThe exponential function series verifies normalization. At $\lambda =0$, define $N=0$ with certainty. The equality of mean and variance is a distinguishing model property, though it does not by itself characterize the law.

When counting events in a homogeneous Poisson process, distinguish its rate $r$ from the distribution parameter: an interval of duration $t$ has $\lambda =rt$. A rate measured per minute times a duration measured in minutes gives a dimensionless expected count. This process assumes independent increments, a constant rate, and a negligible chance of multiple events relative to interval length in a sufficiently short interval. A rate alone does not establish a Poisson distribution. See Bertsekas and Tsitsiklis, chapter 6 summary, pages 46-47.

For a constructed arrival model with $r=2$ events per minute over $t=1.5$ minutes, $\lambda =3$. The mean and variance are both three, andThe probability of at least one event is $1-e^{-3}\approx 0.9502129$. These results describe the specified process, not every arrival mechanism with the same average rate.

The Poisson limit theorem, often called the law of rare events, connects this law to the binomial distribution. Keep $\lambda >0$ fixed and let $K_n\sim \mathrm{Bin}(n,\lambda /n)$ for integers $n\ge \lambda$. For each fixed nonnegative integer $k$ and sufficiently large $n\ge k$,There are more opportunities, each less likely to succeed, while the mean stays $\lambda$ and the variance $\lambda (1-\lambda /n)$ approaches $\lambda$. For $\lambda =3$, the probability of three successes is $0.2668279$ at $n=10$, $0.2310569$ at $n=50$, and $0.2257398$ at $n=200$, approaching the Poisson distribution value $0.2240418$.

The panels show the same count scale and the same Poisson distribution reference. At $n=10$, $p=0.3$ is not especially small and the difference is visible. No universal trial-count cutoff guarantees an adequate approximation for every event.

The independence assumption matters even when every success is rare. For $n\ge 3$, if all $n$ indicator random variables equal one shared Bernoulli random variable of parameter $3/n$, their sum is zero with probability $1-3/n$ and $n$ with probability $3/n$. Its mean is three, yet its zero-count probability tends to one, not $e^{-3}$. Some dependent counts still admit approximation through the Stein-Chen method, but this requires additional control of dependence.

For historical orientation, the BnF records Poisson's original probability treatise, Recherches sur la probabilité des jugements, as published in Paris in 1837. This bibliographic fact alone does not establish when the modern distribution terminology arose. See BnF record of Poisson's 1837 treatise.

A model with $Y\mid \Lambda \sim \mathrm{Poisson}(\Lambda )$ for a nonnegative random intensity. Conditional expectation and total variance give $\mathbb{E}Y=\mathbb{E}\Lambda$ and $\mathrm{Var}Y=\mathbb{E}\Lambda +\mathrm{Var}\Lambda$, explaining overdispersion. A Poisson-gamma mixture gives the negative binomial distribution.

For independent photon detections with a fixed mean rate, counts in a fixed exposure have a Poisson distribution, so the variance equals the mean. Multiplying by a representative photon energy gives ${\sigma }_E=h\nu \sqrt{N}$. A source-only signal-to-noise threshold $q$ therefore needs an expected $N=q^2$ photons. Background, detector noise, variable illumination and photon correlations can invalidate that source-only formula.

A Homogeneous Poisson process models arrivals at a constant rate $\lambda >0$. Its counting process $N_t=N(t)$ counts arrivals in $(0,t]$, with $N_0=0$. Its paths are right-continuous nondecreasing functions, take nonnegative integer values, and have unit jumps almost surely. The defining Poisson distribution of each increment isIt has stationary increments: the probability distribution above depends on the duration $t$, not the starting time $s$. It also has independent increments: counts in any finite collection of disjoint time intervals are independent random variables. Together these conditions specify the joint count laws, not merely the Poisson distribution at a single time.^[ref]

The rate has units of arrivals per unit time; $\lambda t$ is dimensionless. The Poisson distribution givesFor a toy rate $\lambda =2$ per minute and a duration of $1.5$ minutes, the expected value and variance of the count are both $3$. The probability of no arrival is $e^{-3}\approx 0.049787$, and the probability of exactly three is $3^3{e}^{-3}/3!=\frac{9}{2}{e}^{-3}\approx 0.224042$. In two consecutive one-minute intervals, the probability of exactly two arrivals in the first and exactly one in the second is $(2e^{-2})(2e^{-2})=4e^{-4}\approx 0.073263$, using independent increments. These are exact model calculations with rounded decimals, not measured frequencies.

The diagram's nine hand-chosen arrivals illustrate a path; the line $\lambda t$ is an expected value, not a prescribed trajectory. Constant rate does not imply evenly spaced arrivals. Poisson interarrival times connect the count description to waiting, while Poisson thinning explains how independently routing arrivals produces separate streams. This makes the Poisson process useful as a simple arrival model for a queue, provided constant rate and independent increments are plausible. For a deterministic changing rate, use an Inhomogeneous Poisson process; bunching or history-dependent arrivals need additional assumptions. Conditional on $N_t=n$, the arrival times are the order statistics of $n$ independent uniform distributions on $[0,t]$, as developed in Poisson process conditional arrival times.

With ${\mu }_i=e^{\beta x_i}$, define $U={\sum }_i{x}_i(Y_i-{\mu }_i)$ and $I={\sum }_i{x}_i^2{\mu }_i$. The estimating equation and a Taylor expansion give $\hat{\beta }-{\beta }_0\approx I({\beta }_0{)}^{-1}U({\beta }_0)$. If responses are independent with correct means and $\mathrm{Var}(Y_i)=\varphi {\mu }_i$, then $EU=0$ and $\mathrm{Var}U=\varphi I$. Consequently the first-order estimator variance is $\varphi /I$, even if the working Poisson distribution is wrong. Regular asymptotics require growing information and control of leverage and the Taylor remainder.

In Poisson regression with log means $a+\beta x_i$, the Fisher information matrix is $\left(\begin{array}{cc}{S}_0& S_1\\ S_1& S_2\end{array}\right)$, where $S_j={\sum }_i{x}_i^j{\mu }_i$. Inverting this matrix gives slope variance $S_0/(S_0{S}_2-S_1^2)$. Writing ${\bar{x}}_{\mu }=S_1/S_0$ reduces it to the displayed reciprocal weighted sum of squares. Thus a common intercept consumes information about the overall rate, while slope information comes from the spread of the covariates under the mean-count weights. Constant covariates make the two parameters unidentifiable.

Subtracting an exactly known mean sky level leaves source-plus-sky variance of a Poisson distribution $S+B$. If the sky per source pixel is estimated from $n_b$ equal-area, equally exposed independent sky pixels, its additional variance is $B/n_b$, so $R=S/\sqrt{S+B+B/n_b}$ before read and dark terms. The variance of removed background photons does not disappear with their mean.

The population principal components are centered linear compounds along an orthonormal eigenbasis of the covariance matrix, ordered by decreasing eigenvalue. They are uncorrelated and have variances ${\lambda }_j$. Each successive direction maximizes variance subject to unit length and orthogonality to earlier directions. Uncorrelatedness does not imply independence unless additional distributional assumptions hold.

Spreading a portfolio across assets can reduce risk that is not shared by them. For centered asset risks with covariance matrix $\Gamma$, the variance of the weighted risk is $w^T\Gamma w$. Equal weights across $n$ independent risks with common variance $s^2$ give variance $s^2/n$, whereas equal weights across risks that are all the same random variable leave variance $s^2$. Thus small individual weights do not by themselves remove common risk; the covariance criterion for diversification of factor residuals identifies sufficient dependence and weight conditions.

The position uncertainty of a quantum state is the standard deviation of its measured position, provided the second moment is finite. It is computed from the position operator with the Born rule. An oscillator coherent state has $\Delta x=\sqrt{\hslash /(2m\omega )}$ independently of its mean position; translating the state changes the mean without changing this variance.

A posterior variance is the variance of a parameter under its Bayesian posterior, namely $\mathbb{E}[(\theta -\mathbb{E}[\theta \mid y]{)}^2\mid y]$ when finite. It describes remaining uncertainty after observing data and is distinct from the repeated-sampling variance of a fitted estimator.

A power spectral density distributes the mean square of a signal over frequency. Its integral over the specified frequency domain gives that mean square. State whether the convention is one-sided or two-sided and whether frequency is measured in hertz or as angular frequency. For a zero-mean stationary signal, the mean square is its variance; spectral density of a stationary process gives the discrete-time convention.