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For equal independent binomial samples, the displayed normal-approximation size per group targets a two-sided significance level $\alpha$ and power $1-\beta$ at specified alternative rates, with $\bar{p}=(p_1+p_0)/2$. Its first variance term sets the null critical value and its second the alternative distribution. A numerical sample size is not determined by an effect ratio alone; power, sidedness, allocation and baseline rates must also be specified. Rare-event discreteness and clustered designs can materially change the achieved power.

The sample standard deviation is the nonnegative square root of the usual sample variance for $n\ge 2$. The displayed convention uses divisor $n-1$. Although the sample variance is an unbiased estimator of the population variance under independent identically distributed sampling with finite variance, taking its square root does not in general give an unbiased estimator of the population standard deviation.

The sampling fraction is the sample size divided by the finite population size. In simple random sampling, it determines how much the finite population correction reduces the variance relative to $S^2/n$.

Each Cartesian Brownian increment has variance $2Dt$. Summing their normalized squares gives a chi-squared distribution with $3n$ degrees of freedom. Hence $\hat{D}$ is unbiased and has relative standard deviation $\sqrt{2/(3n)}$. Its inverse is not unbiased: for $3n>2$, $\mathbb{E}[1/\hat{D}]=[3n/(3n-2)]/D$. Thus the direct inverse-diffusivity radius estimate has appreciable finite-sample bias when only a few trajectories are observed. Measurement noise or drift adds separate systematic error.

For the state-space model $S_t=G{S}_{t-1}+w_t$, $X_t=F{S}_t+v_t$, assume independent Gaussian process noise and measurement noise of variances $W,V$, independent of the prior state and previous observations. A normal distribution for the filtered state, with mean $m$ and variance $P$, predicts mean $Gm$ and variance $R=G^{2}P+W$. Here $R$ denotes predicted state variance, rather than the observation-noise covariance notation used in the vector equations above. Gaussian conditional expectation then gives posterior mean $Gm+K(X-FGm)$ and posterior variance $RV/(F^{2}R+V)$, where the Kalman gain is $K=FR/(F^{2}R+V)$.

Mere weak white noise assumptions do not imply normal posterior distributions: independent process noise with a Rademacher distribution makes the next state a mixture of two shifted normal distributions. The Kalman filter can still describe a best affine state estimate under appropriate second-order assumptions, but that is distinct from an exact Gaussian conditional law.

With query and key dimension $d_k$, form the score matrix $Q{K}^T/\sqrt{d_k}$, apply the softmax function row by row, and multiply by the values $V$. If query and key coordinates have independent unit variance, their dot product has variance $d_k$; division by $\sqrt{d_k}$ keeps that score scale of order one. Optional additive masks exclude forbidden keys before the softmax function.

For $d\sigma /d\Omega =K/[p^2{\sin }^4(\theta /2)]$ between angular cutoffs, the leading total cross section is $16\pi K/(p^2{\theta }_{\min }^2)$. Multiply its mean count ${\rho }_{n}x\sigma$ by the one-scatter second moment $2{\theta }_{\min }^2\ln ({\theta }_{\max }/{\theta }_{\min })$ to obtain the displayed total angular variance. A single-plane projected variance is half as large. Comparing the scattering angle to a magnetic bend explains the approximately momentum-independent scattering contribution to relative track uncertainty.

The second moment of a real random variable is $\mathbb{E}[X^2]$. Its variance is $\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$ when these quantities are finite.

For a nonnegative random variable $N$ with positive expected value and finite second moment, write $N=N{1}_{\{N>0\}}$. The Cauchy-Schwarz inequality giveswhich proves the bound in second moment method. It is the zero-threshold case of the Paley-Zygmund inequality. Equality occurs when $N$ is a fixed positive value on the event $N>0$ and zero elsewhere, almost surely. For $N=10B$ with $B$ a Bernoulli random variable of parameter $1/10$, $(\mathbb{E}N{)}^2/\mathbb{E}{N}^2=1/10$ equals the exact positivity probability.

The variance identity $\mathbb{E}{N}^2=(\mathbb{E}N{)}^2+\mathrm{Var}(N)$ turns the lower bound into a statement about variance relative to the squared mean. A small ratio forces positivity to be likely. A positive expected value alone does not do so: keeping $\mathbb{E}N=1$ while taking $N=1/q$ with probability $q$ and zero otherwise makes the positivity probability as small as any chosen $q>0$.

The second moment method uses a random variable's expected value and second moment to establish that a counted or measured quantity is positive with substantial probability. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the variance rules out too much of that concentration in rare outcomes.

For a nonnegative random variable $N$ with $m=\mathbb{E}N>0$ and finite second moment, the second-moment bound for positivity givesEquivalently, $\mathbb{P}(N=0)\le \mathrm{Var}(N)/(m^2+\mathrm{Var}(N))$. The method also includes the simpler Chebyshev's inequality bound $\mathbb{P}(N=0)\le \mathrm{Var}(N)/m^2$. Thus a sequence of such random variables with $\mathrm{Var}(N)/(\mathbb{E}N{)}^2\to 0$ has $N>0$ with probability tending to one. These statements require no independence assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. ^[ref].

For an original toy count, let $N$ count successes in ten independent Bernoulli trials with success probability $p=1/10$. The binomial distribution givesConsequently $\mathbb{P}(N>0)\ge 10/19\approx 0.526316$. The exact probability is $1-(9/10{)}^{10}\approx 0.651322$. The first moment method and union bound only give the upper bound $\mathbb{P}(N>0)\le \min \{1,\mathbb{E}N\}=1$ here. First and second moments therefore answer complementary existence questions.

For a sum of indicator random variables, the variance of a sum isIndependence makes the covariances vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single Bernoulli random variable $B$: then $N=10B$, with mean one, variance nine, and $\mathbb{P}(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive covariances would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. ^[ref].

A probability distribution is self-decomposable if, for every $0<c<1$, an independent remainder $R_c$ can be chosen so that the displayed equality in distribution holds. A normal distribution $\mathcal{N}(\mu ,{\sigma }^2)$ admits a normal distribution remainder with mean $(1-c)\mu$ and variance $(1-c^2){\sigma }^2$. For an exponential distribution of rate $\lambda$, choose $R_c=0$ with probability $c$ and otherwise an independent exponential distribution of rate $\lambda$. Its moment-generating function is $c+(1-c)\lambda /(\lambda -t)=(\lambda -ct)/(\lambda -t)$; multiplication by $M_X(ct)$ proves the claim using the uniqueness theorem for moment-generating functions.

Let $B=0$ with probability $\pi$, and otherwise let $B$ have a Gamma distribution with mean one and variance $\tau$. Given $B$, counts $Y_j$ are conditionally independent Poisson random variables with means $B{\mu }_j$. For $q=1-\pi$, the law of total variance and law of total covariance giveEach component has a zero-inflated negative binomial distribution, but the components are not independent after marginalizing $B$. The zero component is shared by the whole profile. At $\tau =0$, interpret the positive Gamma component as a point mass at one.

For fixed $0<\lambda <1$ in $G(n,\lambda /n)$, the largest component of a graph has the displayed order with high probability. A dominating Galton-Watson process gives the tail bound $\mathbb{P}(T\ge j)\le e^{-(\lambda -1-\log \lambda )(j-1)}$ by the exponential Markov inequality. For the lower bound, the tree-component expectation in the Erdős-Rényi model diverges at $j=\lfloor (1-\eta )\log n/(\lambda -1-\log \lambda )\rfloor$. The disjoint-set covariance factor and the exclusion of overlaps give relative variance tending to zero, so the second moment method supplies a tree component of that order.

For a long potential approximately constant over the short-scale region, local-type primordial non-Gaussianity gives ${\phi }_s=(1+2f_{\mathrm{NL}}{\phi }_{G,L}){\phi }_{G,s}$, to first order in the long mode and ignoring short-mode squares. A linear cosmological Poisson equation therefore givesto the same order. The variance is conditional on the long mode.

Shot noise is fluctuation associated with discrete transfer events. For an electric current, individual charge carriers contribute finite amounts of electric charge, even when a meter displays a smooth mean. The simplest model is a stream of equal, independently arriving charges with constant rate: its Poisson process predicts both the mean electric current and the variance around it. Discreteness alone does not imply a Poisson distribution; the timing and correlations matter.

The useful surprise is that noise can reveal something a mean hides. Two devices can carry the same mean electric current yet have different fluctuations because one transfers charges independently and the other groups or regulates them. A calibrated shot noise photocurrent measurement can connect a macroscopic voltage variance to the elementary charge. Thermal electrical noise instead exists in an unbiased resistor at thermal equilibrium; a measured background can contain both mechanisms.

Walter Schottky's 1918 paper Über spontane Stromschwankungen in verschiedenen Elektrizitätsleitern treated amplifier sensitivity limits. Part II considered independent discrete transfers, including thermionic emission through a vacuum. On p. 563 he proposed amplifying fluctuations from a separate discharge tube to determine the elementary charge; those experiments were prospective, not reported results.^[ref]

Let $q>0$ be the magnitude of each transferred electric charge, and let arrivals form a Homogeneous Poisson process of rate $\lambda$. Extend the stream over past and future times when considering stationary filtering. For electrons, $q=e$, the elementary charge. Choose the positive electric current direction to follow these transfers, so $I=q\lambda >0$ is its mean magnitude. If $N_T$ counts arrivals in a gate of duration $T>0$, thenThe gate's transferred electric charge is $Q_T=q{N}_T$, and its average electric current is ${\overline{i}}_T=Q_T/T$. Scaling the expected value and variance givesMore arrivals increase the absolute fluctuation but decrease its size relative to the mean. For an original count example with $\lambda T=4$, the mean count and its variance are both $4$, the standard deviation is $2$, and the relative standard deviation is $1/2$. The probability of no transfer is $e^{-4}\approx 0.01832$; the probability of four is $4^4{e}^{-4}/4!\approx 0.19537$. A constant arrival rate does not prescribe four arrivals in every gate.

These are fluctuations of a time average, not the variance of an ideal instantaneous electric current. A model of point transfers has arbitrarily sharp impulses; it acquires a finite measured variance only after specifying a gate or linear filter. Actual collection pulses have finite duration and electrical circuits have finite response times.

The Fano factor $F_T$ compares count variance with its mean over a specified gate. For a stationary unidirectional stream with fixed charge $q$, finite count second moments, and a generalized autocovariance with finite integral so that the long-gate limit exists,The zero-frequency relation concerns the continuous fluctuation spectrum, excluding the mean spectral line. It follows by writing $\mathrm{Var}(Q_T)$ as the double integral of the current autocovariance; for long $T$, its leading term is $T{S}_I^{(2)}(0)$. A finite-gate $F_T$, a frequency-dependent ratio $S_I^{(1)}(f)/(2qI)$, and a factor inferred from one filtered variance are not interchangeable. The last is weighted by the instrument's $|K(f){|}^2$.

Two original counting counterexamples show why electric charge discreteness is insufficient. In $n$ independent time slots, transfer one charge with probability $p$ in each slot: the binomial distribution gives mean $np$, variance $np(1-p)$ and Fano factor $1-p$. A random common origin for the slots removes a preferred time origin; the long-gate factor remains $1-p$. Alternatively, if independent Poisson process bursts each transfer exactly two charges, then $N_T=2M_T$ gives $\mathrm{Var}(N_T)=2\mathbb{E}[N_T]$ and factor $2$. Both count individual charges; only the first model allows at most one per slot, and the second groups them in pairs.

Consequently $2qI$ is a reference for independent unidirectional arrivals, not a universal noise law for all electrical conductors. Correlations, charge multiplication, and bidirectional transfer require a more specific model. For example, independent opposite streams with positive current magnitudes $I_{+}$ and $I_{-}$ have net mean $I=I_{+}-I_{-}$ but one-sided fluctuation spectrum $2q(I_{+}+I_{-})$. Using $2q|I|$ would miss their fluctuations when the net current is small. At thermal equilibrium, the full transport model must also respect thermal electrical noise; adding unrelated phenomenological formulas can double count the same underlying transfers.

MIT's guide uses an illuminated vacuum photodiode, with collecting anode positive relative to the emitting cathode. A transimpedance amplifier gives $V_1=R_{F}I$; a multimeter reads its mean. The fluctuation path rejects below $100\mathrm{Hz}$, adds roughly tenfold voltage gain, then nominal hundredfold preamplification and a $1$–$50\mathrm{kHz}$ band-pass filter; the readout measures root mean square voltage. With diode bias off, a sinusoid at the test port calibrates $g(f)$. Restore bias, retain gain settings, vary lamp brightness, and repeat both readings. The guide warns about overload; it supplies a procedure, not measured results.^[ref]

The inference can be expressed without relying on nominal passband edges. Let $g(f)$ denote the calibrated voltage transfer after the $R_F$ conversion, so $K(f)=R_{F}g(f)$ and define $\mathcal{K}={\int }_0^{\infty }|K(f){|}^{2}df$. Under independent arrivals and an unchanged additive background,The slope of voltage variance against mean electric current can therefore estimate $q$. A free intercept accounts for a constant background; forcing the line through zero generally biases the slope. The root mean square readings must have their mean removed and then be squared. Subtracting two root mean square amplitudes does not subtract independent variances. Estimate the variance from successive records; averaging the waveforms first would suppress the fluctuation being measured.

Additional controls sharpen this interpretation. Block the light with detector bias, temperature, and gain unchanged; this preserves dark current and electronics while removing photocurrent. Repeat this dark comparison between illumination settings to check drift. If independent dark arrivals have mean magnitude $I_d$ and illuminated arrivals add $I_{\mathrm{ph}}$, their spectra add: $2q(I_d+I_{\mathrm{ph}})$. Subtracting the unchanged dark variance isolates $2q{I}_{\mathrm{ph}}\mathcal{K}$. This subtraction model assumes unchanged transfers and independence; switching off bias for gain calibration is a different control and need not preserve the dark background.

Vary measured $I_{\mathrm{ph}}$ rather than assuming lamp knob position is proportional to photocurrent. Check collection versus bias to identify a range where incomplete collection is negligible; a space charge limited tube can correlate transfers. Verify that the transimpedance amplifier remains linear and that its frequency response is unchanged over the illumination range. Changing signal bandwidth should change the excess variance through $\mathcal{K}$; compare inferred input spectra in several bands. Correlated fluctuations of the light source can add a contribution scaling approximately as $I_{\mathrm{ph}}^2$, while an electronics floor can hide the linear term at small electric currents. A fitted line over a short range alone does not establish independent transfers.

The gain calibration must include the same detector-to-readout path used for noise, or explicitly supply any omitted transfer. Detector capacitance, finite collection time, amplifier voltage and current noise, thermal electrical noise, pickup, clipping, and finite recording precision limit the inference. Current-dependent electrical impedance or heating can make a background current-dependent too. Since $\mathcal{K}$ contains squared gain, a small fractional error in gain produces approximately twice that fractional error in the predicted variance scale. Knowing a nominal gain and two cutoff settings is insufficient.

For an original model, set $q=1.602176634\times 10^{-19}\mathrm{C}$, $I=1\mu \mathrm{A}$, $R_F=100\mathrm{k}\Omega$, $G=10$, and the exponential low-pass filter's $\tau =100\mu \mathrm{s}$. Assume an ideal detector without charge multiplication and omit all additive noise for this calculation. ThenDoubling $I$ doubles the voltage variance and multiplies its standard deviation by $\sqrt{2}$. Halving $\tau$ also doubles the variance, provided the detector remains approximately white throughout the enlarged passband. Using $f_c$ in place of $B_{\mathrm{n}}$ underestimates the variance by the factor $2/\pi$. These are calculated predictions, not measured values or settings claimed for the laboratory below.

To read the dimensionless figure, set $a=\lambda \tau$ and $x=\tau i_{\mathrm{filtered}}/q$, with frequency $\nu =f\tau$ in cycles per unit $t/\tau$. A gate of duration $\tau$ has count mean and variance $a$. The continuously filtered $x$ instead hasThe count and the filtered electric current are different observables. The factor $1/2$ comes from the exponential weighting, not from non-Poisson arrivals. The dimensionless count illustration is a separate choice of rate; it does not assert that the $1\mu \mathrm{A}$ example has only four electrons per $100\mu \mathrm{s}$.

For a periodic signal, the mean squared magnitude over a period. Its units are the square of the signal's units: a voltage signal has mean-square power in ${\mathrm{V}}^2$. Physical electrical power in a resistive load additionally divides this quantity by electrical resistance. Subtracting the squared mean gives the variance of the signal over a uniformly sampled phase; the two quantities coincide for a zero-mean signal.