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Write $m_k=\mathbb{E}{S}^k$, $m_0=1$, and ${\mu }_j=\mathbb{E}{X}^j$. Apply the Poisson size-bias identity with $h(s)=s^{k-1}$, and expand $(S+X{)}^{k-1}$ using the binomial theorem and independence. A finite $k$th severity moment suffices. In particular, the aggregate expected value, variance, and third central moment are $\lambda {\mu }_1$, $\lambda {\mu }_2$, and $\lambda {\mu }_3$, respectively; the latter two use raw severity moments.

Using $u=r^2/(2{\sigma }^2)$ in the defining expected value integral givesThe last integral is the gamma function. Near zero, the integrand before substitution behaves as $r^{q+1}$, explaining the threshold $q>-2$; at infinity the Gaussian tail controls every power. In particular,^[ref]

For the chosen $\sigma =2\mathrm{cm}$, the mean radius is about $2.5066\mathrm{cm}$, its median is $2\sqrt{2\log 2}\simeq 2.3548\mathrm{cm}$, and its mode is $2\mathrm{cm}$. Its root mean square is $2\sqrt{2}\simeq 2.8284\mathrm{cm}$. Different summaries answer different questions; the component scale $\sigma$ is neither the radial mean nor the radial standard deviation.

The recombination fraction is the probability that a transmitted gamete carries a recombinant haplotype between two specified genetic loci, relative to the parent's known haplotype phase. Write it as $r$ here; the symbol $\theta$ is also common in genetic linkage analysis. In the ordinary symmetric two-locus model, $0\le r\le 1/2$, with $r=0$ for complete genetic linkage and $r=1/2$ for independent assortment.

For phase-known genotype $AB/ab$, fair Mendelian segregation and equal reciprocal gamete probabilities giveFor example, 1,000 independent informative transmissions with observed counts $400,400,100,100$ in this order have empirical recombination fraction $(100+100)/1000=0.2$. Those illustrative counts match the expected values in the toy model, but actual counts fluctuate. The genetic linkage diagram shows both the parental and recombinant classes, so their separate probabilities are not confused with the total $r$.

The recombination fraction measures the probability of an odd number of origin switches along the transmitted chromatid. It is neither the expected value of the exchange count nor the probability of at least one exchange: even positive counts can be invisible at the two endpoints. Converting an observed recombination fraction to a genetic map distance therefore needs an explicit model such as the Haldane mapping function. The distinction between transmitted-chromatid counts and four-chromatid exchange counts is discussed in Weeks, Tang and Kwon (2009), Casares' map function: No need for a 'corrected' Haldane's map function, Genetica 135, 305–307.

A spatial image after a chosen dark interval is a different readout from continuously recording arrivals at a fixed probe. Here is an original idealised fluorescence imaging model: release a dilute cloud, let it move freely, then apply a short uniform probe pulse. Neglect collisions, residual forces, attenuation and motion during the pulse. Use each image's cloud centre to remove mean drift. Uniform gravity shifts that centre without changing the centred variance.

For each atom let $x(t)=x_0+v_{0}t$ along a horizontal imaging axis. Expanding the square about the expected values gives exactlyNo Gaussian distribution is needed for this variance identity. A jointly Gaussian distribution of initial positions and velocities does make the later density Gaussian. For a separable three-dimensional cloud viewed along $z$, integration along the viewing direction gives the column densityHere $N$ is atom number. Integrating this image along $y$ leaves a one-dimensional Gaussian distribution with standard deviation ${\sigma }_x(t)$. This is a projected density measurement, not a three-dimensional radius or a direct velocity histogram. Under the uniform, optically thin probe assumptions, expected fluorescence is proportional to this column density. A known, independent point spread function with finite variance adds its own variance to the measured width; probe duration and intensity gradients need separate treatment.

For a Maxwell-Boltzmann distribution, $T_x=m{\sigma }_v^2/k_B$, where $m$ is atomic mass and $k_B$ the Boltzmann constant. For a nonthermal distribution this defines a directional second-moment temperature, which need not describe every feature of the distribution. A covariance between position and velocity can initially narrow the cloud even though ${\sigma }_v^2$ is positive. The physical covariance matrix requires $C_{xv}^2\le {\sigma }_{x0}^2{\sigma }_v^2$.

Choose ${\sigma }_{x0}=0.60\mathrm{mm}$, ${\sigma }_v=0.060\mathrm{m}{\mathrm{s}}^{-1}$ and $C_{xv}=-0.018\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$, giving correlation coefficient $-1/2$. With $t$ in milliseconds, the numerical variance in square millimetres is $0.36-0.036t+0.0036t^2$. Its minimum occurs at $5\mathrm{ms}$, with standard deviation $\sqrt{0.27}\mathrm{mm}$. These chosen values satisfy the covariance matrix constraint; early contraction is compatible with free motion.

Generate thirteen synthetic variance observations at equally spaced times from $0$ to $30\mathrm{ms}$, adding independent Gaussian noise of standard deviation $0.012\mathrm{m}{\mathrm{m}}^2$. Linear least squares fits all three coefficients in $a+bt+c{t}^2$. The example yields $a\simeq 0.37026\mathrm{m}{\mathrm{m}}^2$, $b\simeq -0.036413\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-1}$ and $c\simeq 0.0035820\mathrm{m}{\mathrm{m}}^2\mathrm{m}{\mathrm{s}}^{-2}$. With the chosen approximate sodium mass $m=3.82\times 10^{-26}\mathrm{kg}$, the true $T_x$ is $9.96\mu \mathrm{K}$ and the fitted value is $9.91\mu \mathrm{K}$. This is a fit to generated width data, not historical observations.

If the same data are fitted to $a+c{t}^2$, forcing $C_{xv}=0$, the inferred $T_x$ falls to $6.79\mu \mathrm{K}$. The omitted term is partly absorbed into both fitted coefficients. In noiseless linear least squares with a free intercept, its precise effect on the quadratic coefficient iswhere the subscript denotes equal-weight moments over the chosen observation times, using one consistent unit system. Since the ratio is positive for distinct nonnegative times, negative $C_{xv}$ biases this estimate downwards. An incorrect release-time origin can also generate a linear term, so fitting it does not by itself identify a physical initial covariance. Timing calibration, spatial resolution, residual magnetic fields and probe-induced recoil energy remain separate controls. Neither a cold-looking image nor a single long expansion time identifies all three coefficients.

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.

Connect one physical resistor $R$ across an ideal lossless capacitor $C$, with no applied source and an ideal readout that draws no electric current. Represent the resistor by its noisy Thévenin equivalent. The resulting RC low-pass frequency response gives the stationary capacitor voltage power spectral densityIts mean square follows by the effective noise bandwidth calculation:This is also the equipartition theorem: the mean capacitor energy is $\frac{1}{2}C⟨v_C^2⟩=\frac{1}{2}{k}_{\mathrm{B}}T$. Raising $R$ makes the low-frequency spectrum taller and its width narrower in exactly compensating proportions. It changes the equilibration time $RC$, not the final variance.

The integral to infinity belongs to the ideal classical model. For a physical electrical circuit, assume thermal equilibrium, linear components with negligible extra dissipation, and a corner far below the frequencies where parasitic effects, frequency-dependent electrical resistance, or $hf\sim k_{\mathrm{B}}T$ invalidate the model; the omitted high-frequency tail must be negligible. A finite-bandwidth instrument measures only part of this variance. Its input capacitance contributes to the total $C$, and its finite input electrical resistance introduces both loading and its own thermal electrical noise.

For $T=300\mathrm{K}$ and $C=1\mathrm{nF}$, the predicted equilibrium variance is $4.141947\times 10^{-12}{\mathrm{V}}^2$ and the root mean square voltage is $2.0352\mu \mathrm{V}$. At $R=10\mathrm{k}\Omega$, $RC=10\mu \mathrm{s}$ and $B_{\mathrm{n}}=25\mathrm{kHz}$. At $R=100\mathrm{k}\Omega$, $RC=100\mu \mathrm{s}$ and $B_{\mathrm{n}}=2.5\mathrm{kHz}$; the equilibrium root mean square voltage is identical.

To obtain the growth curve, prepare $v_C(0)=0$ exactly and connect the resistor at $t=0$ while maintaining its temperature. The RC low-pass filter's causal impulse response is $h(t)=e^{-t/(RC)}/(RC)$ for $t\ge 0$. Classical ideal white noise with two-sided power spectral density $2k_{\mathrm{B}}TR$ givesThe factor $2k_{\mathrm{B}}TR$ here is the two-sided time-domain noise intensity; the equilibrium spectral integral above used the one-sided $4k_{\mathrm{B}}TR$. They give the same long-time limit. The variance approaches equilibrium with time constant $RC/2$, although a deterministic initial voltage decays with time constant $RC$. If the capacitor already starts in thermal equilibrium, its initial variance supplies the missing exponentially decaying term and the total variance stays constant.

Applying a measurable per-claim retention $g$ to independent claim sizes preserves the compound Poisson distribution form. The transformed severity is the pushforward measure of the original severity law. With count parameter $\lambda$, the retained aggregate has expected value $\lambda \mathbb{E}g(X)$ and variance $\lambda \mathbb{E}[g(X{)}^2]$. Zero retained marks may be removed by Poisson thinning.

If $S$ has exponential distribution with expected value $\mu$, the tail integral formula for moments gives $\mathbb{E}\min (S,M)=\mu (1-e^{-m})$ and $\mathrm{Var}(\min (S,M))={\mu }^2(1-2m{e}^{-m}-e^{-2m})$, where $m=M/\mu$. At matching retained expected value, the excess variance under quota share reinsurance is $2{\mu }^2{e}^{-m}(m-1+e^{-m})\ge 0$.

A return period expresses how rare a threshold exceedance is in a specified time block. Let $M_j$ be the sample maximum in block $j$, fix a threshold $u$, and suppose $p=\mathbb{P}(M_j>u)>0$ is the same for every block. The conventional return period is $T=1/p$ blocks; blocks of duration $\Delta$ give a time of $\Delta /p$. Thus an annual exceedance probability $p=0.01$ means a 100-year return period. This names a frequency scale, not a schedule or a guarantee of one exceedance in each century. The annual convention and the misleading “once every 100 years” interpretation are discussed in USGS: the 100-year flood.

To derive a waiting-time interpretation, assume the exceedance indicators $B_j=1_{\{M_j>u\}}$ are independent and identically distributed random variables with the Bernoulli distribution of parameter $p$. Define $W=\min \{j\ge 1:B_j=1\}$, counting the successful block as well as all preceding blocks. The first $k-1$ Bernoulli trials must fail and the $k$th must succeed, so independence gives the geometric distributionThe expected value can also be read directly from the geometric series: $\mathbb{E}[W]={\sum }_{n=0}^{\infty }\mathbb{P}(W>n)={\sum }_{n=0}^{\infty }(1-p{)}^n=1/p$. This counts blocks from the starting boundary, not the exact time of an exceedance within its block. Counting only failed blocks instead would give expected value $T-1$. The waiting law and its memorylessness of the geometric distribution are developed in Kyle Siegrist, Random Services: the geometric distribution.

For a horizon of $n$ complete years, at least one exceedance is the complement of no exceedances. Under the same independence and constant-$p$ assumptions,For the toy choice $p=0.01$, this gives $9.6\%$ in 10 years, $26.0\%$ in 30 years, $39.5\%$ in 50 years, and $63.4\%$ in 100 years. Even after 100 years there is a $36.6\%$ chance of no exceedance. These numbers are exact model calculations rounded to one decimal place, not measured frequencies. The median waiting count is 69 years, since $1-0.99^{68}<1/2\le 1-0.99^{69}$; the expected value is 100 years. A mean and a median answer different questions.

The exceedance count $N_n={\sum }_{j=1}^n{B}_j$ has the binomial distribution under independence, with expected value $np$. Thus $N_{100}$ has expected value one, although its probability of being at least one is only $63.4\%$: some realizations contribute several exceedances. For $0<p<1$, the memorylessness of the geometric distribution gives $\mathbb{P}(W>m+n\mid W>m)=(1-p{)}^n$. A long quiet spell does not make the next independent year's exceedance overdue, and an exceedance this year does not force a quiet interval. Two specified consecutive years both exceed with probability $p^2=0.0001$ in this toy model.

A reproducible tabletop experiment makes the assumptions tangible. Put 100 identical slips numbered 1 to 100 in a bag; after thoroughly mixing, draw one and record whether it is 100. Replace the slip and mix again before every draw. Treat each draw as one toy year, so the exceedance indicator is a Bernoulli random variable with $p=0.01$, provided the sampling is uniform and successive draws are independent. Repeat complete 30-draw runs and record the fraction containing at least one 100; separately repeat runs stopped at the first 100 and record their waiting counts. The first fraction targets $1-0.99^{30}$; the waiting-count histogram targets $0.99^{k-1}(0.01)$. Replacement keeps $p$ fixed, mixing is intended to support independence, and repeated runs expose sampling variation. This is a proposed demonstration of the toy model; no experimental results are asserted here.

Stationarity supports a time-invariant annual probability distribution, but does not imply independence. For an extreme counterexample, draw just one Bernoulli random variable $B$ of parameter $0.01$ and set $B_j=B$ for every year. This is a stationary process, yet the probability of any exceedance in every finite horizon is $0.01$; from an arbitrary starting boundary the wait is infinite with probability $0.99$. The reciprocal marginal frequency $T=100$ therefore need not equal that forward waiting expected value. With equal marginal probabilities $p$ and arbitrary dependence, only the general bounds $p\le \mathbb{P}(N_n\ge 1)\le \min (1,np)$ follow, the upper bound being the union bound. Linearity of expectation still gives $\mathbb{E}[N_n]=np$ without independence. Mere zero correlation does not restore the product formula; even pairwise independence does not imply joint independence. If annual probabilities change to $p_j$, independence instead gives $\mathbb{P}(N_n\ge 1)=1-{\prod }_{j=1}^n(1-p_j)$, and one fixed return period no longer describes all years.

In a finite-state discrete-time financial market, a replicable terminal financial payoff $X$ has the displayed value under any equivalent martingale measure $Q$ for the bank account numéraire. Here ${\mathcal{F}}_t$ represents current information and conditional expectation averages over future states. The self-financing portfolio's discounted portfolio wealth is a martingale, so its current value equals the discounted expectation of its terminal value. In an incomplete market, nonreplicable claims need not have a unique price; continuous-time extensions require further conditions on admissible trading strategies and integrable discounted wealth.

For the European call option in the example,Discounting the physical expected value instead gives $200/21\approx 9.52381$ when $p=1/2$, or $320/21\approx 15.23810$ when $p=4/5$. Those amounts describe discounted physical expectations, not the replicating portfolio's cost. Holding $S_0,u,d,R$ fixed is crucial: changing beliefs can influence traded prices in a richer financial market model.

The martingale condition for the discounted stock is $qu{S}_0+(1-q)d{S}_0=R{S}_0$. Therefore the up risk-neutral probability isWith down factor $d=1+a$, up factor $u=1+b$ and $R=1+r$, this is equivalently $(r-a)/(b-a)$. For $d=0.8$, $u=1.2$, $R=1.05$, it gives $q=0.25/0.4=0.625$. Under the risk-neutral measure, the terminal stock price has expected value $(5/8)120+(3/8)80=105$, whose discounted value is 100. The physical probability $p$ need only satisfy $0<p<1$; it does not enter this calculation.

The wait for the requested sector to reach the disk head after positioning on its track. For a uniformly distributed arrival phase and rotation period $T_{\mathrm{rot}}$, its expected value is half a rotation. At $r$ revolutions per minute, this is $30/r$ seconds.

Fix $N>1$ real population values $y_1,\dots ,y_N$. Define the finite population mean and two normalizations of the finite population variance byThe values are fixed; only the selected subset $A$ is random. For its sample mean ${\bar{y}}_A=n^{-1}{\sum }_{i\in A}{y}_i$, the exact design expectation and variance areThe sample mean is an unbiased estimator of the finite population mean. At $n=N$, every unit is observed and ${\bar{y}}_A=\bar{Y}$ exactly, so the variance is zero. The $N=1$ case is deterministic and is handled directly, since $S^2$ is undefined then.^[ref]

The finite population correction depends on the baseline. With the $N-1$ normalization it is $1-n/N$ multiplying $S^2/n$. With the $N$ normalization it is $(N-n)/(N-1)$ multiplying ${\sigma }_N^2/n$, the exact variance for independent uniform sampling with replacement. The standard error uses the square root of the relevant variance factor. Mixing the two normalizations gives a wrong answer even at $n=1$, when replacement cannot matter. For the six-value example, $\bar{Y}=7$, ${\sum }_i(y_i-7{)}^2=70$, $S^2=14$ and ${\sigma }_N^2=35/3$, reproducing $14/3$ without replacement and $35/6$ with replacement.

A short proof exposes the statistical dependence. Let $X_r$ be the value at position $r$ in uniform sequential sampling without replacement, so each $X_r$ has expected value $\bar{Y}$ and variance ${\sigma }_N^2$. Write $d_i=y_i-\bar{Y}$. Every ordered pair of distinct labels is equally likely, and ${\sum }_i{d}_i=0$, hence for $r\ne s$,Linearity of expectation proves unbiasedness without requiring independence. The variance of a sum includes all $n(n-1)$ ordered covariance terms, givingFor the original population, each pair of distinct draw positions has covariance $-7/3$. Drawing a high value leaves a lower mean among the remaining values, balancing subsequent draws. Equal marginal probability distributions therefore coexist with negative covariance.

When $2\le n\le N$, the sample variance $s_A^2=(n-1{)}^{-1}{\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2$ is an unbiased estimator of $S^2$. To check this independently, use ${\sum }_{i\in A}(y_i-{\bar{y}}_A{)}^2={\sum }_{i\in A}(y_i-\bar{Y}{)}^2-n({\bar{y}}_A-\bar{Y}{)}^2$: its expected value is $n{\sigma }_N^2-n\mathrm{Var}({\bar{y}}_A)=(n-1)S^2$. Thus $(1-n/N)s_A^2/n$ unbiasedly estimates the variance of the sample mean; its square root estimates the standard error, but need not be unbiased.^[ref]

The diagram compares variance with the expected fraction of distinct population labels observed. Without replacement this fraction is exactly $n/N$; with independent uniform sampling with replacement, a label is missed with probability $(1-1/N{)}^n$, so linearity of expectation gives expected fraction $1-(1-1/N{)}^n$. At six draws, replacement covers only about $66.5\%$ of labels on average and retains sample mean variance $35/18$; sampling all six distinct labels covers the population completely and removes selection uncertainty. This population coverage is distinct from confidence interval coverage probability.

The correction matters when the sampling fraction is appreciable. It calibrates uncertainty about this fixed population under the specified selection mechanism. It does not correct unequal selection weights, missing observations or measurement errors; observing every label also does not eliminate uncertainty about a larger unobserved population. Moreover, an exact variance alone does not guarantee the coverage probability of a proposed confidence interval: a small discrete sampling distribution can make a normal approximation inaccurate.

The sample median is a central value of an ordered statistical sample. The usual numerical convention selects the middle order statistic for odd sample size and the arithmetic mean of the two central order statistics for even sample size:NIST uses this convention. ^[ref]. For $(0.7,0.1,0.9,0.4,0.2)$, the sample median is $0.4$; for $(0.1,0.2,0.4,0.7)$, it is $(0.2+0.4)/2=0.3$, which need not be an observed value.

More generally, every value between the two central order statistics in an even statistical sample is a median of its empirical distribution function. The generalized-inverse sample quantile at $p=1/2$ selects the lower central order statistic in that case, so it can differ from the midpoint convention. Ties are allowed in all these definitions.

The sample median measures the center using sorted positions. Changing the largest observation from $0.9$ to $9$ in the five-value example leaves the sample median at $0.4$, while the sample mean rises from $0.46$ to $2.08$. Its sampling distribution still depends on the population probability distribution: the mean $1/2$ found for five observations from a uniform distribution is a special calculation, and a sample median's expected value need not equal the population median in general.

For a concrete counterexample, three independent and identically distributed random variables with rate-one exponential distribution have sample median $X_{(2)}$ with expected value $1/3+1/2=5/6$, by independent exponential order-statistic spacings. The population median instead solves $1-e^{-m}=1/2$, giving $m=\log 2\approx 0.693147$. Thus even the middle order statistic can have mean different from the population median.

A scoring rule turns a probability distribution forecast and an observed outcome into a number. For example, a forecast of rain with probability $0.7$ should express uncertainty more usefully than a bare prediction of rain. A scoring rule lets us compare such forecasts after observations arrive, and its choice determines whether exaggerating confidence pays. A proper scoring rule rewards an honest account of uncertainty in expected value.

There are two sign conventions. A reward $S(P,x)$ is maximized; a loss function $L(P,x)$ is minimized. The conversion $L=-S$ reverses every comparison. Here the linear probability score is a reward, while the binary Brier score and the negative logarithmic scoring rule are losses. Calling all three a "score" without specifying the direction would hide the central difference.

Throughout the binary examples, the random variable $Y$ has a Bernoulli distribution with actual success probability $q=0.7$, and the report $p\in [0,1]$ assigns probability $p$ to $Y=1$ and $1-p$ to $Y=0$. The value $q=0.7$ is stipulated by this statistical model; it is not inferred from data or a claim that a forecasting system is calibrated. In practice the true predictive distribution is unknown, and the relevant $q$ is the conditional probability of success given the information available when the forecast is issued. Propriety evaluates incentives for reporting that predictive distribution; it does not supply the unknown probabilities.

For positive claims with finite nonzero expected value $\mu$, the function $(M_X(r)-1)/r$ is continuous and strictly increasing on the positive finite-transform domain, starting at $\mu$. If $M_X$ diverges at a finite upper endpoint, or is finite for all positive arguments, the secant slope tends to infinity. In the latter case use $M_X(r)\ge \mathbb{P}(X\ge a)e^{ar}$ for some $a>0$ of positive tail probability. Every target $(1+\theta )\mu$ with positive relative safety loading therefore has exactly one positive root.

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].

Take identical rigid spheres of radius $a$, particle density ${\rho }_p$ and displaced liquid density ${\rho }_f$, at uniform absolute temperature $T$. Let $z$ increase upwards from a reference layer and write $\Delta \rho ={\rho }_p-{\rho }_f>0$. The buoyancy correction gives the effective gravitational massThe actual grain mass is ${\rho }_{p}V$; the displaced liquid subtracts ${\rho }_{f}V$ from the effective gravitational mass. In a dilute, noninteracting particle suspension, the Boltzmann distribution therefore givesHere $n$ is number density and $k_B$ is the Boltzmann constant. Equivalently, the dilute osmotic pressure $\Pi =n{k}_{B}T$ supports the grains above it: $d\Pi /dz=-n{m}_{b}g$. Dividing by $n$ and integrating yields the same exponential function. No dynamic viscosity appears in this equilibrium argument. If ${\rho }_p<{\rho }_f$, the population instead increases upwards; at equal densities this gravitational model predicts uniform $n$.

A microscope counts a finite optical slab, not an infinitesimal plane. For equal viewed area $A$ and slab thickness $\delta$, an ideal uniform detection window centred at $z$ givesThus equal optical volumes preserve the exponential function count ratio exactly, even for finite $\delta$, provided the windows lie wholly inside the accessible chamber and have the same detection response. Unequal depth response or overlapping a boundary can corrupt this inference. Repeated observations estimate the mean count; closely spaced visits need not be independent because grains remain in view.

For an original chosen model, set $a=0.250\mu \mathrm{m}$, ${\rho }_p=1200\mathrm{kg}{\mathrm{m}}^{-3}$, ${\rho }_f=1000\mathrm{kg}{\mathrm{m}}^{-3}$, $T=298\mathrm{K}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. These are illustrative inputs, not Jean Baptiste Perrin's measurements. Using $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$ givesIf equal observation effort gives expected count $400$ in the reference layer, the expected counts $20$, $40$ and $60\mu \mathrm{m}$ higher are approximately $214.3$, $114.8$ and $61.49$. Fractions describe expected values over observations, not fractional grains. Using the actual grain mass instead of its buoyancy-corrected value would make the inferred Avogadro constant six times too small for these chosen densities.

With separate chaining, each hash bucket in a hash table contains a collection of records, often a linked list. Lookup computes the bucket index, then follows that chain until a key equality comparison succeeds or the chain ends. A hash collision therefore affects the search work without merging unequal keys. Deletion searches the same chain and unlinks the matching record; in a singly linked list, remembering the preceding record during the search permits the unlink.^[ref]

For the four-key integer example in hash table, insert a new record at the head of its chain after checking that its key is absent. Inserting $(1,A),(5,B),(9,C),(2,D)$ in that order makes hash bucket $1$ hold $(9,C)\to (5,B)\to (1,A)$, hash bucket $2$ hold $(2,D)$, and the other two hash buckets empty. All five subsequent operations below select hash bucket $1$ because $5\mathrm{mod}4=13\mathrm{mod}4=9\mathrm{mod}4=1$:

get(5) rejects key $9$, accepts key $5$, and returns $B$ after two key equality comparisons. get(13) rejects $9,5,1$ and reports absence after three comparisons. put(5,B*) examines $9,5$ and replaces the value without changing $n=4$. delete(9) examines just the head record, removes it, and gives $n=3$. Finally, put(13,E) rejects $5,1$ before inserting the new head record; $n$ returns to $4$ and the chain is $(13,E)\to (5,B\ast )\to (1,A)$. The successive hash table load factors are $1,1,1,3/4,1$. These counts follow from the stipulated chain order; a different insertion policy changes the successful-search counts.

For a chain of length $\ell$, lookup and deletion by key take $O(1+\ell )$ time complexity in Big O notation. The same applies to put before any resizing, since replacement semantics require checking for an existing equal key. Head insertion alone is $O(1)$ if absence is already known; that is a different precondition. These bounds assume $O(1)$ bucket-index hashing, array access, key equality comparisons, allocation and link updates. A long string or large integer can make hashing or comparison depend on its encoded length, so counting record visits alone does not establish constant running time.

Under simple uniform hashing, each distinct key's bucket index is an independent random variable with a uniform distribution on a finite set of $m$ indices. The keys and query are fixed independently of this random assignment. Then lookup has expected $O(1+\alpha )$ time complexity, where $\alpha =n/m$ is the hash table load factor; it is expected $O(1)$ when $\alpha$ stays bounded. These are model assumptions, not an automatic consequence of using a deterministic hash function.^[ref]

Here is the short calculation. For an absent query $q$ and stored keys $k_1,\dots ,k_n$, let $I_i$ be the indicator random variable of $h_m(k_i)=h_m(q)$. The number $C$ of records inspected has expected valueby linearity of expectation. For a present fixed query, scanning its entire chain inspects at most itself plus all other colliding records, with expected value $1+(n-1)/m$; stopping at the match only decreases this upper bound. Adding constant hashing and bucket access yields expected $O(1+\alpha )$ for either query. The argument needs pairwise collision bounds; it does not require all the indicator random variables to be independent. A universal hash family can provide a related bound by random choice of a hash function for a fixed key set, as explained in that article.

The worst case remains $O(n)$, and an absent query takes $n$ comparisons if all $n$ records occupy its hash bucket. For instance, $1,1+m,1+2m,\dots ,1+(n-1)m$ all collide under $h_m(k)=k\mathrm{mod}m$. Separate chaining remains correct, but a sequence of insertions that each checks the existing chain can take quadratic total work. Amortized analysis of resizing does not remove this search cost, and an adaptive choice of keys based on a known hash function is outside the stated simple uniform hashing model.