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The binomial distribution counts successes in a fixed number $n$ of independent Bernoulli trials with common success probability $p$. Equivalently,Here $n$ is a nonnegative integer and $0\le p\le 1$. The binomial coefficient $\left(\genfrac{}{}{0ex}{}{n}{k}\right)=n!/[k!(n-k)!]$ counts placements of the $k$ successes; each particular arrangement has probability $p^k(1-p{)}^{n-k}$. The binomial theorem makes the masses sum to $[p+(1-p){]}^n=1$. At $p=0$ all mass is at zero, and at $p=1$ all mass is at $n$; these endpoint laws can be read directly without interpreting $0^0$.

The linearity of expectation and addition of variances for independent random variables giveThe fixed trial count, common success probability, and joint independence are substantive model assumptions. Sampling uniformly without replacement from a finite population instead leads to a hypergeometric distribution. Independent trials with different success probabilities generally give a Poisson binomial distribution. Common marginal success probabilities alone are insufficient: if all ten indicators equal the same Bernoulli random variable with $p=0.3$, their sum is zero with probability $0.7$ or ten with probability $0.3$. Its mean is still three, but its variance is $100(0.21)=21$. The standard construction is given in Penn State STAT 504, binomial distribution.

For an exact constructed example, let each of ten independent trials choose uniformly from ten labelled tokens, with replacement, and mark three tokens as successes. Then $p=3/10$ and $K\sim \mathrm{Bin}(10,0.3)$. Exactly three successes have probabilityThe mean is $10(0.3)=3$, the variance is $10(0.3)(0.7)=2.1$, and the standard deviation is $\sqrt{2.1}\approx 1.4491$. Zero successes have probability $0.7^{10}=0.0282475249$, so at least one success has probability $1-0.7^{10}=0.9717524751$. A single count calculation replaces enumeration of all $10^{10}$ equally likely token sequences.

Each bar's height is one probability mass. Three is the most likely count, but its mass is only about $26.7\%$: a most likely outcome need not occur most of the time.

A capture-recapture model is a statistical model for repeated observations of identifiable individuals, including those missed on some occasions. A closed-population capture-recapture model estimates a fixed population cardinality; models allowing individuals to enter or leave can instead describe survival and recruitment. The sampling assumptions determine which likelihood function is appropriate.

For the two-sample closed-population capture-recapture model, condition on the sample sizes $M$ and $C$. If the second sample selects $C$ distinct individuals from $N$ by simple random sampling, then the overlap $R$ has the hypergeometric distribution:Here $r$ is an integer and the probability mass function is zero outside these bounds. The binomial coefficients count choices of $r$ marked and $C-r$ unmarked individuals, divided by all equally likely size-$C$ subsets. The hypergeometric distribution has expected valueEquating the observed overlap with this expected value gives the Lincoln-Petersen estimator. The underlying sampling without replacement, probability mass function, and expected value are developed in Raymond P. Guiteras, MIT 14.30 hypergeometric distribution handout, §§1–4 and 8.

For the toy counts $M=20$, $C=25$, and $R=5$, hold the observations fixed and vary the unknown integer $N$. The likelihood function isValues below forty cannot contain the observed set union. Exact comparison of consecutive likelihoods gives two tied maximum-likelihood estimates, $N=99$ and $N=100$. The Lincoln-Petersen estimator returns $100$; the Chapman estimator returns $90$ and is a different rule for estimation.

The vertical scale is $L(N)/{\max }_{n\ge 40}L(n)$. This relative likelihood compares how well candidate statistical parameters explain the same observations; it is not a posterior probability. The displayed range $40\le N\le 300$ is only a plotting window. The likelihood function remains positive for every integer $N\ge 40$ and tends to zero as $N$ increases, so the curve has a continuing right tail. The data do not justify truncating statistical inference at three hundred. The conditional hypergeometric distribution formulation also appears in Louis-Paul Rivest and Mamadou Yauck, Small Sample Estimators for Two-way Capture Recapture Experiments, §3, equation (5).

Zero recaptures require care. With $R=0$ and positive $M,C$, the Lincoln-Petersen estimator has no finite ratio, but the capture-recapture model still gives a valid likelihood function:This likelihood function increases toward one as $N$ grows, so there is no finite maximum-likelihood estimate. For the same sample sizes twenty and twenty-five with zero overlap, forty-five distinct individuals would be observed. Lack of overlap supplies a lower bound but very weak information about large population cardinalities.

In more elaborate capture-recapture models, a first-recapture array $m_{ij}$ counts individuals released at occasion $i$ and next observed at $j$. A probability of survival ${\varphi }_j$ and a probability of detection $P_{j+1}$ contribute a no-observation continuation factor ${\varphi }_j(1-P_{j+1})$. The resulting likelihood function depends on both survival and detection; these probabilities need not be separately identifiable at the last observation interval. Missing-count EM for capture-recapture describes one way to fit such a model, and identifiability remains a separate issue from computing an estimator output.

A closed-population capture-recapture model assumes the same individuals remain throughout the observations: no births, deaths, immigration, or emigration. Its elementary two-sample version additionally requires retained, correctly recognized marks and a second sample chosen by simple random sampling independently of first-sample membership. Mixing and unchanged capture behavior help make this plausible. Equal individual capture probabilities alone do not guarantee that every size-$C$ subset is equally likely; that stronger sampling assumption gives the hypergeometric distribution. See Barker, §§1.1.1 and 1.1.3: closure, marks, and catchability assumptions.

Unequal catchability, mark loss, and population change can invalidate this capture-recapture model. For example, if especially catchable individuals appear repeatedly, overlap can be inflated and the Lincoln-Petersen estimator can underestimate population cardinality. Lost marks can decrease recorded overlap and increase the same ratio estimator, all else equal. Two samples alone cannot establish that the sampling assumptions hold.

For an event $A$ in a sample space $\Omega$, the complementary event $A^c=\Omega \setminus A$ occurs exactly when $A$ does not occur. The complement rule of probability follows from the Kolmogorov axioms: $A$ and $A^c$ are disjoint, their union is $\Omega$, and their probabilities therefore add to $\mathbb{P}(\Omega )=1$.

This rule is useful when an event is easier to describe through its failure. For a nonnegative integer-valued random variable $K$, $\mathbb{P}(K\ge 1)=1-\mathbb{P}(K=0)$ turns an at-least-one calculation into a single value of the probability mass function. In sampling without replacement, the hypergeometric distribution gives the chance of missing all marked objects, so its complement gives the chance of finding at least one.

For independent uniformly placed cyclic intervals of length $k<d$, an overlap is possible at at most $2k-1$ starting-point offsets, or at all $d$ offsets if that number exceeds $d$. Thus the overlap probability is at most $\min \{1,(2k-1)/d\}$. If $R=|S\cap T|$, then $0\le R\le k$ and $\mathbb{E}(e^{aR/k}-1)\le (2k/d)(e^a-1)$ for $a\ge 0$. The overlap law is not the hypergeometric distribution for unrestricted random subsets.

For two independent binomial counts, form the Fisher exact test rejection region from their conditional null hypergeometric distribution. Summing alternative joint binomial probabilities over that region gives its exact power. For equal group sizes, the null allocation is symmetric, so the two-sided probability-ordering test agrees with doubled lower-tail probabilities. Conservative test discreteness can make a normal-approximation sample size underpowered even when both samples are large but event counts are low.

For uniform sampling without replacement from a population of $n>1$ objects, the finite population correction accounts for the reduced spread caused by drawing distinct objects. For the marked-object count $K$ in a hypergeometric distribution, put $p=c/n$. ThenRelative to the corresponding binomial distribution, the variance is multiplied by $f^2=(n-r)/(n-1)$ and the standard deviation by $f$. The square-root convention for the correction is used in Philip Stark's discussion of sample sums.^[ref]

The factor has a direct interpretation. For $r=1$, replacement makes no difference and $f=1$. For $r=n$, the entire population is observed and the marked count is exactly $c$, so $f=0$. In the ten-object, five-draw example, $f^2=5/9$ and $(6/5)(5/9)=2/3$. The exceptional population size $n=1$ is deterministic and should be handled directly rather than inserting it into a zero denominator.

A short explanation uses an indicator random variable $I_j$ for a marked object at draw $j$. Each has expected value $p$ and variance $p(1-p)$. For distinct draw positions $i,j$,Since $K={\sum }_{j=1}^r{I}_j$, the variance of a sum includes $r$ individual terms and $r(r-1)$ ordered covariance terms, giving the displayed correction. Linearity of expectation still gives $rp$ despite this dependence.

For arbitrary fixed real population values, use $N>1$ for population size and $1\le n\le N$ for sample size. Write the finite population mean $\bar{Y}=N^{-1}{\sum }_i{y}_i$, the finite population variance $S^2=(N-1{)}^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$ and ${\sigma }_N^2=N^{-1}{\sum }_i(y_i-\bar{Y}{)}^2$. Under simple random sampling without replacement, the sample mean satisfiesThus $1-n/N$ is the variance factor relative to $S^2/n$, while $(N-n)/(N-1)$ is the exact variance ratio to independent uniform sampling with replacement, whose sample mean has variance ${\sigma }_N^2/n$. The standard error uses the square root of the corresponding factor. These formulas agree because ${\sigma }_N^2=(N-1)S^2/N$; the denominator convention must be specified. The sample mean under simple random sampling gives a derivation and an exact six-value example.

Conditional on the pooled number $s$ of successes in $m+n$ exchangeable random variables that are binary, all placements of those successes are equally likely under a common success-chance mixture. The count in a chosen batch of size $n$ therefore has a hypergeometric distribution, with mass $\left(\genfrac{}{}{0ex}{}{s}{y}\right)\left(\genfrac{}{}{0ex}{}{m+n-s}{n-y}\right)/\left(\genfrac{}{}{0ex}{}{m+n}{n}\right)$. Under the uniform-count predictive property, $s$ is uniform on $0,\dots ,m+n$. Dividing the joint pooled-and-batch probability by the first batch's uniform marginal gives the posterior predictive distribution, with factor $(m+1)/(m+n+1)$. Substituting $s=x+y$ in the hypergeometric expression does not give a single normalized hypergeometric mass as $y$ varies, since its population success total then varies too.

The hypergeometric distribution counts marked objects in a uniformly chosen sample of fixed size from a finite population, using sampling without replacement. Let $n\ge 1$ be the population size, let $0\le c\le n$ objects be marked, and sample $0\le r\le n$ distinct objects. If $K$ counts the marked objects selected, its probability mass function isThe bounds describe the support of a probability distribution and $k$ must be an integer; the mass is zero elsewhere. The denominator counts all size-$r$ subsets, while the numerator counts choices of $k$ marked and $r-k$ unmarked objects using binomial coefficients. This counting argument requires every size-$r$ sample to have the same probability.^[ref]

For a constructed example, a box contains ten labelled objects, four red and six blue. Select five without replacement. There are $\left(\genfrac{}{}{0ex}{}{10}{5}\right)=252$ possible samples. The probabilities for $K=0,1,2,3,4$ areIn particular, $\mathbb{P}(K=2)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)\left(\genfrac{}{}{0ex}{}{6}{3}\right)/252=120/252=10/21$. Five red objects are impossible because the box only contains four. The expected value is $r(c/n)=2$, and the variance is $2/3$.

The count law also answers a detection question: what is the chance that five draws find at least one red object? By the complement rule of probability, it is $1-\mathbb{P}(K=0)=41/42$, approximately $0.9762$. This can model inspecting five distinct objects from a known finite collection, with a mark indicating the property we want to detect. If repeated inspections can select the same object again, the replacement model instead gives $1-0.6^5=0.92224$. Equal expected values therefore do not imply equal chances of detecting a marked object.

The corresponding binomial distribution counts red draws when each selected object is put back before the next independent draw: $K_{\mathrm{rep}}\sim \mathrm{Bin}(5,0.4)$. It has the same expected value $2$, but variance $5(0.4)(0.6)=6/5$, and it permits five red draws. Without replacement, a red first draw leaves red probability $3/9$ for the next draw; a blue first draw leaves $4/9$. This statistical dependence pulls the total toward the fixed composition of the population, producing the finite population correction.

The useful modelling question is whether previously selected objects remain available. A fixed-size sample of distinct items calls for a hypergeometric distribution; fresh independent trials with constant success probability call for a binomial distribution. When the population grows with $c/n\to p$ while $r$ stays fixed, the hypergeometric masses tend to the corresponding binomial distribution masses: removing finitely many objects has an increasingly small effect on the next draw's probability.

Suppose a finite population is split into $m$ categories with fixed sizes $c_1,\dots ,c_m$ summing to $n$. A uniform size-$r$ sample drawn by sampling without replacement has category counts $K_1,\dots ,K_m$ with the displayed joint probability mass function, for integers $0\le k_i\le c_i$ summing to $r$. It extends the two-category hypergeometric distribution; replacing the sampled objects instead gives a multinomial distribution.

For a box with category sizes $(5,3,2)$ and a sample of four distinct objects, the count pattern $(2,1,1)$ has probabilityPut $p_i=c_i/n$. When $n>1$, its covariance matrix is the corresponding multinomial distribution covariance matrix multiplied by the squared finite population correction:This connects the fixed-total dependence between categories with the further reduction from sampling without replacement. X. G. Duan gives a symmetry-based explanation of this shared covariance structure.^[ref]

Using the upper stated scrapie-test-positive fraction, its value among slaughtered sheep is $2000/4000000=0.0005$. The stipulated conditional BSE positivity then gives marginal test-positive probabilitybecause the other stratum contributes zero. If the sample consists of independently tested, representative animals, the count $D$ has a binomial distribution with parameters $n,q$. ThereforeThe two results areThe Poisson distribution approximation uses mean counts $0.5$ and $5$, respectively. If fewer than 2000 animals belong to the test-positive stratum, these probabilities of finding nothing are higher; the calculation at 2000 is not a guaranteed detection probability.

There is also a sampling-design distinction. If exactly $K$ positive animals in a fixed annual population of $N$ are known and the sample is without replacement, use the hypergeometric distribution instead:The given expected prevalence does not assert an exactly fixed $K$, so the independent prevalence model is the natural interpretation of the requested calculation.

The log-rank test compares event counts with their conditional expectations under equal group hazard functions, using the individuals currently in each risk set. Assume the usual independent censoring conditions and independent strata. At pooled event time $t_j$, let $n_{Aj},n_{Bj}$ be the group numbers at risk, $n_j=n_{Aj}+n_{Bj}$, and let $d_{Aj},d_{Bj}$ be the event counts with total $d_j$. Under the null hypothesis, conditioning on the risk counts and total events gives the hypergeometric distributionThus the null expected A count and its conditional variance areWhen there is only one person at risk, the event allocation is deterministic and its variance is zero. The unstandardized log-rank statistic is $U={\sum }_j(d_{Aj}-E_j)$, with variance estimate $V={\sum }_j{V}_j$. Under standard large-sample conditions, $U/\sqrt{V}$ is approximately standard normal under the null; the test detects systematic group differences in event rates. It is particularly effective for proportional-hazards alternatives, while crossing effects can cancel in the score.

For a stratified log-rank statistic, construct these risk sets and observed-minus-expected differences separately within each stratum $s$, then sum:This allows different baseline hazard functions between strata and avoids comparing an event with individuals in another stratum. The null is equal A and B hazards within each stratum, together with a censoring mechanism that does not invalidate the within-risk-set comparison.

In a paired stratum, both individuals initially form the risk set. A first event while both are still observed gives one nontrivial event allocation. After either individual leaves, every subsequent event is certain to belong to the only remaining group and has observed count equal to expected count. Therefore each pair supplies at most one informative table. A pair supplies none if the first exit is censoring before either observed event, if neither has an event during their joint follow-up, or if both fail simultaneously so that the tied event allocation is fixed. An event-censoring tie needs a specified convention; with events processed before same-time censoring, the other individual remains at risk for that event and the pair is informative. The untied continuous-time case avoids this ambiguity.

If A has the informative first event, the table at that time isThe phrase in the second column refers only to this event time; it does not mean that B can never fail. The expected A count is $1\cdot 1/2=1/2$, so this table contributes $1-1/2=1/2$ to $U$, with variance $1/4$. An informative B-first pair contributes $-1/2$. All other tables contribute zero. Therefore the requested unnormalized score isA positive score means more observed early events on A. There is no need to standardize it to answer the question.

To obtain an exact conditional law, put $m=d_A+d_B$. Under the within-pair equal-hazard null, conditional on an event while both are at risk, either subject supplies it with probability $1/2$. Equivalently, before the pair's first exit, the cause-specific event intensities for A and B are equal; integration over the common period of joint observation gives equal probabilities for the two informative outcomes. This argument requires noninformative observation of the failure process, not just equality of unadjusted marginal survival curves. Different independent pairs then have independent fair signs when informative. Conditioning on the set of informative pairs gives $m$ independent Bernoulli outcomes with probability $1/2$; conditioning only on their count mixes identical binomial laws and gives the same result:This proves the paired log-rank reduction to a sign test. The conditional mass function is $\left(\genfrac{}{}{0ex}{}{m}{k}\right)2^{-m}$, $0\le k\le m$, so an exact two-sided p-value sums the masses for $k$ with $|2k-m|\ge |d_A-d_B|$. If $m=0$, there is no information for a treatment comparison. Failure-time distances beyond their ordering play no role in this paired score.

For the uniform prior, write $s=x+y$ for the pooled success count in $m+n$ trials. Conditional on $s$, exchangeability makes all placements of the successes equally likely. The successes in the second batch therefore have a hypergeometric distribution, and the factor in question isThis is hypergeometric allocation of exchangeable Bernoulli counts: sample $n$ positions without replacement from a population of $m+n$ positions containing $s$ successes. The uniform-count predictive property gives $\mathrm{Pr}(X+Y=s)=1/(m+n+1)$ and $\mathrm{Pr}(X=x)=1/(m+1)$, soAs $y$ varies with $x$ fixed, $s=x+y$ also varies. Therefore $B$ alone is not a normalized hypergeometric distribution over $y$; its normalization factor is essential.

Define $P_{\theta ,n}=d^{-1}{\sum }_{S\in \mathcal{S}}{P}_{\theta ,S}^{\otimes n}$. Expanding its chi-squared divergence by the second moment of a mixture likelihood ratio, and using the supplied one-observation cross moment and independence, givesHere $S,T$ are independent uniform members of the specified family of cyclic intervals. Their overlap $R=|S\cap T|$ does not have the hypergeometric distribution of arbitrary uniform $k$-subsets. The family's geometry must be retained.

Let $x={\theta }^2{R}^2/k^2<1/4$. Since $-\log (1-x)\le x/(1-x)\le 2x$ and $R^2/k^2\le R/k$, we obtain $(1-x{)}^{-n/2}\le e^{nx}\le e^{n{\theta }^{2}R/k}$. Taking the expected value proves the upper bound. The factor $n$ in the last exponent is present in the PDF and missing in the TeX aid.

At event time $t_j$, let $n_{Aj},n_{Bj}$ be the two risk set sizes, $n_j=n_{Aj}+n_{Bj}$, and let $d_{Aj},d_{Bj}$ be the event counts with $d_j=d_{Aj}+d_{Bj}$. Under the null hypothesis of equal hazards, conditioning on the risk set and total number of events gives a hypergeometric distribution, soandThe log-rank statistic and its estimated null variance areUnder the null, $U/\sqrt{V}$ is asymptotically standard normal, or $U^2/V$ is asymptotically chi-squared with one degree of freedom.

Cross-classify the binary exposure $A_i=1_{\{Z_i\ge 1\}}$ against the binary outcome $Y_i$. The four cells count placebo survivors or deaths and active-treatment survivors or deaths.

Under the sharp causal null hypothesis, every observed outcome equals the fixed value $Y_i(0)=Y_i(1)=Y_i(2)$, regardless of assignment. With fixed arm sizes, the active set is a uniformly chosen subset of size $n_1+n_2$. Under independent assignment, $A_i$ are independent Bernoulli variables with probability ${\pi }_1+{\pi }_2$, and conditioning on their total again makes the active set uniform. Conditional on the table margins, the active-group outcome count therefore has exactly the hypergeometric distribution used by Fisher's exact test.

The conditional p-value is super-uniform for every set of margins. The law of total probability then givesHence the test is valid under either assignment mechanism.