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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.

A semi-log plot uses a logarithmic scale on one axis and a linear scale on the other. Which axis is logarithmic matters. Plotting $x$ against $\log y$ makes positive exponential functions appear as straight lines; plotting $\log x$ against $y$ instead makes an affine function of a logarithm appear straight. The plotting operations change positions while retaining the original tick values.^[ref]

Take the original dimensionless mathematical model $y(t)=3e^{0.4t}$ for $0\le t\le 8$. It has exponential growth, with $y(0)=3$, $y(4)\approx 14.8591$, and $y(8)\approx 73.5976$. Taking a base-ten logarithm givesThe slope in these transformed positions is $0.4/\ln 10\approx 0.173718$, although the original relative growth rate $y^{\prime }/y$ is $0.4$. Equal steps of two in $t$ multiply $y$ by $e^{0.8}\approx 2.22554$. The corresponding ordinary gains increase, so the left plot bends while the right plot is straight.

One practical use is checking a proposed exponential distribution for failure times. For rate $\lambda >0$ and $t\ge 0$, its cumulative distribution function is $F(t)=1-e^{-\lambda t}$, so the reciprocal survival function $y=1/[1-F(t)]$ satisfies ${\log }_{10}y=\lambda t/\ln 10$. Estimated probabilities can therefore be placed on a logarithmic vertical axis and compared with a straight line. Agreement supports compatibility with the chosen statistical model; a roughly straight finite sample does not establish a unique physical mechanism.^[ref]

If $X$ is a semimartingale under $P$ and $Q\ll P$, then it is a semimartingale under $Q$. The Bichteler-Dellacherie theorem gives a short proof: terminal elementary integrals associated with uniformly small elementary predictable processes with stopping-time intervals converge to zero in $P$-probability. Uniform convergence on compacts in probability under an absolutely continuous measure change, or its single-random-variable proof, transfers convergence to $Q$. The same good-integrator characterization now applies under $Q$. Adaptedness and càdlàg paths are preserved because $P$-null sets remain $Q$-null; if necessary use the usual $Q$-completion of the same filtration. If the integrands are bounded only outside $Q$-null sets, clip their coefficients to the specified deterministic bounds before applying the $P$-criterion; this preserves their $Q$-versions. Completion adds only equivalent versions of measurable coefficients. This theorem does not preserve the local martingale property.

Under an equal-variance Gaussian evidence model, $d^{\prime }$ is separation of signal and noise means in units of their common standard deviation. A yes/no task gives $d^{\prime }={\Phi }^{-1}(H)-{\Phi }^{-1}(F)$, where $H$ and $F$ are hit and false-alarm probabilities. This formula assumes the model; it does not define every possible sensory-discrimination measure.

Complete separation means that a predictor direction $v$ strictly separates the binary outcomes as displayed. In logistic regression, replacing $\beta$ by $\beta +tv$ increases each observed-outcome probability as $t$ increases, so the log-likelihood has no finite maximizer. Weak inequalities with some equalities can similarly produce quasi-complete separation. Finite coefficient estimates and ordinary information-matrix inference therefore require an existence check rather than relying only on the formal score equations.

Consider an original balanced-readout model, rather than a fit to the experiment. Let $n\in \{0,\dots ,N_{\max }\}$, with a normalized prior probability $p_0(n)\ge 0$ and ${\sum }_n{p}_0(n)=1$. A detected chemical atom has outcome $s\in \{e,g\}$. Define the conditional probabilitiesHere $\chi$ is the accumulated quantum phase per photon, $\varphi$ the controlled analysis quantum phase, and $C$ the fringe contrast. The two probabilities sum to one for every $n$. Actual calibration may also require an offset and asymmetric detection errors; $C$ alone does not specify a quantum instrument.

Assume no photon number change during the record and fresh independently prepared atomic meters. For observed outcomes $s_1,\dots ,s_k$ at recorded settings ${\varphi }_1,\dots ,{\varphi }_k$, the Born rule and Bayes' theorem giveThe posterior probability is defined only for $Z_k>0$. Summing its numerator over $n$ proves normalization. This is also the update of diagonal elements of a field density operator under a measurement in quantum mechanics; it does not assume that an initial quantum superposition was an unknown pre-existing integer. The observed settings are part of the record, especially when selected adaptively.

For an exact finite example take $p_0=(1,1,1,1)/4$, $C=4/5$, and $\chi =\pi /2$. An $e$ outcome at $\varphi =0$ has likelihood $(9,5,1,5)/10$, giving $p_1=(9,5,1,5)/20$. Another $e$ at $\varphi =-\pi /2$ gives $p_2=(9,9,1,1)/20$. A subsequent $g$ at $\varphi =0$ givesThus three atomic records raise the posterior probability of $n=1$ from $1/4$ to $45/68$, while leaving finite uncertainty. No one of these outcomes identifies $n$ with certainty.

The periodic likelihood function imposes an identifiability limit. If $\chi =2\pi /q$ for an integer $q>0$, then $L_s(n+q;\varphi )=L_s(n;\varphi )$ for every setting: the posterior odds between $n$ and $n+q$ never change. Varying $\varphi$ can resolve a cosine reflection ambiguity within a period, but cannot remove this exact period. Changing $\chi$ can supply additional information. A finite cutoff is therefore a physical assumption to justify, not evidence that excluded high photon numbers were absent. At $C=0$ no outcome changes the prior probability; at $0<C<1$ neither binary outcome has an exactly zero likelihood.

The no-loss assumption also has a time limit. In an original zero-temperature loss model, let $\kappa =1/T_c$. A Fock state $|n⟩$ loses its first photon at rate $n\kappa$, so its survival probability is $\exp (-n\kappa t)$ and its mean lifetime is $T_c/n$ for $n>0$. Several records must be acquired within that lifetime to identify a plateau. When losses matter, first predict $p_k^{-}(n)={\sum }_m{T}_{nm}(\Delta t)p_{k-1}(m)$ using a normalized transition matrix, then apply the atomic Bayesian inference update to $p_k^{-}$. Thermal excitation requires upward transitions too. A delayed or broadened inference after a jump need not indicate that an atom caused the jump.

Prepare $I/2$, measure a Pauli observable $N=\mathbf{n}\cdot \mathbf{\sigma }$, apply a fixed unitary operator $U$, then measure $N$ again. If the first and second eigenvalue labels are $j,k\in \{+1,-1\}$, their joint probability is $\frac{1}{2}|⟨n_k|U|n_j⟩{|}^2$. ConsequentlyThe first outcome has probability $1/2$ and its conditional state is $|n_j⟩$ by the Lüders rule. Summing $jk$ times the joint probabilities and using $N={\sum }_{j}j|n_j⟩⟨n_j|$ proves the trace formula. This temporal correlation differs from the spatial correlation $⟨\psi |N\otimes N|\psi ⟩$ of two simultaneously measured qubits.

For otherwise outbred opposite-sex first cousins, the probability that the mother's randomly transmitted X copy matches the father's X by descent depends on which of their parents are siblings. It is zero when the male cousin is related through his father, $1/8$ when the female cousin is related through her father and the male through his mother, and $3/16$ when their mothers are sisters. Daughters can be X-autozygous; sons have one X and a Y and are not homozygous for a paired X chromosome. Recombination and pseudoautosomal regions require a locus-based formulation instead of whole-chromosome blocks.

For a finite source alphabet with at least two symbols and strictly positive weights, every ternary prefix code minimizing weighted codeword length has all internal vertices of outdegree three, except possibly one of outdegree two whose tree children are deepest tree leaves.

To prove this, let $L$ be the greatest tree leaf depth. A missing tree child at any depth $n<L$ would permit moving a deepest tree leaf into that vacant branch, strictly reducing its positive weighted cost. Thus all internal vertices above depth $L-1$ have three tree children. Tree parents at depth $L-1$ have only tree leaves as tree children; one tree child would permit contraction and strict improvement. If two such tree parents each had two tree children, move one tree child of the second into the first tree parent's third slot, preserving its depth, and contract the second tree parent's remaining tree child. This again strictly improves cost. There is therefore at most one binary tree parent. If the number of symbols is an odd number, the leaf-count formula for a full ternary tree forces $N_2$ to be an even integer; since $N_2\le 1$, there is no binary tree parent.

Strict positivity matters. For probabilities $(1,0,0)$, the prefix code $(0,10,110)$ minimizes expected codeword length at one trit, yet has unary vertices and a deepest tree leaf without a sibling codeword. Removing or rearranging zero-weight branches can select a reduced minimizer, but arbitrary minimizers need not have the positive-weight shape.

The Shapley value averages each player's marginal contribution over uniformly random player orderings. Exactly $|S|!(n-|S|-1)!$ orderings have $S$ immediately before player $i$, giving the formula. The values sum to $v(N)$ by telescoping each ordering. In a simple cooperative game this is the probability of being pivotal. For a convex cooperative game, Shapley value belongs to the core of a convex game guarantees a stable allocation as well.

In a uniformly random player permutation, edge $\{i,j\}$ contributes $w_{ij}$ to $i$'s marginal contribution exactly when $j$ precedes $i$, which occurs with probability $1/2$. Sum these expected contributions over incident graph edges to obtain the displayed Shapley value. This proof works for arbitrary real edge weights; nonnegativity is not required.

An entrepreneur is indispensable and a coalition containing the entrepreneur and $k$ of $n$ workers earns $p(k)$; a coalition without the entrepreneur earns zero. A worker's marginal contribution is nonzero only when the entrepreneur precedes them. The probability that the entrepreneur and exactly $k$ other workers precede that worker is $(k+1)/(n(n+1))$, giving the displayed common Shapley value wage. The entrepreneur receives $p(n)-nw$. Convex nondecreasing $p$ makes all marginal contributions increase with predecessor sets, so the whole Shapley value allocation belongs to the core.

Draw one random variable $R$, taking values zero and two with equal probabilities. Conditional on $R$, generate a Poisson point process on the real line with constant rate $R$. This is a Cox process directed by the random measure $Rdx$: the same rate is shared by every window. It is a locally finite simple point process and a stationary point process. The law of total expectation gives $\mu (A)=\mathbb{E}[R]|A|=|A|$; the law of total variance and law of total covariance give, for bounded Borel sets $A,B$ with $A\cap B=\varnothing$,The conditional covariance vanishes by independence, but the shared conditional means fluctuate together. This distinguishes conditional from unconditional independence.^[ref]

For a window of length $L$, the exact probability mass function isIn particular, its void probability functional is $(1+e^{-2L})/2$, compared with $e^{-L}$ for the unit-rate Poisson point process. Whole realizations can be empty or comparatively crowded, while their mean density remains one. This example demonstrates shared-rate statistical dependence, rather than a model of short-range offspring clusters.

Choose one random variable $U$ with the continuous uniform distribution on $[0,1]$, and reuse it in every event:Each probability is $1/(n+1)$, so their harmonic series diverges, just as in summable and harmonic event probabilities. But the events are decreasing: $A_{n+1}\subseteq A_n$. For every outcome $U=u>0$, all indices with $n+1>1/u$ fail. Only finitely many events occur on that outcome. The limsup of events is exactly $\{U=0\}$, a zero-probability event. This is a concrete closed-interval version of the shrinking-interval dependence example.^[ref]

The missing assumption is visible already in two events. For $i<j$, their intersection is $A_j$, henceMoreover, the whole tail union is just $A_m$, with probability $1/(m+1)\to 0$. This calculation explains directly why divergence alone fails: overlapping events keep spending their individual probabilities on the same outcomes.

Let $N={\sum }_{n\ge 1}{1}_{A_n}$ be the total number of occurrences, a sum of indicator random variables. Although $N<\infty$ almost surely, Tonelli theorem gives $\mathbb{E}[N]={\sum }_{n\ge 1}1/(n+1)=\infty$. More explicitly, $\mathbb{P}(N\ge k)=1/(k+1)$ for $k\ge 1$: very small positive values of $U$ give very long, but finite, initial runs. Infinite expected value does not imply a positive probability of an infinite count.

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.

Let a random vector $V$ have a finite actual range of $q$ values. Its nonempty fibres partition the sample space, and every event in $\sigma (V)$ is a union of those fibres. Thus each fibre is an atom of a sigma-algebra and $\sigma (V)$ has exactly $2^q$ elements. This assertion concerns the actual range; additional values attained only on null sets must still be counted when working with an uncompleted sigma-algebra literally.

To verify the claim, the fibres belong to $\sigma (V)$ as preimages of singleton Borel sets. Their unions form a finite sigma-algebra making $V$ measurable, so minimality of the generated sigma-algebra gives the reverse inclusion. Distinct subcollections have distinct unions because the fibres are nonempty and disjoint.

Consequently a finite history of finite-valued observations generates a finite natural filtration at each fixed time. It may be strictly smaller than the filtration of its driving continuous observations: a single uniform random variable already generates infinitely many distinct events $\{U\le t\}$ with different probabilities.

For $m\ge 1$ independent valid null p-values, the displayed inequality follows by applying the Benjamini-Hochberg procedure with all nulls true. Its false discovery rate then equals the probability of at least one rejection. Valid means $\mathrm{Pr}(p_i\le t)\le t$ for $0\le t\le 1$; independence is a sufficient hypothesis, and arbitrary dependence is not covered by this proof.

A simple cooperative game is a monotone transferable utility game in which each coalition either loses, with value zero, or wins, with value one. In the usual nontrivial convention the empty coalition loses and the grand coalition wins. The Shapley value is then the probability that a player changes a losing coalition into a winning one when players enter in uniformly random order.

Here simple random sampling means a random unordered subset $A$ of exactly $n$ distinct labels from $\{1,\dots ,N\}$, with every size-$n$ subset equally likely. The sample size is fixed in advance, $1\le n\le N$, and sampling without replacement prevents duplicate units. This is SRSWOR; some texts also use "simple random sampling" for independent uniform draws with replacement, which is a different design.^[ref]

Every label has inclusion probability $n/N$, but equal inclusion probabilities alone are insufficient. For example, from six labels choose either $\{1,2,3\}$ or $\{4,5,6\}$ with probability $1/2$. Each label is included with probability $1/2$, yet eighteen of the twenty size-three subsets are impossible: this is a two-group cluster sampling design. Unequal probability sampling deliberately changes inclusion probabilities, often to improve precision for a particular estimand, and generally requires an appropriately weighted estimator rather than the plain sample mean.

Consider the original mathematical population with labelled values $(2,4,6,8,10,12)$, and let $n=2$. The fifteen equiprobable subsets, followed by their sample means, areTheir sample means sum to $105$, giving expected value $105/15=7$. Their squared deviations from $7$ sum to $70$, giving variance $70/15=14/3$. This is an enumeration over hypothetical selections from six stipulated numbers, not measured survey data.

With independent uniform sampling with replacement, the $6^2=36$ ordered outcomes each have probability $1/36$. The two orders $(2,4)$ and $(4,2)$ are distinct outcomes, and $(2,2)$ is now possible. Forgetting order does not make the resulting unordered multisets equiprobable: $\{2,4\}$ has probability $2/36$ but $\{2,2\}$ has probability $1/36$. The thirty-six sample means still have expected value $7$, but their squared deviations sum to $210$, giving variance $210/36=35/6$. Avoiding duplicate labels improves precision for this fixed population.