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For $\varphi =f+h$, with $f$ a convex function having an $L$-Lipschitz gradient and $h$ a proper convex function with a well-defined proximal operator, the proximal gradient method with step size $1/L$ satisfies the displayed bound whenever a global minimizer $x^{\ast }$ exists. The one-step subgradient inequality and descent lemma giveThe objective function values are nonincreasing. A telescoping sum over $j=0,\dots ,k-1$, followed by comparison of the last error with the arithmetic mean of the errors, proves the result. The bound controls the last objective function error without requiring a strongly convex function or a unique minimizer.

Target: convex-optimization.bigb, under Proximal gradient method.

Quantization maps numerical values to a specified discrete collection of representable values. In lossy block averaging, $Q(m)=\lfloor m+1/2\rfloor$ maps block arithmetic means in $[0,255]$ to eight-bit integers. Thus $Q(40.25)=40$ and $Q(40.5)=41$. Different means can share a stored value, making this step irreversible.

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.

Scientific reproducibility concerns whether the basis of a scientific result can be reconstructed and checked. It requires enough information about the measurement, data handling and reasoning to distinguish repeating a procedure from merely obtaining a similar-looking answer.

Terminology varies between fields. The 2019 National Academies report distinguishes computational reproducibility, using the same data and computational procedure, from scientific replication, which collects new data to address the same scientific question. These test different links in the argument. Repeating a calculation can expose a software or transcription error; new observations can test whether the finding persists beyond the original sample.^[ref]

Consider an original toy measurement of three lengths, $10$, $11$ and $12$ units. Recomputing their arithmetic mean as $11$ checks the recorded calculation. Measuring three new objects tests a different question: whether the estimated mean describes further observations, allowing for measurement uncertainty and sampling variation. Neither exercise alone establishes that the instrument was correctly calibrated.

Equivalent chemical bonds in an intact molecule need not have equal successive bond-dissociation enthalpies. Once one chemical bond is removed, the next reactant has a different electronic quantum state and may relax to a different molecular geometry. The successive steps compare different species. This distinction also explains why a tabulated average chemical bond strength is not necessarily the cost of the first cleavage.^[ref]

For an invented species $A{B}_2$, stipulate the following gas-phase steps, with all enthalpies at the same temperature and the same standard states:The symbols stand for hypothetical species with consistently chosen electronic states. They are not values for water or another measured molecule. By Hess's law, the total enthalpy of atomization for $A{B}_2(g)\to A(g)+2B(g)$ is $700\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$; the arithmetic mean per broken chemical bond is $350\mathrm{kJ}\mathrm{mo}{\mathrm{l}}^{-1}$. Neither individual step equals that mean. Nor must the second step be cheaper: here it is deliberately more expensive. An exact sequence sums exactly; substituting averages taken from other molecules generally gives only an estimate.

Shared bias in an interlaboratory consensus occurs when a common systematic error shifts several laboratories' results together. Their interlaboratory consensus value can then be highly reproducible while displaced from an independent reference. Shared calibration standards, a common correction or the same inadequately modeled measurement step can create such dependence. The GUM explicitly warns that using the same instrument, standard or reference datum can correlate input quantities.^[ref]

A simple original model iswhere $\theta$ is the target quantity value, $b$ a common fixed offset and ${\epsilon }_i$ a laboratory-specific deviation. The arithmetic mean is $\bar{x}=\theta +b+\bar{\epsilon }$. Increasing the number of independent laboratories reduces the fluctuations in $\bar{\epsilon }$ but leaves $b$ in place.

The obstacle is also one of identification. For any constant $c$, replacing $(\theta ,b)$ with $(\theta +c,b-c)$ leaves all predicted results unchanged. Pairwise differences $x_i-x_j$ cancel both terms. Thus the comparison's internal agreement cannot determine the absolute common offset without additional information.

To represent residual common-reference uncertainty, use an uncertain zero-centered quantity $B$ with standard uncertainty $u_B$, independent of mutually independent laboratory errors of variance ${\sigma }^2$. ThenThis is a stated uncertainty model, distinct from asserting that an unknown fixed bias has been observed to vary randomly. The common contribution is counted once in the mean's uncertainty; it is not divided by the number of laboratories.

For stipulated $u_B=0.50$, $\sigma =0.20$ and $n=25$, treating everything as independent would retain only $0.20/5=0.04$. The stated correlated model instead gives $u(\bar{x})=\sqrt{0.25+0.0016}\approx 0.502$, in the same units as the measurement. The appropriate covariance terms prevent the implausible conclusion that enough copies of a shared reference can remove its uncertainty.^[ref]

An independently characterized certified reference material, a different measurement principle or an independently calibrated reference can help test the suspected common shift. Independence must itself be checked: a nominally separate reference derived from the same calibration chain may retain the same dependency. Recognized biases should be corrected where appropriate, with the correction's uncertainty retained.

Social group size is the number of individuals assigned to a social group under a stated membership rule and observation period. A headcount at one moment, a count of regular members and the number present during an experiment can differ. Report which quantity was measured before comparing groups.

The way groups are sampled changes their apparent typical size. Consider two groups containing 2 and 8 animals. Their arithmetic mean size is $(2+8)/2=5$. If an animal is instead sampled uniformly from the ten individuals, the expected size of its group isMore animals experience the larger group. For group sizes $n_1,\dots ,n_k$, this individual-weighted mean group size is ${\sum }_i{n}_i^2/{\sum }_i{n}_i$, rather than ${\sum }_i{n}_i/k$. These are original illustrative calculations, not measured animal data.

Social group size also affects the interpretation of replication. A finch experiment used 70 Bengalese finches in seven groups and 58 white-rumped munias in six groups. The groups contained ten birds each except one eight-bird munia group. Those totals describe birds and groups, not interchangeable counts of independent treatment assignments.^[ref] The experimental unit must follow the design; a larger headcount alone does not solve pseudoreplication.

A speech timing lag is the time difference between corresponding events in two speech recordings. Define ${\Delta }_i=t_{B,i}-t_{A,i}$ for landmark $i$. Positive values mean speaker B's event occurs later; negative values mean B leads. The landmark might be a specified syllable onset or phrase boundary, so its definition is part of the measurement.

In synchronous speech experiments, Fred Cummins recorded paired readings and compared corresponding events. Typical differences were about $40\mathrm{ms}$ on average, with larger mismatches at phrase beginnings.^[ref]

Different summaries answer different questions. For illustrative lags $10,20,-30\mathrm{ms}$, the arithmetic mean is zero, the mean absolute value is $20\mathrm{ms}$, and the root mean square isZero average leadership therefore does not imply close alignment at every event. The square gives larger mismatches more influence on the final statistic.

Recording delays also matter. If channel B adds a known $12\mathrm{ms}$ delay relative to channel A, then ${\Delta }_{\mathrm{recorded}}={\Delta }_{\mathrm{speech}}+12\mathrm{ms}$ under this sign convention. Subtracting that calibrated offset separates equipment delay from speaker timing. Consistent landmark annotation and synchronized recording are consequently necessary for interpreting the measured lag.

Every finite Markov chain has a stationary distribution, without requiring Markov chain irreducibility or aperiodicity. For a row-stochastic matrix $P$, put $B=P^T$ and choose a column probability vector $q$. Every $B^{j}q$ lies in the probability simplex. Their arithmetic averages $x_N=N^{-1}{\sum }_{j=0}^{N-1}{B}^{j}q$ therefore lie in this compact set, while telescoping givesA convergent subsequence has a limit $x$ in the probability simplex. Continuity of matrix multiplication gives $Bx=x$, so $\pi =x^T$ is a stationary distribution. The individual iterates need not converge: a deterministic two-state swap alternates its point masses forever.

Synchronous speech is a task in which speakers try to produce the same speech together in time. In paired reading, each speaker can hear a live partner and adjust the ongoing utterance. This provides an experimental form of joint speech with a specified text and measurable timing.^[ref]

In a 2003 study, Fred Cummins recorded 27 pairs under different conditions. Corresponding points in the two speech channels were typically about $40\mathrm{ms}$ apart on average; larger differences occurred at phrase beginnings. The study compared practice and access to visual information and found reciprocal timing adjustment relative to individual readings.^[ref]

A constructed measurement example shows why the statistic matters. Let a speech timing lag be ${\Delta }_i=t_{B,i}-t_{A,i}$ at landmark $i$. Offsets of $-30,+30,-30,+30\mathrm{ms}$ have an arithmetic mean of zero, although no landmark is simultaneous. Their mean absolute value is $30\mathrm{ms}$. Signed lag measures which voice leads; absolute lag measures temporal mismatch. Keeping both prevents alternating leadership from being mistaken for perfect alignment.

The measured landmark and aggregation rule are therefore part of the result. Close agreement at phrase endings does not imply equally close agreement at every consonant or phrase beginning.

A type species is the species serving as the nomenclatural type of a genus under a naming code that uses this form of typification. It anchors application of the genus name. It is not necessarily an average, an ancestor or the most familiar member of the genus.

In prokaryotic nomenclature, the Code explicitly identifies the type of a genus or subgenus with a type species and specifies how that species is fixed. If a genus originally included only one species when validly published, that species is its type even without an explicit designation. These are naming rules, not measurements of evolutionary distance.^[ref]

An original hypothetical example separates the reference from a summary statistic. Suppose a genus contains species T, U and V, with T the designated type species. A measured trait has values1,10 and11, whose arithmetic mean is $22/3\approx 7.33$. U lies nearer that mean, but this does not make U the type. The designation and the statistical comparison answer different questions.

Now suppose new evidence from phylogenetics divides the hypothetical group into T+U and V. The existing reference is associated with the group containing T; how the remaining group is named requires the relevant naming rules and publication history. Merely discovering that V has an unusual trait does not transfer the reference to it.

The anchor has another level beneath it. In the ordinary prokaryotic case a species itself has a type strain, whereas the genus has a type species. A stored culture accession, a species reference and a genus reference therefore do not all denote the same rank or the same physical object.^[ref]

A type species stabilizes the connection between a name and its reference while taxonomy remains open to revised evidence. The detailed code governing the organisms concerned must be stated; one nomenclatural example should not silently impose every prokaryotic rule on other branches of biology.

A unimodal distribution, under the density or mass-function convention, has probabilities that rise toward one central region and then fall away. For an absolutely continuous probability distribution, this means it has a probability density function that is nondecreasing up to a mode and nonincreasing afterward. This is the single-maximum meaning of unimodality; the chosen density is a one-dimensional quasiconcave function. A plateau of maximizing points is allowed.

For a discrete original example on outcomes0,1,2and3, choose probabilities0.1,0.4,0.4and0.1. The probability mass function rises, has two adjacent maximizing values, then falls. It is unimodal under the weak monotonicity definition despite not having a unique mode. In contrast, separated high-probability regions with a genuine dip between them are not one mode plateau.

Unimodality does not imply symmetry or equality of arithmetic mean, median and mode. For the exponential distribution with density $p(x)=\lambda e^{-\lambda x}$ on $x\ge 0$, $\lambda >0$, the mode is0, the mean is $1/\lambda$ and the median is $\log (2)/\lambda$. The density decreases from its boundary maximum. The NIST handbook gives the family and these summary characteristics, using a scale parameter reciprocal to this rate parameter.^[ref]

Nor is unimodality generally preserved by a mixture distribution. Equal mixing of uniform distributions on $[0,1]$ and $[3,4]$ gives density1/2 on those two intervals and zero in between. Each component has a flat unimodal density, while the mixture has separated peaks.

A density is defined only up to changes on sets of measure zero. Thus the distributional statement concerns existence of a representative with the required shape, rather than artificial isolated spikes added to one displayed formula. In higher dimensions, several notions of unimodality are used; one convex-level-set formulation requires convex superlevel sets, as in the quasiconcave function definition.^[ref] The shape convention should be specified before extending a one-dimensional conclusion to another setting.

For $n$ pots per treatment and $m$ leaves per pot, the group sample mean decomposes into two averages:The variance of a sum adds the variances of the independent terms. Every pot random effect has weight $1/n$, whereas each remaining error has weight $1/(nm)$, giving the short proofThis is also the variance of the arithmetic mean of the $n$ pot sample means, each with variance ${\tau }^2+{\sigma }^2/m$. Equal numbers of leaves per pot make the two averaging procedures identical.

If all $nm$ leaf measurements were falsely treated as independent, the predicted variance would instead be $({\tau }^2+{\sigma }^2)/(nm)$. The ratio of correct to false variance isThus pseudoreplication can underestimate a standard error by the square root of this ratio. Equality holds when $m=1$ or ${\tau }^2=0$; sharing a pot creates an error here precisely when the shared influence has positive variance.

Choose ${\tau }^2=4$, ${\sigma }^2=1$, $n=4$ and $m=3$ as constructed model parameters in squared length units. These are assumptions rather than fitted experimental values. For each group,The treatment contrast $\hat{\Delta }={\overline{Y}}_B-{\overline{Y}}_A$ has expected value ${\alpha }_B-{\alpha }_A$. Since the group errors are independent conditional on the allocation, its variance is twice the group variance, namely $13/6$, whereas the false calculation gives $5/6$. Its correct standard error is $\sqrt{13/6}\approx 1.472$ length units; the false value is $\sqrt{5/6}\approx 0.913$. The variance ratio is $13/5=2.6$.

With four pots fixed, increasing $m$ drives the error contribution ${\sigma }^2/(4m)$ towards zero, but the limit of the group variance remains ${\tau }^2/4=1$. Measuring twelve leaves on each of four pots gives $49/48$. Using the same forty-eight measurements on sixteen pots, with three leaves per pot, gives $13/48$. More independent experimental units reduce both contributions. Extra nested measurements are useful for characterizing each experimental unit, but cannot average away its shared random effect. This is why sample collection effort and the number of independent treatment replicates answer different questions.^[ref]

For nonnegative real numbers $a_i$ and positive weights $w_i$ with ${\sum }_i{w}_i=1$, the weighted geometric mean does not exceed the weighted arithmetic mean:For positive $a_i$, this follows from Jensen's inequality and the concavity of the natural logarithm; a limit extends it to zero inputs. For two terms it gives $a^s{b}^{1-s}\le sa+(1-s)b$ when $0<s<1$.

Linearizing weakly coupled locked phases gives $0=\xi -\Omega 1-kL\theta$, where $L$ is the directed graph Laplacian. Multiplying by a normalized left null vector ${\pi }^{\mathsf{T}}$ eliminates coupling and gives the displayed weighted frequency. In a balanced network, including symmetric reciprocal coupling, $\pi =1/N$ and the frequency is the arithmetic mean. A zero diagonal in the coupling matrix alone does not imply balance. For two directed weights $p,q>0$, the weights are $(q,p)/(p+q)$.