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The arithmetic mean of memory access latencies. Suppose each read completes before the next begins, every cache hit costs $t_h$, and every cache miss costs $t_h+p$, with the same additional cache miss penalty $p$. If $h$ is cache hit rate and $m=1-h$ is cache miss rate, the weighted arithmetic mean isThe common lookup cost is paid on both outcomes. This additive assumption matches the simple model in Beyond Physical Memory: Policies (§22.1, equation 22.1, pp. 1–2), where main computer memory acts as a memory cache for memory pages backed by disk; here the same accounting is applied to a toy data memory cache.

Assign synthetic values $t_h=2\mathrm{ns}$ and $p=30\mathrm{ns}$. The twelve-read, two-slot fully associative cache example has five cache hits and seven cache misses, givingWith four cache slots, its four compulsory misses instead give $\mathrm{AMAT}=12\mathrm{ns}$. This comparison holds the timing costs fixed as capacity changes, which is a model assumption rather than a hardware promise.

If a problem supplies a total miss time $t_m$, use $h{t}_h+m{t}_m$; adding $t_h$ to $t_m$ again would count lookup twice. Variable cache miss penalties, multiple consulted levels, or overlapping requests require a richer timing model. Real CPU work can overlap a pending read, so this arithmetic mean cannot by itself predict program speed or stall time. ^[ref]

An expected value is the probability-weighted arithmetic mean of a real-valued random variable. On a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, its general definition is the Lebesgue integralwhere ${\mu }_X$ is the probability distribution of $X$. For an integrable random variable, this is a finite real number. With a probability mass function $p_X$, use ${\sum }_{x}x{p}_X(x)$; with a probability density function $f_X$, use ${\int }_{\mathbb{R}}x{f}_X(x)dx$. These are two ways of computing the same weighted average, rather than different definitions for different experiments.

For example, suppose $X$ takes the values $-1,2,8$ with probabilities $1/2,3/10,1/5$. ThenThe most likely outcome is $-1$, and $1.7$ is not a possible outcome at all. The expected value summarizes all outcomes and their weights; it need not predict the result of one trial. Geometrically, placing masses equal to the probability masses at the outcome positions gives a center of mass at $1.7$, since ${\sum }_x{p}_X(x)(x-1.7)=0$. This balance-point interpretation also explains the connection to a weighted arithmetic mean. ^[ref].

For the dimensionless random variable $Y$ with probability density function $f_Y(y)=2y$ on $[0,1]$, the expected value is $\mathbb{E}[Y]={\int }_0^{1}y(2y)dy=2/3$. Integrating the probability density function itself gives one; to compute the expected value, multiply by the outcome first. ^[ref].

Why is this useful? Linearity of expectation lets us compute an average total by adding average contributions, even for dependent random variables. And for independent and identically distributed integrable random variables, the strong law of large numbers says their sample mean approaches their common expected value almost surely. The theorem supplies the link to repeated observations; the weighted-average definition itself does not promise that any particular finite run will be close to the expected value.

There is a useful historical connection to valuing uncertain rewards. In his 1657 De ratiociniis in ludo aleae, Christiaan Huygens assigns the value $(a+b)/2$ to equally likely rewards $a,b$, and $(pa+qb)/(p+q)$ when they have $p,q$ equally likely chances. These are the finite weighted-average formulas for expected value. His opening discussion asks how to divide the stakes when a game stops before completion, connecting probability to a concrete question about an unfinished contest. ^[ref].

An isotopic abundance is an atom fraction within a specified population of one chemical element.^[ref] If $n_i$ counts its chemical atoms of isotope $i$, useThe denominator includes every isotope of that chemical element included in the stated population, not every chemical atom of every chemical element in the sample. Thus an oxygen atom fraction among all chemical elements is a different quantity.

For an illustrative neutral neon population, take the displayed central isotopic abundances $f_{20}=0.9048$, $f_{21}=0.0027$, and $f_{22}=0.0925$ from the reference table, so $0.9048+0.0027+0.0925=1$. Its mean atomic mass is the weighted arithmetic meanThese are rounded reference inputs with uncertainties, not exact universal isotopic abundances; the long decimal shows their arithmetic result, not measurement precision.^[ref]

The population convention matters. If instead $w_i$ is the fraction of total neutral-atom mass supplied by isotope $i$, then $w_i=f_i{m}_i/\bar{m}$ andThis is a weighted harmonic mean in the mass fractions. Using ${\sum }_i{w}_i{m}_i$ with those mass fractions would answer a different question. The conversion follows directly by counting chemical atoms and adding their masses.

Mixing calorimetry determines heat capacities by bringing bodies at different temperatures into thermal contact. The striking feature is that the final temperature is a weighted arithmetic mean with heat capacities as weights, rather than weights given by mass. The vessel participates in the exchange even though it is not the sample being studied.

For a hot nonreacting sample of mass $m_x$, initial temperature $T_x$ and unknown specific heat capacity at constant pressure $c_x$, let receiving water have mass $m_w$, initial temperature $T_w$ and known specific heat capacity at constant pressure $c_w$. Let the vessel and immersed parts start at $T_w$ and have combined heat capacity $C_{\mathrm{cal}}$, excluding that water. Assume a common final temperature $T_f$, approximately constant pressure and heat capacities, no phase transition, no chemical reaction and negligible external nonexpansion work. For signed outward heat loss $Q_{\mathrm{loss},\mathrm{out}}$, the combined enthalpy balance isHenceThe loss is positive if the room receives net heat, negative if the room supplies it. If the vessel starts at a different temperature $T_{\mathrm{cal},i}$, replace its warming term by $C_{\mathrm{cal}}(T_f-T_{\mathrm{cal},i})$; do not assume the vessel shared the water's initial temperature. For variable material properties, replace each $mc\Delta T$ by the corresponding integral $m\int c_p(T)dT$ and each vessel term by $\int C_{\mathrm{cal}}(T)dT$. The inferred constant $c_x$ then represents an average over the sample's cooling interval.^[ref]

For a synthetic no-loss example, mix $0.100\mathrm{kg}$ of hot water at $80{}^{\circ }\mathrm{C}$ with $0.200\mathrm{kg}$ of cold water at $20{}^{\circ }\mathrm{C}$ in a vessel initially at $20{}^{\circ }\mathrm{C}$. Stipulate $c_w=4180\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$ and $C_{\mathrm{cal}}=100\mathrm{J}{\mathrm{K}}^{-1}$. The enthalpy balance givesThe hot water releases $17.337\mathrm{kJ}$: the cold water receives $15.485\mathrm{kJ}$ and the vessel $1.852\mathrm{kJ}$. Omitting the vessel predicts $40{}^{\circ }\mathrm{C}$, despite perfect insulation. These are computed numbers from stipulated properties, not laboratory observations.

For an unknown-sample example using the same cold water and vessel, stipulate $m_x=0.100\mathrm{kg}$, $T_x=80{}^{\circ }\mathrm{C}$ and $c_x=450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With no loss, $T_f\approx 22.752{}^{\circ }\mathrm{C}$. Inverting the full balance returns $450\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$; ignoring vessel storage returns about $402\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{K}}^{-1}$. With a positive outward loss, ignoring that loss also biases the inferred $c_x$ downward under these hot-sample conditions. A small water temperature rise makes sensor offsets and loss estimates relatively influential; propagation of uncertainty must include the measured temperatures, masses and calibrated vessel term.

Dividing two extensive quantities with the same replication factor produces an intensive quantity:Division by volume gives a density, such as mass density $\rho =m/V$ or energy density $U/V$. Division by mass gives a specific quantity, such as specific internal energy $u=U/m$ or specific volume $v=V/m$. Division by amount of substance gives a molar quantity, such as molar volume $V_m=V/n$. Their units are respectively $\mathrm{kg}{\mathrm{m}}^{-3}$, $\mathrm{J}{\mathrm{m}}^{-3}$, $\mathrm{J}\mathrm{k}{\mathrm{g}}^{-1}$, ${\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}$ and ${\mathrm{m}}^3\mathrm{mo}{\mathrm{l}}^{-1}$. The denominator must be stated: “per particle”, “per mole” and “per unit mass” are different normalizations.

For a hypothetical uniform parcel with mass $2\mathrm{kg}$, volume $0.002{\mathrm{m}}^3$ and internal energy $40\mathrm{kJ}$ relative to a specified reference, $u=20\mathrm{kJ}\mathrm{k}{\mathrm{g}}^{-1}$ and $\rho =1000\mathrm{kg}{\mathrm{m}}^{-3}$. Three independent copies have $6\mathrm{kg}$, $0.006{\mathrm{m}}^3$ and $120\mathrm{kJ}$; both ratios remain unchanged. This permits comparison of materials without confusing the amount tested with the state tested.

If $X$ and $Y$ are additive but the parcels have different $r_i=X_i/Y_i$, their composite ratio is a weighted arithmetic mean:The denominator supplies the weights: specific internal energy uses mass weights, while mass density uses volume weights. This identity is accounting for the specified subsystems; predicting a new state after mixing also needs the relevant equation of state.

This deliberately synthetic differential equation model illustrates response direction and lag. Every plotted number is chosen for teaching; none is a biological measurement or a fitted human parameter. Let externally prescribed, dimensionless $g>0$ represent a glucose stimulus relative to a reference. Let $I$ and $H$ be normalized insulin and glucagon response variables, with independently chosen time constants ${\tau }_I=4$ and ${\tau }_H=6$ in arbitrary time units:Here $I_{\infty }$ rises and $H_{\infty }$ falls with $g$. Their targets add to one by construction; the evolving $I+H$ need not equal one because their time constants differ. Start with $g=1$ and $I=H=1/2$. Set $g=1.6$ from $t=10$ to $40$, $g=0.65$ from $t=40$ to $70$, and $g=1$ otherwise. Within a constant-input interval beginning at $t_0$, the exponential decay solution isEach response is a continuous function of time even when the imposed stimulus jumps. The chosen starting values stay between zero and one, since each solution is a weighted arithmetic mean of its interval's starting value and target.

For an illustrative downstream comparison, define release tendency $R=0.3+0.6H/(1+I)$ and uptake tendency $U=(0.25+0.5I)g$, in the same arbitrary flux units. At the reference target, $R=U=0.5$. At the high-stimulus targets, $R\approx 0.343$ and $U\approx 1.094$; at the low-stimulus targets, $R\approx 0.742$ and $U\approx 0.212$. These chosen functions encode greater uptake with insulin and greater release with glucagon, moderated by insulin; they are not established enzyme kinetics. The $g$ trace is imposed externally, so the figure does not demonstrate self-correction of blood glucose. A closed model would also need a conservation of mass equation, nutrient input, finite glycogen stores, and additional tissue interactions. The qualitative restoring loops are shown separately in the insulin circuit.

For a real random variable $X$, an upper-tail probability has the form $\mathbb{P}(X\ge t)$ and a lower-tail probability has the form $\mathbb{P}(X\le t)$. Each is the probability of an event lying beyond a chosen threshold in one direction. A tail probability need not be small: the threshold determines how much of the probability distribution it includes. A concentration inequality bounds such probabilities, often for deviations from an expected value.

Endpoint conventions matter. Write $F(t)=\mathbb{P}(X\le t)$ for the cumulative distribution function and $F(t^{-})={\lim }_{s\uparrow t}F(s)=\mathbb{P}(X<t)$. The complement rule of probability gives four distinct formulas:The strict upper-tail convention is the survival function $S(t)=\mathbb{P}(X>t)$. Its value differs from the inclusive upper tail probability by exactly the probability mass at the threshold:A singleton with positive probability mass is an atom of a measure. Thus $1-F(t)$ excludes an atom of a measure at $t$, whereas $1-F(t^{-})$ includes it. These endpoint identities follow directly from the cumulative distribution function properties in Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails.

As the threshold increases, both upper tail probabilities are nonincreasing functions. The strict survival function is a right-continuous function; the inclusive upper tail probability takes the upper endpoint value at a downward jump, equal to the left limit of the survival function. Both tend to zero as $t\to \infty$ for a real random variable, but the speed of that decay differs across probability distributions. The cumulative distribution function properties and their proof from continuity of probability measures are given in MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties.

For a fair six-sided die $D$, the discrete uniform distribution assigns probability mass $1/6$ to each of $1,\dots ,6$. At threshold four,Between four and five, both upper tail probabilities equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb{P}(D\ge 6)=1/6$, but $\mathbb{P}(D>6)=0$. The diagram shows exact probabilities for this mathematical model; its dots specify which endpoints belong to each step function.

If a probability distribution has a probability density function $f$, then every singleton has zero probability mass andFor an exponential distribution of rate two, $f(x)=2e^{-2x}$ for $x\ge 0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\ge 0$ and $S(t)=1$ for $t<0$. Hence $\mathbb{P}(X\ge 1)=e^{-2}\approx 0.135335$. This tail probability is an integral of the probability density function, not its height $f(1)=2e^{-2}\approx 0.270671$. A mixed probability distribution retains endpoint jumps even with a smooth component: give zero probability mass $1/4$ and choose this exponential distribution with probability $3/4$. Then $\mathbb{P}(X\ge 0)=1$, $\mathbb{P}(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\ge 0$.

A two-sided tail probability can measure the absolute value of displacement from a specified center $c$. For $a>0$, the two extreme events are disjoint events, soFor strict displacement $|X-c|>a$, replace the last expression by $F((c-a{)}^{-})+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any probability mass at $c$. For a uniform distribution on $[-2,2]$, take $c=0$ and $a=1$: the two tail probabilities are $1/4$ each and their sum is $1/2$. Equal tails follow from this symmetric probability distribution, not from the phrase two-sided alone; for a uniform distribution on $[0,4]$ with $c=1,a=1$, the two tail probabilities are zero and $1/2$.

The tail-sum formula for expectation turns a whole curve of tail probabilities into an expected value. For any nonnegative random variable $Y$, without assuming a probability density function or a finite expected value,For a proof outline, write $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise and exchange the expected value and Lebesgue integral using Tonelli theorem; $1$ is an indicator function. The inclusive version has the same Lebesgue integral: positive probability masses occur at only a countable set of thresholds, which has Lebesgue measure zero. For a nonnegative integer-valued random variable $N$, the tail-sum formula instead gives $\mathbb{E}N={\sum }_{k=1}^{\infty }\mathbb{P}(N\ge k)={\sum }_{k=0}^{\infty }\mathbb{P}(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its indicator function proof, and the integer version appear in Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails.

For the fair die, this tail-sum formula yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct weighted arithmetic mean $(1+2+3+4+5+6)/6$. For the rate-two exponential distribution, the area under the survival function is ${\int }_0^{\infty }{e}^{-2t}dt=1/2$, its expected value. The first panel below shades an integral of the probability density function for a single threshold; the second shades an integral of the survival function across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.

Nonnegativity is essential to interpreting the upper tail probability area as $\mathbb{E}Y$: for a general real random variable that area equals the expected value of its positive part. An integrable random variable satisfies $\mathbb{E}X=\mathbb{E}{X}^{+}-\mathbb{E}{X}^{-}$, subtracting the expected value of its negative part; infinite terms cannot be canceled. Even when $Y$ is finite almost surely, its expected value can be infinite. For the shape-one Pareto distribution on $[1,\infty )$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\ge 1$, so the area is $1+{\int }_1^{\infty }dt/t=\infty$. The finite versus infinite expected value distinction is illustrated in Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values.

Why are tail probabilities useful? They answer threshold questions directly, and a concentration inequality can bound them without knowing the full probability distribution. For nonnegative $Y$ with finite expected value, Markov inequality gives $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$ for $t>0$: a large threshold relative to the expected value forces rarity. The quantile function reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha )$ with $0<\alpha <1$, then $\mathbb{P}(X>q)\le \alpha \le \mathbb{P}(X\ge q)$; an atom of a measure may prevent equality. For the die, $Q(0.9)=6$, with strict upper tail probability zero and inclusive upper tail probability $1/6$.

A tail probability describes rarity under a specified probability distribution; it is not automatically a conditional probability after extra information. For $u\ge t$ with $\mathbb{P}(X\ge t)>0$,because the threshold events are nested. Thus the die has $\mathbb{P}(D\ge 5)=1/3$ but $\mathbb{P}(D\ge 5\mid D\ge 4)=2/3$. Conditioning changes the sample space and renormalizes its probability measure; the same extreme outcomes can become common in that restricted population.

The weighted arithmetic mean combines values according to how much each contributes to the quantity being summarized. For a finite list of real numbers $x_i$ and nonnegative weights $w_i$ with $W={\sum }_i{w}_i>0$, defineZero weights ignore an entry; at least one weight must be positive. Multiplying every weight by the same positive number leaves the result unchanged. Writing $p_i=w_i/W$ makes the weighted mean a convex combination, since $p_i\ge 0$ and ${\sum }_i{p}_i=1$. It therefore lies between the smallest and largest input values. ^[ref]

For example, give the values $90$ and $60$ weights $2$ and $8$:The larger weight pulls the result closer to $60$. The balancing identityexplains a geometric interpretation: if the weights were masses placed at positions $x_i$, the weighted mean would be their center of mass. Here $8(66-60)=2(90-66)=48$, so the two sides balance even though their distances from the mean differ.

Weights need an interpretation. They can represent numbers of observations, masses, or probabilities in a finite probability distribution. These meanings lead to the same formula, while answering different questions. For instance, combining means of groups uses group sizes as weights to recover the mean over individual observations. For a function $f$ and nonnegative weight function $w$, the analogous integral is $\int wf/\int w$, provided $wf$ is integrable and the total weight $\int w$ is positive and finite.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted geometric mean isIt takes a weighted mean of logarithms. A zero weight excludes that input; equal weights give the ordinary geometric mean. If $x_i$ occurs $m_i$ times in a list of total length $M$, its correct weight is $w_i=m_i/M$. For the list $(12,3,3,3)$,and replacing all four entries by $G_w$ preserves the product $324$.

There is also an optimization interpretation. A logarithmic transformation turns multiplicative discrepancies into additive ones. For a positive representative $c$, the weighted mean minimizes squared deviations identity givesThus $c=G_w$ uniquely minimizes weighted quadratic loss in logarithmic coordinates. The weighted arithmetic mean minimizes squared discrepancies in the original coordinates. These two objectives explain why different choices of what constitutes an error select different summaries.

The strictly concave function $\log x$ and the reversed Jensen's inequality give $G_w\le A_w$, where $A_w={\sum }_i{w}_i{x}_i$ is the weighted arithmetic mean. Apply the same reasoning to the multiplicative inverses to get $H_w\le G_w$ for the weighted harmonic mean. Thus $H_w\le G_w\le A_w$; equality holds exactly when all positive-weight inputs agree. For this list, $H_w=48/13$, $G_w=3\sqrt{2}$, and $A_w=21/4$. The weights specify which inputs are represented and must stay the same across the comparison.

For positive real numbers $x_i$ and nonnegative weights with ${\sum }_i{w}_i=1$, the weighted harmonic mean isEqual weights give the harmonic mean, while a zero weight excludes that input. For travel at positive constant speeds $v_i$ over positive leg lengths $d_i$, the weights are fractions of total distance:For $30\mathrm{km}$ at $30km/h$ and $90\mathrm{km}$ at $60km/h$, the weights are $1/4,3/4$, elapsed time is $5/2$ hours, and the average speed is $48km/h$. Weighting the speeds by their time fractions instead gives the same result as a weighted arithmetic mean: $(2/5)30+(3/5)60=48$. The two weight systems describe the same journey in different coordinates and must not be interchanged. ^[ref].