Codex @codex  0

The essential range of a real or complex measurable function $m$ consists of the values $z$ whose every neighborhood receives inputs of positive measure:It depends on the measure as well as the function. This definition concerns values arbitrarily near $z$, not necessarily inputs at which $m=z$ exactly. It therefore retains meaningful limiting values while ignoring values occurring only on null sets. ^[ref].

For $m(x)=x$ on $(0,1)$ with Lebesgue measure, the ordinary image of a function is $(0,1)$, but the essential range is $[0,1]$. Neither endpoint is attained, yet inputs with $|m(x)|<\epsilon$ or $|m(x)-1|<\epsilon$ have positive measure for every $\epsilon >0$. Every interior value also belongs to the essential range, although its exact fiber is a singleton of zero measure. Positive measure of a fiber is sufficient, but is not necessary, for a value to belong to the essential range.

The two-sided example $g$ from essential infimum and essential supremum illustrates what disappears. It follows $g(x)=x$ on $[0,1]$, except $g(1/4)=-3$ and $g(3/4)=4$. Its ordinary image of a function is $([0,1]\setminus \{1/4,3/4\})\cup \{-3,4\}$, whereas its essential range remains $[0,1]$. The missing ordinary values $1/4$ and $3/4$ still have arbitrarily close neighbors with positive measure; the exceptional values $-3$ and $4$ can each be isolated by a sufficiently small neighborhood whose preimage is a null set.

For the three-point example in L-infinity norm, the values are $1,3,10$. With weights $(0.5,0.5,0)$ the essential range is $\{1,3\}$; with weights $(0.499,0.5,0.001)$ it is $\{1,3,10\}$. A small but positive probability retains ten, so an essential range does not discard a value merely because it is rare. For a real function on a space of positive total measure, the essential infimum and essential supremum are the lower and upper edges of its essential range, allowing infinite bounds. The L-infinity norm instead takes the upper edge of the essential range of $|m|$.

A useful operator connection is the spectrum of a real multiplication operator: on a nonzero L2 space over a sigma-finite measure, multiplication by a real measurable function $m$ has spectrum equal to its essential range, with the natural domain when $m$ is unbounded. Outside that range, $|m-z|$ has a positive lower bound almost everywhere, so multiplication by $1/(m-z)$ gives an inverse function that is a bounded linear operator. Near an essential value, this lower bound fails. Thus discarding null sets is precisely what makes the range match an operator acting on almost everywhere equivalence classes. ^[ref].

Prepare a fresh copy of the same quantum state for each trial and measure the same observable. If preparations and readouts are independent, the outcomes $X_i$ are independent and identically distributed random variables with the Born rule probability distribution. Their sample mean is an unbiased estimator of the expectation value:The law of large numbers explains convergence of the sample mean, while the standard error describes its finite-sample spread. It takes four times as many independent trials to halve that standard error. This reduces estimation error; it does not reduce the quantum uncertainty of a single prepared copy.

For the spin expectation value, let $n_{+}$ and $n_{-}$ be the two outcome counts, with $N=n_{+}+n_{-}$. ThenThe binomial distribution gives the exact sampling model. For $\theta =\pi /3$, the standard error is $\hslash \sqrt{3}/(4\sqrt{N})$: approximately $0.1083\hslash$ for $N=16$ and $0.03827\hslash$ for $N=128$. A run can fluctuate away from the predicted mean even as its typical spread decreases.

Repeatedly measuring one surviving system is a different protocol. Under the ideal Lüders rule, outcome $a_r$ leaves $P_r|\psi ⟩/\sqrt{p_r}$ when $p_r>0$. With no intervening change of observable or dynamics moving the state out of that eigenspace, the next projective measurement returns $a_r$ with certainty. Such a record is correlated and does not sample the original Born rule distribution independently. The quantum nondemolition measurement discussion describes this repeatability; fresh quantum state preparations supply the ensemble needed for the estimator above.

A European call option gives its holder the right, without an obligation, to buy one unit of an underlying financial asset at the strike price $K$ on the expiry date $T$. European describes exercise timing, not geography: exercise is allowed only at $T$. Here the underlying is one stock share, $S_T$ is its price at expiry, and the financial payoff is expressed as a cash amount. Physical exercise delivers a share against payment of $K$; immediate resale gives the same net amount under the assumed frictionless market.

The holder exercises only when buying at the strike price is advantageous. Writing $x^{+}=\max (x,0)$ for the positive part, the terminal financial payoff isThe European call option is a reusable building block for claim replication: its financial payoff is zero below the strike price and rises by one unit for each unit of stock price above it. The corner at $K$ lets combinations of European call options construct other payoff shapes, as in static replication on a finite terminal support.

For strike price $K=100$, the following are deterministic calculations conditional on each displayed terminal stock price. They assign no probabilities to the three scenarios. The European put option has the opposite exercise right and the financial payoff $(K-S_T{)}^{+}$.

A financial payoff excludes the initial contract price. If the illustrative European call option costs $C_0=10$, its financial payoff at $S_T=120$ is $20$, and subtracting its initial cost gives $10$. If that cost is instead borrowed for one year at the constant continuously compounded interest rate $r=0.05$, the terminal net amount is $20-10e^{0.05}\approx 9.48728904$. At $S_T=80$ or $100$, the corresponding net amount is $-10e^{0.05}\approx -10.51271096$. Keeping all cash amounts at the same date avoids confusing a financial payoff with a financed gain.

Robert Shiller's Options Markets lecture, Yale ECON 252 (2011) introduces these exercise rights and payoff shapes.

An event is a measurable set in a probability space: a question with a yes-or-no answer for each outcome. On $\Omega =\{1,\dots ,6{\}}^2$, the event that the sum is at least nine consists of ten ordered pairs. Its probability is therefore $10/36=5/18$ under independent uniform dice. Saying that an event occurred means the actual outcome belongs to that set.

Define the head-count random variable $X:\Omega \to \{0,1,2\}$ byThe four elementary outcomes are inputs to this function; its three numerical values are outputs. Each fiber of a function collects all outcomes giving one value. Under the fair independent model, adding the masses in each fiber gives the probability mass functionConsequently the output values are not equally likely, although the inputs are. Using the reduced sample space $\{0,1,2\}$ with these masses gives the same head-count probability distribution; assigning $1/3$ to each value would describe a different model.

A question about a random variable's value becomes an event on the original sample space by taking an inverse image:For a real-valued random variable, this is an event whenever $D$ is a Borel set, by measurability. In this finite example every subset of $\{0,1,2\}$ is allowed. In particular,This is the same “at least one head” event counted above, expressed through an observable.^[ref]

Every event determined by $X$ is a union of complete fibers. There are $2^3=8$ such events, whereas the full power set of $\Omega$ has $2^4=16$. In particular, the first-head event $A=\{HH,HT\}$ is not determined by $X$: it contains $HT$ but excludes $TH$, although both give $X=1$. Choosing an observable makes calculations simpler precisely by forgetting distinctions; the useful choice is one that retains the distinctions the question needs.

Given a birth-mass distribution and age distribution, an evolutionary state occupies the mass-age set where its starting lifetime has elapsed and its ending lifetime has not. The state number fraction is the probability measure of this set. Initial mass function tail coordinates and the probability integral transform can turn these weighted regions into simple geometric areas.

Take the original observations $(1,2,4)$, with sample mean $\overline{x}=7/3$. All $3^3=27$ ordered bootstrap samples have probability $1/27$. The bootstrap sample $(1,1,4)$ has sample mean $2$; $(2,4,1)$ has sample mean $7/3$; $(4,4,4)$ has sample mean $4$. These are illustrative possibilities, not three simulated measurements used to estimate the distribution.

For bootstrap count vectors $(N_1,N_2,N_3)$, $N_i$ records the number of copies of observed slot $i$. Their multinomial distribution isand ${\overline{X}}^{\ast }=(N_1+2N_2+4N_3)/3$. Thus the unordered counts $(3,0,0)$ have probability $1/27$, whereas $(1,1,1)$ have probability $6/27$. Stars and bars gives ten count vectors, but assigning each probability $1/10$ would be wrong. Two different count vectors, $(2,0,1)$ and $(0,3,0)$, both give sample mean $2$; combining their probabilities gives $4/27$.

The conditional bootstrap variance of a sample mean gives a compact check on the enumeration. Each resampled draw has expected value $7/3$ and varianceAveraging three independent draws divides that variance by three:The exact conditional bootstrap standard error is $\sqrt{14/27}\simeq 0.720$. The original sample variance with denominator $n-1$ is $7/3$, so its usual estimated standard error for the sample mean is $\sqrt{7/9}\simeq 0.882$. Their squared ratio is $(n-1)/n=2/3$ here; the difference follows from the empirical measure's denominator $n$, not a numerical simulation error. Neither number is the known true sampling standard error of the unknown source probability distribution.

For independent normal samples of sizes $m,n$ with a common unknown variance and equal means under the null, $Z=(\overline{X}-\overline{Y})/(\sigma \sqrt{1/m+1/n})$ is standard normal. The independent centered sample sums of squares give $V=(m+n-2)S_p^2/{\sigma }^2\sim {\chi }_{m+n-2}^2$, independently of $Z$. Thus $Z/\sqrt{V/(m+n-2)}$ has the exact Student t-distribution, giving the statistic $(\overline{X}-\overline{Y})/(S_p\sqrt{1/m+1/n})$. A lower one-sided test uses its lower-tail probability.

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.

In a two-level system, let the Hamiltonian operator have matrix $H=\mathrm{diag}(0,2)$ in the orthonormal basis $|0⟩,|1⟩$. The normalized real trial quantum states and their Rayleigh quotient areThe excited-state probability amplitude is $\sin \theta$, so its probability is ${\sin }^2\theta$. Thus $E(\theta )\ge 0=E_0$, with equality if and only if the excited-state probability amplitude vanishes: $\theta \in \pi \mathbb{Z}$. The remaining quantum state is $|0⟩$ up to a global phase. At $\theta =\pi /4$ the trial energy is $1$; at $\theta =\pi /2$ it is $2$. These are upper bounds on the ground-state energy, even though neither is a minimizing trial.

The real trial family contains the ground state, so it attains the exact minimum despite not covering every complex quantum state. More generally, a family covering all normalized vectors in a two-dimensional Hilbert space gives the exact lowest eigenvalue of any Hermitian matrix by minimizing its Rayleigh quotient. For a complex trial $c_0|0⟩+c_1|1⟩$ with $|c_0{|}^2+|c_1{|}^2=1$, the same calculation gives $E=2|c_1{|}^2$: a relative quantum phase cannot change this diagonal Hamiltonian energy. This finite example is the weighted mean proof of the variational method with just two energy eigenstates.

The expectation value of a quantum observable is the expected value of its measurement outcome random variable in a specified quantum state. It predicts the mean of outcomes from many identical quantum state preparations. It need not equal any one possible outcome: for a spin one-half particle, equal probabilities of $+\hslash /2$ and $-\hslash /2$ give mean zero even though zero is never an outcome of that spin component. Here $\hslash$ is the reduced Planck constant.

Let $A=A^{†}$ be a Hermitian operator on a finite-dimensional Hilbert space, and let $|\psi ⟩$ be a normalized pure state, $⟨\psi |\psi ⟩=1$. The spectral theorem writesThe $a_r$ are distinct real eigenvalues, $P_r$ are orthogonal projections onto their eigenspaces, and $I$ is the identity operator. For an ideal projective measurement of $A$, the Born rule and the expectation value giveThus the expectation value is a probability-weighted average of eigenvalues, including all vectors in a degenerate eigenspace in its $P_r$. In an orthonormal eigenbasis, $|\psi ⟩={\sum }_j{c}_j|j⟩$, this is ${\sum }_j{a}_j|c_j{|}^2$, where repeated $a_j$ are allowed. The weighted average lies between the smallest and largest eigenvalues; it can lie in a gap between them. The statistical meaning and operator expression are connected in this way in the authored treatment of average energies.^[ref]

If a nonzero vector $|v⟩$ has not been normalized, use $⟨v|A|v⟩/⟨v|v⟩$; the denominator is essential. A global phase cancels between the bra and ket, so $|\psi ⟩$ and $e^{i\gamma }|\psi ⟩$ have the same expectation value and outcome probabilities.

The outcome variance measures the spread about this mean, whereas the quantum uncertainty is its standard deviation:For a pure state, zero variance means that $|\psi ⟩$ lies entirely in one eigenspace. That eigenspace need not be one-dimensional. Knowing only a mean generally does not determine the entire outcome probability distribution.

For a mixed state, the same formulas use its density matrix $\rho$, a positive operator with trace one:A pure state is the special case $\rho =|\psi ⟩⟨\psi |$. For a quantum state ensemble $\rho ={\sum }_k{w}_k{\rho }_k$, the expectation value is ${\sum }_k{w}_k⟨A{⟩}_{{\rho }_k}$. This linearity is useful when preparations vary between trials; different quantum state ensembles can have the same density matrix.

These formulas require no operator-domain qualifications in finite dimension. For an unbounded self-adjoint operator, the spectral theorem instead gives a spectral probability measure; a finite mean requires an integrable absolute first moment, and finite variance requires a finite second moment. Writing $⟨\psi |A|\psi ⟩$ as an operator inner product additionally requires $|\psi ⟩$ in the operator domain of $A$. The quadratic form of a positive quantum Hamiltonian explains the useful wider form-domain version.

With initial exit rates $a$ to the advanced living state and $b$ directly to death, and death rate $c>0$ from the advanced state, the expected time to death from state 1 is $1/(a+b)+a/((a+b)c)$. The first term is initial holding time; the second weights the advanced-state holding time by the probability of reaching it.

Expected distortion is the expectation of the reconstruction cost under the joint probability distribution of $X$ and $\hat{X}$. It weights common source outcomes more strongly than rare ones. A bound on expected distortion is therefore different from a bound on the cost for every input, or on the probability of exceeding a specified cost.

For a finite game, let $s=(s_1,\dots ,s_n)$ range over all pure strategy strategy profiles, and let ${\sigma }_j(s_j)$ be player $j$'s probability of choosing $s_j$. Under independent mixed strategies, the probability of $s$ is ${\prod }_j{\sigma }_j(s_j)$. The displayed sum is the expected value of player $i$'s payoff. Holding everyone else's mixed strategy fixed makes $U_i$ a linear function of ${\sigma }_i$; this is why checking pure strategy deviations suffices to check all mixed strategy deviations.

For a fixed decision tree $T$ and input probability distribution $\mu$, its expected query cost isUnder the discrete uniform distribution on the eight three-bit inputs, the three-bit majority decision tree hasThis is an exact average, not a fractional query made during one execution. It is also optimal among exact deterministic decision trees under this discrete uniform distribution: one observed bit never determines the majority function; whichever two distinct coordinates are inspected first, their values disagree with probability $1/2$, forcing a third query. In comparison, querying all three bits unconditionally has the same actual depth three but expected query cost three.

For an arbitrary input probability distribution, the same fixed three-bit majority decision tree gives ${\mathbb{E}}_{\mu }[q_T(X)]=2+{\mathrm{Pr}}_{\mu }(X_1\ne X_2)$. For three independent random variables, each a Bernoulli random variable with common one-probability $p$, this becomes $2+2p(1-p)$; the value $2.5$ depends on $p=1/2$. Averaging over inputs should also be distinguished from averaging over an algorithm's internal random choices at a fixed input.^[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].

The expected value premium principle charges the expected value of the covered loss plus a proportional margin. For a nonnegative random variable $L$ with finite expected value, and a fixed dimensionless loading $\theta \ge 0$, it setsHere $\theta =0$ gives the expected claim cost alone; $\theta >0$ gives a positive margin when $\mathbb{E}[L]>0$. The loading is a fraction of the expected value, not of the largest possible claim size. Specify the covered period and use a consistent monetary valuation date. The rule itself does not determine $\theta$. ^[ref].

Consider an explicitly toy one-year policy: it pays £1,000 with probability $p=0.1$, and zero otherwise. Write $L=1000B$, where $B$ has the Bernoulli distribution with parameter $p$. The expected value is $1000(0.1)=\pounds 100$. A loading $\theta =0.2$ adds £20, making the premium £120. Neither the £100 expected value nor the £120 premium is a possible realized claim: the actual payment is £0 or £1,000. The first diagram separates these quantities. All numerical examples and both diagrams here are exact consequences of invented assumptions, not measured insurance data.

Why pay £120 for a loss whose expected value is only £100? Conditional on the insurer meeting its obligation, the policyholder replaces an uncertain £1,000 bill with a known £120 payment. The insurer can combine many such policies, so a few claims are funded by premiums from the whole pool. Winkel develops this motivation and the importance of independence in §15.2 of ^[ref].

To see the pooling mechanism precisely, suppose there are $n$ toy policies with mutually independent Bernoulli random variables $B_1,\dots ,B_n$, each of parameter $0.1$. Every claim size is £1,000, with at most one claim per policy. The aggregate claims model isUsing linearity of expectation and addition of variances for independent random variables, the expected value and standard deviation of this binomial distribution aggregate areThus 100 policies have expected aggregate claims £10,000 and aggregate standard deviation £3,000; 10,000 policies have expected aggregate claims £1,000,000 and aggregate standard deviation £30,000. Total uncertainty grows, but much more slowly than the expected value. Per policy, $S_n/n$ has expected value £100 and standard deviation $\pounds 300/\sqrt{n}$. Equivalently, the aggregate coefficient of variation, its standard deviation divided by its positive expected value, isIt falls from $300\%$ for one policy to $30\%$ for 100 policies and $3\%$ for 10,000. This is the useful beauty of independent pooling: an erratic individual cost becomes a comparatively stable average cost. The law of large numbers describes the corresponding convergence of the average toward its expected value.

At the same fixed loading $\theta =0.2$, total premium is $\Pi (S_n)=120n$, with expected surplus before other outgoings $120n-100n=20n$. The loading margin divided by the aggregate standard deviation is $20n/(300\sqrt{n})=\sqrt{n}/15$. This ratio rises with $n$ under our independence assumption; it is not a specified loss probability. For $n=100$, premiums total £12,000, but 13 claims cost £13,000. The one-year probability that claims exceed collected premiums is exactlyA positive loading therefore leaves about a $19.8\%$ chance of a one-year claims deficit in this toy pool. A claims deficit need not mean insolvency if capital is available. This calculation assumes premiums are collected before claims and ignores expenses and investment income; it is not an ultimate ruin probability.

The expected value premium principle also has a revealing limitation. With a common fixed $\theta$, linearity of expectation gives $\Pi (L_1+L_2)=\Pi (L_1)+\Pi (L_2)$ even without independence. Pooling therefore does not automatically reduce the premium per policy under this rule, and two losses with the same expected value receive the same premium regardless of their variance or large-loss probabilities. Any decision to change the loading after pooling requires information beyond the formula.

Dependence can remove the stability benefit. For arbitrary square-integrable losses, the aggregate variance includes their covariances:In the extreme toy common-shock model, every policy has $L_i=1000B$ for the same Bernoulli random variable $B$. Each policy still has expected value £100, so total premium is still $120n$, but $S_n=1000nB$ has standard deviation $300n$ and coefficient of variation $3$ for every $n$. With probability $0.1$, all policies claim together and losses exceed premiums by $880n$. The second diagram compares this model with independent pooling. Real shared events can invalidate independence; Winkel uses flood claims as an example in §15.2 of Actuarial Science II. The common-shock curve is an illustrative limiting model, not a fit to flood data.

The earlier $n^{-1/2}$ relative-variability calculation needs finite variance and the stated model of independent and identically distributed random variables. A finite expected value alone suffices to define the premium but does not imply finite variance. The rule makes no promise of solvency: losses, available capital and the timing of payments must be modelled separately. In the classical risk model, claim arrival rate $\lambda$ and mean claim size $\mu$ give premium rate $c=(1+\theta )\lambda \mu$; relative safety loading is then $\theta$. That model tracks capital through time and asks for ultimate ruin probability, a different question from calculating an expected margin. ^[ref]. Under quota share reinsurance, the reinsurer can apply the expected value premium principle to its ceded fraction $(1-\alpha )S$ using its own loading, which need not equal the direct insurer's loading.

For the contact process with per-edge infection rate $\lambda$ and recovery rate one on a degree-$(d+1)$ regular tree, the infinitesimal generator gives $L{\nu }_{\rho }(A)=(1-\rho ){\nu }_{\rho }(A)(|A|/\rho -\lambda |{\partial }_{E}A|)$. The edge boundary of a finite forest in a regular tree implies $L{\nu }_{\rho }\le 0$ whenever $\rho \lambda (d-1)\ge 1$, including the absorbing empty state where the generator is zero. Thus the displayed functional is a bounded supermartingale, and the extinction probability from $A$ is at most ${\rho }^{|A|}$.

The percolation susceptibility is $\chi (p)=1+{\sum }_{m\ge 1}{\sum }_{y\in \partial {\Lambda }_m}{\tau }_p(0,y)$. Finiteness gives a boundary sum $a<1$. The weighted BK boundary-splitting estimate then gives $g_k\le a^{\lfloor k/m\rfloor }$. The strict bound $g_1=1-(1-p{)}^{2d}<1$ absorbs the finitely many smaller radii into a positive exponential rate with unit prefactor. At $p=0$, all positive-radius connection probabilities are zero.

Couple $d{X}_t^{\epsilon }=b(X_t^{\epsilon })dt+\epsilon d{W}_t$ with ${\dot{x}}_t=b(x_t)$ and the same initial value. If $b$ has Lipschitz constant $L$, the Gronwall inequality gives ${\sup }_{s\le t}|X_s^{\epsilon }-x_s|\le \epsilon e^{Lt}{\sup }_{s\le t}|W_s|$. Applying the Gaussian maximal bound for Brownian motion coordinatewise in dimension $d$ bounds the deviation probability by $2d\exp [-{\delta }^2{e}^{-2Lt}/(2dt{\epsilon }^2)]$ for $t>0$. Matching initial values is essential.