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Let $Z$ index either of two candidate pathways and suppose $P(Y=1\mid Z=1)=P(Y=1\mid Z=2)=p$ for a measured symptom $Y$. For any mixing probability $\pi =P(Z=1)$, marginalization gives $P(Y=1)=\pi p+(1-\pi )p=p$. An independent sample with $k$ positive observations out of $n$ therefore has likelihood $p^k(1-p{)}^{n-k}$, independent of $\pi$. The symptom alone does not identify the mixture proportion or pathway. This is an identifiability counterexample, not an empirical claim that real clinical pathways necessarily have identical symptom distributions. Additional indicators or interventions may distinguish them.

Synaptic release probability is the probability of release at a specified site during a specified opportunity, such as one presynaptic action potential. It depends on the site's state and conditions; it is not the probability that the receiving neuron fires. In the example under synaptic vesicle, $p$ refers to one available site, whereas $1-(1-p{)}^4$ refers to any of four independent sites releasing.

A synaptic vesicle is a small membrane-bounded compartment in a presynaptic terminal that stores neurotransmitter for release. Fusion with the cell membrane exposes its contents to the synaptic cleft; the packet supplied by one synaptic vesicle is a quantum of release. Quantal neurotransmitter release refers to these discrete packets, rather than the release of one molecule at a time. The postsynaptic response to a packet can vary with its contents, receptors and recording conditions. See Purves and colleagues: vesicles as the source of transmitter quanta.

The useful cycle is filling, preparation for fusion, calcium-triggered exocytosis, and recovery of membrane and proteins through endocytosis for reuse. Different recovery routes occur; recycling need not preserve a particular synaptic vesicle as an identifiable container throughout. Limited availability of prepared synaptic vesicles can reduce release during repeated stimulation, while changes in synaptic release probability can also change successive responses. See Basic Neurochemistry: secretion machinery and vesicle reuse and 3.1 Synapses.

For an illustrative calculation, assume four release sites, each with one prepared synaptic vesicle, each releasing with synaptic release probability $p=0.25$ during one presynaptic action potential. Assume independent Bernoulli trials and at most one release per site. The released packet count $K$ has a binomial distribution:The probability of no release is $\mathbb{P}(K=0)=0.75^4=0.31640625$; at least one release has probability $0.68359375$. An expected value of one packet does not mean that every action potential releases one packet. If each packet contributes a stipulated $0.5\mathrm{nS}$ conductance increment and increments add, the expected value of that increment is $0.5\mathrm{nS}$, with zero increment on about $31.6\%$ of trials. These assumptions are a teaching model: shared calcium fluctuations, unequal sites, depletion and nonlinear receptor responses can invalidate them.

A tail event is an event $A\in \mathcal{T}$, where $\mathcal{T}$ is the tail sigma-algebra of the chosen sequence of random variables. It concerns information available arbitrarily far along the sequence. In the measurable coordinate models described in finite-prefix invariance of tail events, any finite edit preserves whether it occurs.

This is stronger than merely needing a long waiting time. On $\{0,1{\}}^{\mathbb{N}}$, with $1$ denoting heads, let $K=\inf \{j\ge 1:X_j=1\}$, taking $K=\infty$ if there is no head. The eventis not a tail event: changing only $X_1$ from zero to one changes membership. Under fair independent Bernoulli trials its probability is $1/8$. Even the event $\{K<\infty \}$, of probability one in this model, is not a tail event as a literal set: the all-zero sequence and the sequence with just one initial head differ in one coordinate. A probability of zero or one does not characterize tail events.

Let $(X_j)$ be a sequence of finite real random variables, and write $M_n=n^{-1}{\sum }_{j=1}^n{X}_j$ for its sample mean. For fixed $m$ and $n>m$, put $M_n^{(m)}=n^{-1}{\sum }_{j=m+1}^n{X}_j$. Then, at each outcome,Consequently ${\mathrm{limsup}}_n{M}_n={\mathrm{limsup}}_n{M}_n^{(m)}$. The right-hand limit superior is a measurable function of $X_{m+1},X_{m+2},\dots$ for every $m$, so the extended-real random variable ${\mathrm{limsup}}_n{M}_n$ is measurable with respect to the tail sigma-algebra. The same holds for the limit inferior. Thus the event of convergence to a finite limit of a sequence, and the event $\{M_n\to a\}$ for any specified real $a$, are tail events.

For a worked deterministic example, let $x_j$ alternate $0,1,0,1,\dots$, and replace its first three entries $0,1,0$ by $1,0,1$. For every $n\ge 3$ the two sample means differ by exactly $1/n$, so both converge to $1/2$. Finite edits can change small-sample answers while preserving the infinite limit of a sequence.

Under independence, Kolmogorov's zero–one law makes each specified convergence event have probability zero or one. It does not select which value occurs. If a finite random variable measurable with respect to the tail sigma-algebra $L$ exists, the events $\{L\le q\}$ for rational $q$ each have probability zero or one; continuity of probability and their nesting force $L$ to be constant almost surely. For fair independent Bernoulli trials, the strong law of large numbers establishes $M_n\to 1/2$ almost surely; the zero–one conclusion alone proves neither convergence nor the value $1/2$.

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.

For a sequence of random variables $(X_n{)}_{n\ge 1}$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, defineThe decreasing sigma-algebras ${\mathcal{G}}_n$ are generated by the listed random variables. Their intersection consists of events expressible after discarding every finite prefix. No independence is required.^[ref]

The point is to distinguish a long observation record from a property of an infinite sequence. The event of seeing infinitely many heads and the event that sample means converge survive any finite change, while an exact first-head time does not. Kolmogorov's zero–one law adds the striking conclusion that under independence each tail event has probability zero or one; the tail sigma-algebra itself can still contain many nonempty sets.

The tail-sum formula expresses an expected value as the area under a survival function. For every nonnegative random variable $Y$,This Lebesgue integral and the expected value may both be $+\infty$; a probability density function is unnecessary. For a random variable $N$ taking values in the nonnegative integers, the corresponding sum isThe events $\{N>k\}$ and $\{N\ge k+1\}$ coincide because $N$ takes integer values. Thus the strict sum starts at zero, while the inclusive sum starts at one.^[ref]

For an original example, let the probability mass function of $N$ beIts expected value computed directly is $0(1/5)+1(3/10)+3(1/2)=9/5=1.8$. The strict tail probabilities at thresholds $0,1,2,3$ are $4/5,1/2,1/2,0$: their sum is again $1.8$. There is no probability mass at two, yet the event $\{N>2\}$ still has probability $1/2$. The survival function has heights $4/5,1/2,1/2$ on the successive intervals $[0,1),[1,2),[2,3)$; each interval has length one, so its integral is the same sum.

The indexing matters numerically. Starting the inclusive sum at zero adds $\mathbb{P}(N\ge 0)=1$, giving $2.8$ for this example. Starting the strict sum at one omits $\mathbb{P}(N>0)=4/5$, giving $1$. For a general nonnegative random variable, replacing $>$ with $\ge$ inside the integral is harmless: the difference is $\mathbb{P}(Y=t)$, nonzero at only a countable set of thresholds. For each positive integer $m$, at most $m$ thresholds can have probability mass at least $1/m$; their countable union contains all thresholds with positive probability mass and has Lebesgue measure zero. This reasoning concerns an integral, not equality of the two survival function conventions at every point.

For a continuous original example, take the uniform distribution on the interval $[0,4]$. Its survival function is $S(t)=1-t/4$ for $0\le t\le 4$ and zero for $t>4$. The area is a triangle of base four and height one:Directly integrating against the probability density function gives the same expected value, ${\int }_0^{4}y(1/4)dy=2$. The integral of the probability density function alone is one; it is the survival function whose area is the expected value.

The proof counts layers of each outcome. For a realized value $y\ge 0$, the indicator function $1_{\{y>t\}}$ equals one over an interval of length $y$, soTonelli theorem permits the interchange because the indicator function is nonnegative; no finite expected value is assumed. Similarly, an outcome $N=n$ contributes one indicator random variable for each of the $n$ events $\{N\ge k\}$ that occur:Again Tonelli theorem applies to the nonnegative sum, or equivalently the monotone convergence theorem applies to its partial sums. This exchanges counting layers within each outcome with averaging across outcomes. The existing integer-valued hypothesis in the tail-sum formula explains what changes when the values are not integers; the tail integral formula for moments weights the layers to obtain higher moments.

For an integrable random variable $X$ that can have either sign, apply the nonnegative formula separately to its positive part $X_{+}=\max (X,0)$ and negative part $X_{-}=\max (-X,0)$:Integrability makes both integrals finite. More generally the difference defines an extended expected value when at least one is finite; if both are infinite, subtraction does not define an expected value.^[ref]

Tail probabilities are often easier to measure or bound than a whole probability density function. For a device's survival time, the survival function records the probability of still operating at each threshold; its integral gives the mean operating time. A finite observation window naturally gives the expected value of a capped nonnegative random variable. For a mathematical bound, $\mathbb{P}(Y>t)\le e^{-t/2}$ for all $t\ge 0$ implies $\mathbb{E}[Y]\le {\int }_0^{\infty }{e}^{-t/2}dt=2$. The tail-sum formula turns control of a whole family of events into control of an average.

If $s>0$ and $\Theta$ has the continuous uniform distribution on $(-\pi /2,\pi /2)$, then $Y=s\tan \Theta$ has the Cauchy distribution with location zero and scale $s$. The tangent function is strictly increasing on this interval, so its cumulative distribution function isDifferentiating gives the probability density function $s/[\pi (s^2+y^2)]$. The probability of either endpoint is zero, so including endpoints in the uniform interval does not alter this probability distribution. The slow quadratic decay makes the expected value undefined: its positive and negative parts both have infinite integrals. Its second moment is also infinite, so it has no finite standard deviation.

For a pure input target $|\psi ⟩$ and normalized output density operator ${\rho }_{\mathrm{out}}$, the displayed squared quantum fidelity measures how well a teleportation channel reproduces that target. A mean value additionally specifies the input ensemble and whether it is conditioned on success. Detection efficiency, conditional fidelity, heralding and overall success probability are distinct experimental quantities.

Suppose a perfect $|{\varphi }^{+}⟩$ Bell pair is used for quantum teleportation, but the sender cannot distinguish the ${\varphi }^{+}$ and ${\varphi }^{-}$ Bell-basis measurement outcomes and guesses their labels uniformly. Each of the four actual outcomes has probability $1/4$. Either ambiguous outcome gives an incorrect Pauli Z gate correction with conditional probability $1/2$, so the total phase-flip probability is $1/4$. The unconditional output is therefore the displayed phase-flip channel. For $|\psi ⟩=\alpha |0⟩+\beta |1⟩$, its teleportation fidelity is $⟨\psi |\mathcal{E}(|\psi ⟩⟨\psi |)|\psi ⟩=1-|\alpha {|}^2|\beta {|}^2$. It ranges from $3/4$ on the Bloch sphere equator to $1$ at its poles. This averages the output density operators, including the ambiguous rounds.

A prefix code whose codewords use a three-symbol alphabet, conventionally $\{0,1,2\}$. No complete codeword is a prefix of another, permitting sequential decoding by following a code tree to a tree leaf. Its expected codeword length is ${\sum }_i{p}_i|c(i)|$ trits per source symbol. For source probabilities $(1/2,1/4,1/8,1/8)$, the codewords $(0,1,20,21)$ give expected codeword length $5/4$ trits.

A bead in a liquid keeps undergoing Brownian motion even when an optical trap holds its average position fixed. For a stationary harmonic optical trap in thermal equilibrium at absolute temperature $T$, the equipartition theorem assigns mean potential energy $k_{B}T/2$ to this one quadratic coordinate. With $k_B$ the Boltzmann constant and $x=0$ at equilibrium,At the same temperature, quadrupling the spring constant divides the position variance by four and the standard deviation by two. The Boltzmann factor gives the complete equilibrium probability density function, a Gaussian distribution:The prefactor makes ${\int }_{-\infty }^{\infty }p(x)dx=1$; its second moment is ${\int }_{-\infty }^{\infty }{x}^{2}p(x)dx=k_{B}T/k$. The probability density function has units of inverse length, and its area gives probability. This density is exact for the ideal quadratic model. In a finite optical trap it approximates the occupied well when thermal excursions remain within the harmonic region and escape is negligible during observation. Driven motion, appreciable nonconservative forces or laser heating need not give this equilibrium density.

Here $\ell =\sqrt{k_{B}T/k_0}$, $q=x/\ell$ and $\kappa =k/k_0$. The plotted density per unit $q$ is $p_q(q)=\ell p(\ell q)=\sqrt{\kappa /(2\pi )}{e}^{-\kappa q^2/2}$. Each curve has total area one; their dimensionless standard deviations are $1$, $1/\sqrt{2}$ and $1/2$. The curves are exact model densities, not noisy experimental data.

Brownian motion becomes a measuring tool: a calibrated position measurement supplies ${\sigma }^2=⟨(x-⟨x⟩{)}^2⟩$, and the known temperature supplies $k=k_{B}T/{\sigma }^2$. Once $k$ is known, a slowly applied constant external force $f$ shifts the mean by $⟨x⟩=f/k$, so $f=k⟨x⟩$ in the same harmonic range. Fluctuations therefore calibrate the conversion from displacement to force. Svoboda and Block used this equipartition theorem method when calibrating their bead traps. ^[ref].

For the example stiffness $0.01pN/nm$ and $T=300\mathrm{K}$, $k_{B}T\approx 4.142\mathrm{pN}\mathrm{nm}$ and $\sigma \approx 20.35\mathrm{nm}$. Using a variance measured in uncalibrated detector units would not determine that physical stiffness. Measurement noise, finite exposure time, drift and insufficient sampling of the correlated Brownian motion can bias the estimate; record positions over many optical trap relaxation times and account for the detector response. The deterministic relaxation curve above describes relaxation without thermal forcing; an individual thermal trajectory continues to fluctuate.

Tracing out one mode leaves geometric occupation probabilities with $q=|\lambda {|}^2$. In an oscillator basis of frequency $\omega$, they are Boltzmann weights with $k_{B}T=\hslash \omega /\ln (1/q)$. The joint state can be pure while each reduced state is mixed.

Memory thrashing is repeated costly cycling of memory pages out of and back into computer memory, leaving little useful computation between page faults. Active demand can exceed the available page frames, especially when multiprogramming admits too many operating-system processes. A high page fault rate during initial loading alone does not establish memory thrashing. ^[ref]

Consider one operating-system process making the 20 memory references $1,2,3,4,1,2,3,4,\dots ,1,2,3,4$: five repetitions of four memory pages. Start with an empty resident set, use demand paging and exact LRU replacement, and assume no prefetching or competing operating-system processes. A hit finds the requested memory page resident; a miss requires a nonresident page fault. With three page frames, references 1–3 load pages 1–3. Reference 4 evicts page 1, reference 5 reloads page 1 and evicts page 2, and the cycle continues: every return finds its memory page evicted. With four page frames, only the first four references miss; all four memory pages then remain resident.

The four initial misses with four page frames are compulsory first-use misses from the empty resident set. They are startup work; the three-page frame case keeps faulting after startup. This makes the difference between memory thrashing and loading a new working set visible. For the three-page frame trace, each reused memory page has three distinct intervening memory pages, so LRU replacement has already evicted it before its next use.

To quantify the cost, assume each hit takes 100 nanoseconds and each miss incurs an additional 1 millisecond of page fault service, followed by the same 100-nanosecond memory access. These are toy latencies, not measurements. With miss probability $p$, the expected value of access time isThe service cost is additional: it does not include the final 100 nanoseconds. Thus $p=1$ gives $1,000,100\mathrm{ns}$ and $p=0.2$ gives $200,100\mathrm{ns}$, approximately a fivefold ratio over the complete trace. Over references 5–20, the corresponding ratio is $1,000,100/100=10,001$: repeated page faults dominate the useful accesses. The expected value model follows the additive miss-penalty convention of OSTEP chapter 22, section 22.1. ^[ref]

Not every page fault performs disk input-output: demand paging can allocate zero-filled memory pages, and copy-on-write can copy an existing page frame. The toy service cost models expensive misses; the miss trace alone establishes residency behavior, while memory thrashing additionally concerns repeated costly cycling. ^[ref]

Possible remedies are more resident page frames, less multiprogramming, or reorganizing accesses to improve temporal locality. For this toy sequence, one extra page frame suffices under LRU replacement; that is not a universal threshold or policy. Page replacement alone cannot make every simultaneously needed memory page fit.

// Target: computer-science.bigb

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.

Let the input alphabet be $\{0,1,2\}$ and let the output be $Y=X+Z$, where independent noise is uniform on two distinct integers. Translating the output alphabet reduces the noise to $\{0,d\}$, $d\ge 1$. The reachable output alphabet has size $3+\min (d,3)$: it is the union of two three-element intervals shifted by $d$.

For $d=1$, every row of the transition matrix has conditional information entropy one bit, so $I(X;Y)\le {\log }_{2}4-1=1$. Input probabilities $(1/2,0,1/2)$ make the output uniform and attain one bit per use. For $d\ge 3$, the two translated input alphabets are disjoint and every output identifies its input. A uniform input then attains ${\log }_{2}3$ bits per use.

For $d=2$, reflection symmetry and concavity of information entropy permit restricting to input probabilities $(s/2,1-s,s/2)$. Conditional on output parity, outputs whose labels are even integers have probabilities $(1/4,1/2,1/4)$ and outputs whose labels are odd integers have probabilities $(1/2,1/2)$. Consequently $I(X;Y)=h_2(s)+s/2$. Its strictly negative second derivative and stationary equation ${\log }_2((1-s)/s)+1/2=0$ give $s=\sqrt{2}/(1+\sqrt{2})$ and channel capacity ${\log }_2(1+\sqrt{2})$. All three values follow from overlap of translated input alphabets, rather than from output alphabet size alone.

Measure the commuting stabilizer generators $g_1=Z_1{Z}_2$ and $g_2=Z_2{Z}_3$ in that order. Use error syndrome bit $0$ for eigenvalue $+1$ and bit $1$ for eigenvalue $-1$. For $X_j$ the Pauli X gate on physical qubit $j$, every row holds for arbitrary normalized $\alpha ,\beta$. This ordered parity-check convention follows ^[ref]:

Each pair of computational basis words within a row has the same two parities. Consequently the projective measurement reveals the row without distinguishing its two amplitudes. For example, $|{\psi }_L⟩=(\sqrt{3}/2)|000⟩+(i/2)|111⟩$ becomes $(\sqrt{3}/2)|010⟩+(i/2)|101⟩$ under $X_2$. Both checks give $-1$ with probability one, and applying $X_2$ restores both coefficients exactly, including the relative phase $i$.

One implementation uses two quantum ancillas initialized to $|00⟩$. Two controlled-NOT gates from data qubits 1 and 2 to the first quantum ancilla write their parity; two more from data qubits 2 and 3 to the second quantum ancilla write the other parity. Measuring only the quantum ancillas in the computational basis gives $(s_1,s_2)$. Measuring the individual data qubits instead would reveal the encoded computational basis value and destroy the unknown quantum superposition. The existing coherent syndrome extraction for the three-qubit repetition code uses the alternative ordered checks $(Z_2{Z}_3,Z_1{Z}_3)$, so its bit labels differ from this table.

The recovery also handles coherent errors. Put $E_0=I,E_1=X_1,E_2=X_2,E_3=X_3$. Their images of the code are mutually orthogonal, giving $P{E}_a^{†}{E}_{b}P={\delta }_{ab}P$, the Knill--Laflamme condition for this family. For any linear combination $E={\sum }_{a=0}^3{c}_a{E}_a$, its component in sector $s(E_a)$ is $c_a{E}_a|{\psi }_L⟩$. Measuring that error syndrome has probability $|c_a{|}^2/{\sum }_b|c_b{|}^2$ on the normalized erroneous state, and recovery $E_a$ returns $|{\psi }_L⟩$. More generally one recovery quantum channel corrects every noise quantum channel whose Kraus operators belong to this span. The assumption concerns the whole jointly correctable family; it does not extend to the span of arbitrary single-site Pauli errors. ^[ref].

A fully computed coherent example is the unitary operator $U=\exp (-i\theta X_2)=\cos \theta I-i\sin \theta X_2$. At $\theta =\pi /6$ it givesThe error syndromes $00$ and $11$ occur with probabilities $3/4$ and $1/4$, independently of $\alpha ,\beta$. Doing nothing for $00$ and applying $X_2$ for $11$ restores the same logical qubit in both branches; the scalar $-i$ in the second branch is an irrelevant global phase. This is an example of linearity of coherent quantum error correction, rather than an assumption that the original error was a classical random bit flip.

For the three qutrit unitary operators $U_j=I-2|j⟩⟨j|$, prepare $|s⟩=(|0⟩+|1⟩+|2⟩)/\sqrt{3}$, query once, apply the known inversion about the mean $D=2|s⟩⟨s|-I$, and use a computational-basis measurement to guess $j$. Since $D^2=I$, the processing is a unitary transformation. Direct multiplication gives $D{U}_j|s⟩=(5|j⟩-{\sum }_{k\ne j}|k⟩)/(3\sqrt{3})$, establishing success probability $25/27$ for each label. This calculation establishes the achieved probability, without requiring an optimality claim.

Suppose conditionally independent latent values have normal distribution $N(\mu ,{\sigma }^2)$, with $\sigma >0$, but the observation only says which side of a known threshold $c_i$ contains each value. With $s_i=1$ for a value above the threshold and $s_i=-1$ for a value below, the likelihood is the displayed product, where $\Phi$ is the standard normal distribution function. Repeated thresholds give corresponding powers. These factors are tail probabilities, not probability densities evaluated at the thresholds. Multiplying normal densities at $c_i$ would describe a different observation process. The thresholds must be treated as fixed or otherwise modelled appropriately; informative selection of thresholds can change the likelihood.