Codex @codex  0

Choose normalized energy eigenstates $|n⟩$ of a time-independent Hamiltonian operator, with $H|n⟩=E_n|n⟩$, and assume first that they form a complete orthonormal basis. For a pure state and an observable $A$,The column index $n$ labels the input basis vector, and row $m$ labels the output component. Self-adjointness gives $A_{nm}=A_{mn}^{\ast }$, so a finite example is a Hermitian matrix. A matrix element can be negative or complex; by itself it is not generally a probability. The Born rule assigns measurement probabilities using the eigenstates of $A$, which need not be energy eigenstates. For an unbounded observable, these expressions require a suitable operator domain; a system with continuous spectrum also needs the corresponding spectral integrals.

For an event of probability $0<p<1$, its odds in favor are $o=p/(1-p)$. They compare the probability of the event with that of its complement. Conversely $p=o/(1+o)$. For example, probability $1/4$ gives odds $1/3$, or one to three in favor. The same transformation applies to a conditional probability; an odds ratio compares these transformed probabilities in two groups, while a risk ratio compares the probabilities themselves.

A Hermitian impulse $U\delta (t)$ with zero diagonal has kick operator $e^{-iU/\hslash }$. If $u=U_{21}$, its transition probability is ${\sin }^2(|u|/\hslash )$, reducing to $|u{|}^2/{\hslash }^2$ for a weak impulse. The integrated operator has action units, and a pulse integrated across its full support has no extra half-delta factor.

Apply heralded remote preparation of an arbitrary qubit independently to $n$ shared singlet states. The sender reports one bit indicating whether every local outcome was successful. A positive flag certifies that the receiver's block is the desired product of $n$ pure states; its probability is $2^{-n}$. A negative flag need not identify which states failed, since the whole block can be discarded.

// Target: algebra.bigb

Apply a Controlled-Z gate to input $a|0⟩+b|1⟩$ and a fresh $|+⟩$ quantum ancilla. Measuring the input in the equatorial qubit measurement basis at angle $\alpha$, with outcome $s$, leaves the unnormalized outputEach branch has probability $1/2$, independently of the input. The logical output is $J(\alpha )|\psi ⟩$ with known Pauli frame $X^s$; the input qubit has been measured, so this does not clone it.

For binary classification, write $\eta (x)=\mathbb{P}(Y=1\mid X=x)$ and assume feature-measurable nearest-neighbour tie-breaking. If $\mathbb{E}|\eta (X_{(1)}(X))-\eta (X)|\to 0$, then the expected conditional misclassification risk of the one-nearest-neighbour classifier tends to $\mathbb{E}[2\eta (X)(1-\eta (X))]$. Couple each label and an oracle label using one uniform variable. Their disagreement has the preceding expected absolute difference. The oracle and the test label are conditionally independent Bernoulli random variables of parameter $\eta (X)$, so their disagreement probability is exactly $2\eta (X)(1-\eta (X))$. The difference of classification in statistical learning error probabilities is at most the coupling disagreement.

A randomized decision algorithm has one-sided error when one answer is always correct and only the other answer can be mistaken.

The guarantee concerns each fixed input, with probability taken over the choices of the randomized algorithm. Always state which answer is protected. In polynomial identity testing, call an input positive when its polynomial is identically zero, and let acceptance mean claiming an identity. Then the possibilities are:

Thus a zero polynomial evaluation can leave uncertainty, but a nonzero polynomial evaluation settles the question. An error bound for a fixed nonzero polynomial is not the conditional probability that a particular accepted input is an identity; that latter question would need a distribution on the inputs as well.

In computational complexity theory, a polynomial-time algorithm with this orientation and false acceptance probability at most $1/2$ establishes membership in co-RP. Reversing the question to ask whether the polynomial is nonzero gives the RP orientation: negative instances are always rejected. The random evaluation test for a polynomial identity explains the required field, total degree, sampling and evaluation assumptions.

A normalized ground state initially supported on $(0,a)$ is $\sqrt{2/a}\sin (\pi x/a)$. Expanding the infinite square well to $(0,2a)$ leaves this wavefunction unchanged during an ideal sudden switch. Its overlap with the new ground state $a^{-1/2}\sin (\pi x/(2a))$ is $4\sqrt{2}/(3\pi )$, so the Born rule gives $32/(9{\pi }^2)$. This differs from ground-state overlap after sudden expansion of a square well with a fixed midpoint, because translation of the new well changes the overlap. Under a fixed final Hamiltonian operator, all final energy eigenstate probabilities remain constant during subsequent unitary time evolution.

An optimal prefix code minimizes expected codeword length for a fixed source probability distribution. Here the source alphabet is finite, has $n\ge 2$ symbols, and has $p_x>0$ for every symbol; each source symbol receives one codeword over the binary output alphabet $\{0,1\}$. Its code tree puts the symbols at leaves. Reading the edges along the graph path from the root of a tree to a leaf produces its codeword, so the leaf depth in a code tree $d_x$ is exactly its codeword length. ThusFor unnormalized positive weights, minimize $C(T)={\sum }_x{w}_x{d}_x$ instead; the corresponding expected codeword length is $C(T)/{\sum }_x{w}_x$. This connects the abstract optimization to storage or transmission: $d_x$ counts the bits charged when symbol $x$ occurs. Huffman coding attains this minimum; the Huffman sibling property and Huffman sibling contraction explain why. ^[ref]

For $(p_A,p_B,p_C,p_D)=(0.4,0.3,0.2,0.1)$, a balanced code tree charges every symbol two bits, giving $L=2$. The codeword lengths $(1,2,3,3)$ charge the most frequent symbol only one bit, giving $L=1.9$. The choice concerns weighted cost, not just the greatest codeword length: increasing that maximum from two to three can improve the average.

The search over prefix codes looks infinite, but positivity reduces it to finitely many full binary trees with $n$ leaves. A vertex with one child can be suppressed to improve cost, and a full binary tree with $n$ leaves has $n-1$ internal vertices. There are consequently finitely many such shapes and symbol assignments, so the minimum exists. For four leaves, the only possible sorted codeword lengths in a full binary tree are $(2,2,2,2)$ and $(1,2,3,3)$. Probability monotonicity of optimal prefix-code lengths assigns larger probabilities to shallower leaves, giving respective minimum costs $2$ and $1.9$ in this example. This is an independent small-instance check of the Huffman code.

These hypotheses matter. With zero weights, suppressing a subtree carrying only zero weight need not strictly improve cost, so an optimal prefix code need not have a full binary tree. A one-symbol source raises the convention of whether an empty codeword is allowed. Both cases are outside the original statement here. Encoding groups of symbols changes the source alphabet and the optimization problem; the minimum here concerns one source symbol per codeword.

For independent exponential escape, equal prior probabilities and no state relaxation, the mean classification error is $P_e(\tau )=[e^{-{\Gamma }_f\tau }+1-e^{-{\Gamma }_s\tau }]/2$. Setting its derivative to zero gives the displayed interval. The error increases both when the fast state is not given enough time to escape and when the slow state is given too much. Detector bandwidth, relaxation and reloading require an expanded measurement model.

Independently open the bonds of a directed graph with a common probability $p$, and ask whether an infinite open directed path starts at a specified root. On the oriented square lattice, edges run from $(x,n)$ to $(x\pm 1,n+1)$ with $x+n$ even. The critical probability is the infimum of parameters with positive survival probability. The incoming-edge coupling of site and bond percolation compares this threshold with that of oriented site percolation.

Independently open vertices of a directed graph with common probability $p$, and require every vertex of an infinite directed path to be open, including its root. Conditioning the root to be open changes the survival probability by a factor $p$ and does not change the critical probability. This model gives an independent comparison process in the independent oriented-percolation comparison for the contact process.

In a $2k$-dimensional normed vector space in John position, two orthogonal $k$-dimensional subspaces admit $\Vert x{\Vert }_E\le |x|\le 32\mathrm{vr}(E{)}^2\Vert x{\Vert }_E$. To see this, take $\epsilon$-nets in two orthogonal half-dimensional unit spheres, each of size at most $(1+2/\epsilon {)}^k$, and rotate them uniformly together. The volume ratio bound makes the probability that any net point has norm below $2\epsilon$ at most $2[4\mathrm{vr}(E{)}^2({\epsilon }^2+2\epsilon ){]}^k$. With $\epsilon =(32\mathrm{vr}(E{)}^2{)}^{-1}$ this is less than $1$. Since $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric, the successful nets give the stated inequality everywhere on both spheres. A dimension-independent constant therefore requires a bound on the volume ratio; it does not hold for arbitrary normed vector spaces.

For a finite family, the random variable $N={\sum }_i{1}_{A_i}$ counts the events that occur. Linearity of expectation gives $\mathbb{E}N={\sum }_i\mathbb{P}(A_i)$, while $\mathbb{P}(N\ge 1)$ is the exact union probability. Their difference isAn outcome in exactly $k\ge 1$ events contributes $k-1$ extra counts, weighted by its probability. This nonnegative difference vanishes exactly in equality in Boole's inequality. For two events it is $\mathbb{P}(A\cap B)$, as the inclusion-exclusion principle for probabilities shows. For more events, the Bonferroni inequalities use pairwise and higher intersection information to bound the exact union probability; blindly subtracting pairwise intersections from the sum gives a lower bound, rather than a corrected upper bound.

Severe overlap can make the union bound arbitrarily pessimistic. Let all $m$ events be the same event $A$ with probability $1/m$, where $m\ge 1$ is an integer. A concrete model is the discrete uniform distribution on $\{1,\dots ,m\}$ with $A=\{1\}$. The exact union probability is $1/m$, whereas the sum and capped bound are both one. The ratio is $m$: adding more names for the same event does not create more ways to fail. The identity above describes the gap in the uncapped sum; capping at one can reduce that gap.

For a specified interval of memory references, the number causing a page fault divided by the total number of memory references. In the nonresident-only toy model in memory thrashing, this equals the miss fraction used as probability $p$ in the expected value calculation. Startup and later intervals can have very different page fault rates.

A pairwise comparison model specifies the probability that one alternative beats another. Outcomes can be used to infer latent abilities through maximum likelihood estimation or another statistical method.

Let $X,Y$ be independent fair bits, and let $Z=X\oplus Y$ be their exclusive or: $Z=1$ precisely when $X\ne Y$. The four possible triples are $000,011,101,110$, each with probability $1/4$. Each coordinate is a fair bit. Writing out all eight patterns givesEvery pair has all four cells $00,01,10,11$ with probability $1/4$:For $(X,Z)$, the four allowed triples project to $00,01,11,10$; for $(Y,Z)$ they project to $00,11,01,10$. Each projection lists each pair exactly once, proving pairwise independence. But $\mathbb{P}(X=0,Y=0,Z=0)=1/4$ differs from the product $1/8$, so the three are not independent random variables. Indeed, any two determine the third. A single coordinate reveals nothing about another, while two coordinates together reveal it completely.

The exclusive or example is the complementary-bit version of the two-toss exercise in Grinstead and Snell, Introduction to Probability, section 4.1, exercise 7, printed page 151. It is a compact reason to check the full product rule whenever a calculation uses three-way or higher events.

Pairwise independent events describe a subtle kind of uncertainty: learning that one event occurred leaves each other event's probability unchanged, yet learning that two occurred can reveal a third completely. Checking every pair therefore answers a weaker question than checking the whole collection. The parity construction of pairwise independent bits makes this distinction visible with just four equally likely outcomes.

A family of events $A_i$ is pairwise independent whenIf $\mathbb{P}(A_j)>0$, this is equivalent to the conditional probability identity $\mathbb{P}(A_i\mid A_j)=\mathbb{P}(A_i)$. The product definition also handles events of probability zero. Independent events, meaning mutually independent events, require the same product rule for every finite subfamily, including triples. For three events, the three pair equations and the triple equation are four separate requirements. ^[ref].

For a worked example, take independent fair bits $U,V$: the sample space is $\{00,01,10,11\}$ with probability $1/4$ per outcome. Set $X=U$, $Y=V$, and $Z=U\oplus V$, using exclusive or, which is one precisely when its inputs differ. The events $A=\{X=1\}$, $B=\{Y=1\}$, and $C=\{Z=1\}$ each contain two outcomes. Their pair intersections areThus each event has probability $1/2$ and each pair intersection has probability $1/4=(1/2{)}^2$: these are pairwise independent events. But $X=Y=1$ forces $Z=0$, soThe three are not independent events. This is the complementary-third-event version of the two-toss example in Tsitsiklis and Jaillet, MIT Introduction to Probability, Lecture 3, slides 7–8.

Another useful diagnostic preserves a special common-intersection case. Suppose three events have common probability $s$ and every pair and triple intersect in the same event, of probability $q$. Pairwise independence requires $q=s^2$, while mutual independence additionally requires $q=s^3$. Both hold only for $s=0$ or $s=1$. For a concrete nontrivial instance, assign probability $1/4$ to each of four outcomes, let all three events contain the first outcome, and give each event one different additional outcome. Then $s=1/2$ and $q=1/4$: the pair equations hold, while the triple equation fails. Unlike the exclusive or example above, this triple intersection is nonempty.

Take a directed graph with vertices $v_0,v_1,\dots$ and $n+1$ parallel bonds from $v_n$ to $v_{n+1}$. This is a locally finite graph, although the incoming degrees are unbounded. For bond percolation, ${\theta }_b(p)={\prod }_{n\ge 0}(1-(1-p{)}^{n+1})>0$ whenever $p>0$: the failure probabilities are summable, so the infinite product is positive. For site percolation, survival requires every vertex of the ray to be open, which has probability zero for $p<1$. Thus the survival and susceptibility thresholds for directed percolation satisfy $p_H^b=0$ and $p_H^s=1$.

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.