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For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.

For two increasing events of equal probability $a$ under a product measure, positive association of their complements gives $1-P(A\cup B)\ge (1-a{)}^2$. Hence $a\ge 1-\sqrt{1-P(A\cup B)}$. For $r$ equal-probability increasing events, the same reasoning gives $a\ge 1-(1-P({\bigcup }_j{A}_j){)}^{1/r}$.

Here $s={\sinh }^{2}r$. The pair moment is $⟨N(N-1)⟩=3s^2+s$. These expressions follow from the single-mode Bogoliubov transformation or by differentiating the even-occupation probability generating function $G(t)=[\cosh r\sqrt{1-t^2{\tanh }^{2}r}{]}^{-1}$.

A lower envelope $h$ can save evaluations of an expensive unnormalized target $\tilde{p}$ in rejection sampling. Draw $Y$ with probability density function $\tilde{q}/Z_q$ and an independent uniform distribution value $U$. Accept immediately if $U\le h(Y)/(M\tilde{q}(Y))$. Otherwise draw a fresh independent uniform distribution value $V$ and accept ifFor a fixed proposal, writing $B=M\tilde{q}(Y)$, $H=h(Y)$ and $P=\tilde{p}(Y)$ gives total acceptance probability $H/B+(1-H/B)(P-H)/(B-H)=P/B$. If $H=B$, the first stage accepts with probability one and the residual test is skipped. Thus the accepted probability density function is $\tilde{p}/Z_p$, and the overall acceptance probability is $Z_p/(M{Z}_q)$. The probability of avoiding a target evaluation on one proposal is $\int h/(M{Z}_q)$.

If $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, then $Z=(X-\mu )/\sigma$ has the standard normal distribution. The change-of-variables formula for a probability density gives $f_Z(z)=\sigma f_X(\mu +\sigma z)=\varphi (z)$, the standard normal density. Thus standardization of a normal random variable removes both parameters without changing the probabilities of corresponding events.

The standard normal distribution is $N(0,1)$: the normal distribution with mean zero, variance one and standard deviation one. Its standard normal density $\varphi$ integrates to one; its standard normal distribution function $\Phi$ records the area accumulated from $-\infty$ to a threshold. The distinction matters: $\varphi (0)=1/\sqrt{2\pi }\approx 0.398942$, $\Phi (0)=1/2$, and $\mathbb{P}(Z=0)=0$ for a standard normal random variable $Z$.

For a toy measurement $X\sim N(100,15^2)$, leave the measurement unit unspecified. The standardization of a normal random variable $Z=(X-100)/15$ converts $x=85,100,115$ into standard scores $z=-1,0,1$. Using the standard normal distribution function,The second area is half the first because the normal distribution is a symmetric probability distribution about its mean. The numerical heights differ: $f_X(115)=\varphi (1)/15$, yet the corresponding shaded areas are equal because $dx=15dz$ in the change of variables.

The standard normal distribution function is the cumulative distribution function of a standard normal random variable $Z$:Here $\varphi$ is the standard normal density and $\mathrm{erf}$ is the error function. The symmetric probability distribution gives $\Phi (-x)=1-\Phi (x)$ and $\Phi (0)=1/2$, while differentiation gives ${\Phi }^{\prime }(x)=\varphi (x)$. For $X\sim N(\mu ,{\sigma }^2)$ with $\sigma >0$, standardization of a normal random variable gives $\mathbb{P}(X\le x)=\Phi ((x-\mu )/\sigma )$; subtracting two values calculates an interval probability.

For a discrete-time Markov chain with transition matrix $P$, these equations, together with ${\pi }_j\ge 0$ and ${\sum }_j{\pi }_j=1$, say that $\pi$ is a stationary distribution. Incoming probability mass equals the current mass at each state. They are global balance equations and do not require detailed balance.

Choose one of two regimes once, with equal probability: a source of independent and identically distributed random variables with one-bit probability $1/10$, or one with one-bit probability $1/2$. Keep the chosen regime forever. The resulting source is a strictly stationary process, because shifting a block does not change either component law. It is not an ergodic stationary process: its limiting empirical frequency identifies the regime.

The source's information rate isConditioning on the single regime bit changes any block's information entropy by at most one bit, so dividing by the block length proves this rate. However, the limiting realized information content per symbol is approximately $0.468996$ in the first regime and one in the second. The alternative component's likelihood ratio decays exponentially along a typical path from the chosen component, at a rate given by the positive relative entropy between the two Bernoulli distributions. The mixture therefore has two possible pathwise limits, not one limit equal to the averaged information rate.

A stationary distribution is a probability distribution $\pi$ preserved by a Markov chain's evolution. For a Discrete-time Markov chain in the row convention,Thus $\pi$ is a normalized nonnegative left eigenvector of the transition matrix for eigenvalue one. Starting with $X_0$ distributed as $\pi$ makes every $X_n$ have that same probability distribution. For a time-homogeneous Markov chain, the path probabilities are also unchanged by shifting all observation times together: the process has strict stationarity. This does not mean its individual paths are constant. See Hao Wu's MIT lecture 2, stationary distributions.

For the sunny/rainy transition matrix, write $\pi =(a,1-a)$. The sunny component of $\pi P=\pi$ gives $a=0.8a+0.4(1-a)$, so $0.6a=0.4$ andMoreover,These equal stationary flows verify detailed balance and show that the Markov chain is reversible. The diagonal detailed balance equations hold automatically. Only $1/3$ of the stationary days are rainy, but $0.6$ of the days following a rainy day are rainy: stationarity permits dependence.

A finite Markov chain that is an irreducible Markov chain has a unique stationary distribution. If it is also an aperiodic Markov chain, ${\mu }_0{P}^n\to \pi$ for every initial probability distribution. The sunny/rainy example satisfies both conditions: both directions are possible and each state has a positive holding probability of a Markov chain. See Hao Wu's MIT lecture 4, convergence theorem. For an infinite state space, the additional positive recurrence hypothesis in the countable-state Markov chain convergence theorem matters.

Existence of a stationary distribution alone does not give this convergence. Consider the deterministic two-cycleHere $\pi C=\pi$, but starting in the first state gives alternating probability distributions $(1,0),(0,1),(1,0),\dots$. Its period of a state in a Markov chain is two. The running fraction of visits to either state nevertheless tends to $1/2$, illustrating the Markov chain ergodic theorem. Starting with $\pi$ instead keeps the one-time probability distributions stationary even though every path alternates.

Statistical inference uses observed data and a statistical model to learn about unknown features of the process that generated them. In a parametric statistical model, write $X\sim P_{\theta }$: the probability distribution depends on an unknown statistical parameter $\theta$. An estimand specifies the target, such as a mean or a difference of means. A statistic is a function of the observations; an estimator is a statistic used to estimate the target. Uppercase $X$ denotes the random data before observation, while lowercase $x$ denotes their realized values.

Three questions lead to different outputs. Statistical estimation asks which target value the observations suggest. A confidence interval asks for a range produced by a procedure with calibrated coverage probability. A hypothesis test asks whether the observations depart sufficiently from a specified null hypothesis to enter its rejection region. These outputs complement one another: an estimated effect size describes magnitude, while a standard error describes the sampling distribution of an estimator. A p-value describes a tail probability under the null hypothesis.

The appeal of statistical inference is that it turns variable observations into reproducible uncertainty statements. In the normal mean confidence interval with known variance below, a sample mean of $1.2$ has standard error $0.5$: both the estimated shift and its precision matter. The calibration depends on the statistical model, including independence and the specified probability distribution. Increasing sample size improves precision under those assumptions; it does not repair a systematically misleading sampling procedure. Bayesian inference provides another framework, using a prior distribution and a likelihood function to obtain a posterior distribution.

Statistical learning constructs predictive rules from data and studies their prediction error, generalization, and computational fitting. In supervised learning, training data contain inputs $X$ and a response variable $Y$: regression analysis predicts a numerical response, while classification in statistical learning predicts a label. Unsupervised learning instead seeks structure without supplied responses, as in cluster analysis and principal component analysis. These tasks connect learning to probability and mathematical optimization. ^[ref]

For a fitted rule $\hat{f}$ and loss function $L$, the predictive goal iswhere $(X^{\ast },Y^{\ast })$ is a fresh observation, independent of the training data ${\mathcal{D}}_n$, from the intended target probability distribution. This prediction error fixes the fitted rule and averages over future observations; averaging again over training data gives the learning procedure's average performance. Its practical beauty is that a rule compresses past observations into useful future predictions. Its central difficulty is that excellent training error need not imply good generalization: cross-validation supports model selection, and a reserved test set assesses the resulting choice.

Statistical physics explains collective behavior using probability distributions over microscopic possibilities. Its central surprise is that incomplete microscopic knowledge can still produce precise predictions for internal energy, entropy and heat capacity. The bridge is statistical mechanics: specify the available microstates, assign their probabilities, then compute the expected values of measured quantities.

A two-level system already displays the idea. Warming it makes its higher energy more likely; its heat capacity first rises, then falls as the two probabilities approach equality. With two distinguishable units, merely counting microstates explains why the middle energy has twice the state degeneracy. These small examples expose the machinery before the much larger sums needed for a photon gas or an Ising model.

The appeal is economical: the same microscopic energies and probabilities explain both an average reading and its variance. A measurement's spread therefore becomes useful physical information rather than an inconvenience to discard.

A statistical test is a measurable rule for deciding whether to reject a null hypothesis. Allowing randomisation, it is represented by a function $\phi (x)\in [0,1]$ giving the conditional rejection probability after observing $x$. A deterministic test takes only values zero and one. Under a simple null, $E_0\phi$ is its size of a statistical test; under an alternative, $E_1\phi$ is its statistical power. Randomisation on a likelihood ratio equality set allows a Neyman-Pearson lemma threshold test to attain a prescribed size even for discrete data.

A phase bounded by mass-dependent lifetimes ${\tau }_a(M)$ and ${\tau }_b(M)$ occupies ${\tau }_a(M)<t<{\tau }_b(M)$, clipped to the population age range. Its boundaries can be mapped into uniform probability coordinates to evaluate evolutionary-state fractions.

The ideal projective measurement of a spin-one-half component along a unit axis $\mathbf{n}$. The two outcomes have Born rule probabilities $⟨\psi |P_{\pm }|\psi ⟩$. For an input with Bloch vector $\mathbf{r}$, they are $(1\pm \mathbf{n}\cdot \mathbf{r})/2$. This model distinguishes a mathematical spin measurement from engineering a literal free-electron beam splitter.

A stochastic computation graph is a Directed acyclic graph combining deterministic transformations, random choices and costs. The changing probabilities of random choices contribute score-function gradient estimators, while differentiable deterministic paths contribute pathwise gradients. This accounts for both sources of a parameter's effect on an expected cost.^[ref]

On a finite partially ordered state space, $\mu ⪯\nu$ means $\int fd\mu \le \int fd\nu$ for every real-valued order-preserving function $f$. Equivalently, every increasing subset has no larger probability under $\mu$ than under $\nu$. A coupling supported on pairs $(X,Y)$ with $X\le Y$ proves this relation. This concerns order of probability laws, rather than stochastic asymptotic order notation.

A row-stochastic matrix has nonnegative entries and each row sums to one. For a time-homogeneous Discrete-time Markov chain, its transition matrix has entriesIf a state has zero probability at time $n$, this row is specified by the model's Markov kernel, rather than recovered by conditioning on that zero-probability event. Rows index departure states and columns arrival states. Writing the current probability distribution as a row vector ${\mu }_n$ givesThe row-sum condition ensures that matrix multiplication preserves total probability. Under a column convention the evolution uses the transpose; the convention must accompany the formula.

For the sunny/rainy transition matrix, direct matrix multiplication givesIn particular, $(P^2{)}_{SS}=0.8\times 0.8+0.2\times 0.4=0.72$. The two terms sum the disjoint paths $S\to S\to S$ and $S\to R\to S$. Requiring both subsequent days to be sunny has probability $0.64$; requiring only the second day to be sunny permits the extra path through rain.

Generally $(P^n{)}_{ij}$ is the $n$-step transition probability. The identity $P^{m+n}=P^m{P}^n$ is the discrete-time Chapman-Kolmogorov equation: sum over the intermediate state. This makes matrix powers useful for forecasting many steps without enumerating every path individually.