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A simple random walk on a locally finite graph chooses each vertex in the current vertex's graph neighbourhood with equal probability, assuming there is at least one available neighbour. On the integer line there are two choices, one unit left or right, so each has probability $1/2$. This is the one-dimensional simple symmetric random walk. Terminology varies: a simple random walk on the integer line is also often allowed unequal step probabilities, with “symmetric” specifying the fair case.

For a simple symmetric random walk starting at $S_0=0$, each increment has the Rademacher distribution, with expected value zero and variance one. Thus linearity of expectation and the variance of a sum of independent random variables giveFor example, after $80$ steps the expected value is zero, the variance is $80$, and the standard deviation is $\sqrt{80}\approx 8.944$. Doubling the typical distance scale takes four times as many steps. This square-root growth explains why a random walk can wander far even when there is no preferred direction.

The colored curves are illustrative sample paths, not averages over all possible walks. The dashed curves show the expected values; the shaded bands show the expected value plus or minus one standard deviation at each time. These bands are a scale of spread, not confidence intervals, and give no guarantee that an entire sample path stays inside them.

Zero expected value does not make this a stationary process: its one-time probability distribution changes with $n$, as the increasing variance already proves. It does have stationary increments and independent increments, because blocks of fresh steps have the same probability distribution at any starting time and disjoint blocks use independent random variables. This separation of position from increment is one reason the random walk is such a useful first example of a stochastic process.

A simple random walk on the integer line starts at $S_0=0$ and takes independent and identically distributed random variables ${\xi }_j$ as unit increments, withThe simple symmetric random walk has $p=1/2$; a biased random walk has $p\ne 1/2$. At $p=0$ or $p=1$ every step is deterministic, giving $S_n=-n$ or $S_n=n$ respectively.

Count the rightward steps using indicator functions:Thus the endpoint is a shifted and rescaled random variable with a binomial distribution. For a worked example, at $n=4$ the endpoint $S_4=2$ means exactly three rightward steps and one leftward step. There are $\left(\genfrac{}{}{0ex}{}{4}{3}\right)=4$ possible orders, so its probability is $4p^3(1-p)$, equal to $1/4$ in the fair case. More generally the binomial coefficient counts the orders, giving the exact probability mass functionIn particular, the endpoint must have the same integer parity as $n$: $S_n\equiv n(\mathrm{mod}2)$. For $0<p<1$ the support of a probability distribution is exactly $\{-n,-n+2,\dots ,n\}$. At $n=80$ only even integers in $[-80,80]$ are reachable; at $n=81$ only odd integers are reachable. This restriction is easy to lose if one draws a smooth curve instead of discrete probability masses. The fair-walk counting argument is also presented in Grinstead and Snell, Introduction to Probability, section 12.1.

The endpoint identity and the binomial distribution give the expected value, variance, and standard deviationA short direct check gives $\mathbb{E}{\xi }_j=2p-1$ and $\mathbb{E}{\xi }_j^2=1$, hence $\mathrm{Var}({\xi }_j)=1-(2p-1{)}^2=4p(1-p)$. Then apply linearity of expectation and the variance of a sum of independent random variables.

The absolute width and the width relative to the number of steps answer different questions. In the fair case the standard deviation grows from $\sqrt{20}\approx 4.472$ to $\sqrt{80}\approx 8.944$, while ${\sigma }_n/n$ shrinks from $0.2236$ to $0.1118$. At $n=80,p=0.65$, the standard deviation is $\sqrt{72.8}\approx 8.532$, or $0.9539$ times the fair width, and ${\sigma }_n/n\approx 0.1067$. Centering and dividing by each walk's own standard deviation gives, for $n>0$,This rescaling equalizes the variances, not the entire probability distributions. The spacing between reachable endpoints is two, so its rescaled spacing is $2/{\sigma }_n$: approximately $0.4472$ for the fair $20$-step walk, $0.2236$ for the fair $80$-step walk, and $0.2344$ for the biased $80$-step walk.

The positions themselves are dependent. For $1\le m\le n$, $S_n=S_m+(S_n-S_m)$, where the second term is independent of $S_m$. Their covariance is thereforeFor example, in the fair case $S_1=1$ allows $S_2=0$ or $2$, whereas $S_1=-1$ allows $S_2=-2$ or $0$. Nevertheless the position process is a Markov chain: given its current position, the next fresh increment has no dependence on its past. Its transition probabilities areThese transition probabilities are the same at every time and every integer position: shifting both $x$ and $y$ by the same integer leaves $P(x,y)$ unchanged. Thus the Markov property and independent increments coexist with dependent positions.

Simulated annealing accepts worse feasible solutions to escape a local minimum. Minimize an objective function $E(x)$, analogous to energy. Effective temperature $T>0$ has the same units as $E$, absorbing the Boltzmann constant; it controls exploration, rather than a material's thermodynamic temperature.^[ref]

Given current feasible point $x$, draw $y$ from a proposal distribution $q(x,y)$. For a symmetric proposal distribution, $q(x,y)=q(y,x)$, use the Metropolis–Hastings acceptance probabilityAccept with this conditional probability; on rejection retain $x$. Every downhill or equal-cost proposal is accepted. Draw a fresh uniform random variable $U$ on $(0,1)$ and accept if $U<a_T(x,y)$. The off-diagonal transition probability $P_T(x,y)=q(x,y)a_T(x,y)$ includes proposing $y$.^[ref]

For a maximization problem with objective function $f$, set $E=-f$: every improving proposed neighbor is accepted, and a worsening move of loss $\Delta =f(x)-f(y)>0$ has conditional probability $\exp (-\Delta /T)$ of acceptance, under the same symmetry assumption. Likelihood-power annealing uses precisely this sign convention for a log-likelihood. Retain the best feasible solution seen, since the current state can worsen even after a global minimum has been visited.

If the proposal distribution is asymmetric, use the full Metropolis–Hastings acceptance probability, for $q(x,y)>0$,A missing reverse proposal gives zero acceptance. Even a downhill proposal need not be accepted with probability one in this case. Choosing uniformly among a variable number of neighbors is generally asymmetric: $q(x,y)=1/d(x)$ and $q(y,x)=1/d(y)$ give the correction $d(x)/d(y)$, where $d$ is vertex degree.^[ref]

The useful idea is to separate exploring from insisting on improvement: an uphill move can open a route to a better feasible solution. Its usefulness depends on the proposal distribution, the scale of the objective function, and the cooling schedule for simulated annealing. Finite practical schedules carry no general guarantee of finding a global minimum.

Fix a failure probability $0<\alpha <1$ and assume an independent and identically distributed statistical sample. A finite-sample confidence band for its population cumulative distribution function isHere $\log$ is the natural logarithm. The DKW inequality impliesThis is simultaneous coverage over the whole real line, valid at each fixed sample size.^[ref]

To obtain the band, solve $2e^{-2n{\epsilon }^2}=\alpha$ for the positive $\epsilon$. Outside a failure event of probability at most $\alpha$, $|F_n(x)-F(x)|\le {\epsilon }_{n,\alpha }$ for every $x$. Since a cumulative distribution function already lies in $[0,1]$, clipping the endpoints to this interval preserves the coverage event. The guarantee is at least $1-\alpha$, and need not equal it.

Choose $\alpha =0.1$, giving at least $90\%$ simultaneous coverage under the stated model. For $n=5,50,500$, respectively, the half-widths before clipping areFor the chosen five observations, the band at $x=1$ is $[0,0.94733]$, while at $x=4$ it is approximately $[0.25267,1]$. At $x=3$ the unclipped band would be $[0.05267,1.14733]$, so clipping gives $[0.05267,1]$. The broad uncertainty is informative: five observations give a precise description of themselves, but a coarse bound for an unknown population cumulative distribution function.

A collection of $90\%$ pointwise confidence intervals only promises coverage separately at each preselected threshold; it does not promise $90\%$ coverage of an entire cumulative distribution function. Here the one simultaneous event permits inspecting any threshold after seeing the data. The probability refers to repeated sampling of the entire statistical sample, with $F$ fixed. It does not assign $90\%$ probability to a fixed $F$ after observing this particular band, and it is not a prediction interval for a future observation. The displayed $n=5,50,500$ comparison gives a guarantee for each fixed $n$ separately, not simultaneous validity over all sample sizes or an optional stopping rule. The chosen observations are an illustration; the coverage claim follows from the DKW inequality, not from their appearance or a simulation. Dependence or observations with different sampling laws require another argument.

At fixed $\alpha$, multiplying $n$ by $100$ divides the half-width by $10$. Conversely, a desired half-width $\delta >0$ is ensured by $n\ge \log (2/\alpha )/(2{\delta }^2)$, rounded up to an integer. For $\alpha =0.1$ and $\delta =0.1$, this sufficient sample size is $150$. This makes the confidence band useful for planning how many observations are needed before estimating threshold probabilities to a chosen simultaneous accuracy.

Offer $k$ independent uniform edges of the complete graph per row and permit any one to be selected from each row. If a chosen graph has no graph component of order at least $2n/3$, the balanced component cut provides an empty cut whose sides both have at least $n/3$ vertices. Each candidate edge crosses that cut with probability at least $4/9$, so a row can avoid crossing with probability at most $1-(4/9{)}^k$. The union bound over at most $2^n$ cuts gives failure probability at most $2^n[1-(4/9{)}^k{]}^{cn}$. Any fixed integer $c$ with $c[-\log (1-(4/9{)}^k)]>\log 2$ suffices. The guarantee is simultaneous even for choices made after seeing every offered edge.

For coefficient weights $w_{\lambda i}$ and independent mean-zero sub-Gaussian random variables of parameter $\sigma$, put $q_{\lambda }^2={\sum }_i{w}_{\lambda i}^2$. Multiplying moment-generating-function bounds shows that the weighted noise has parameter $\sigma q_{\lambda }$. The exponential Markov inequality, optimized in its parameter, gives tail bound $2\exp (-t^2/(2{\sigma }^2{q}_{\lambda }^2))$. At the displayed thresholds, a union bound over $N$ coefficients gives simultaneous control with probability at least $1-\delta$. No coefficient independence is required. Integrated wavelet weights satisfy $q_{\lambda }\le n^{-1/2}$. Finite variance alone only supports the weaker Chebyshev choice ${\tau }_{\lambda }=\sigma q_{\lambda }\sqrt{N/\delta }$.

For N independent equivalent ion channels with open probability $P_O$ and approximately Ohmic open-state electrical conductance $\gamma$, the mean electric current is $I=N{P}_O\gamma (V-E)$. A numerical Electrical conductance cannot be inferred without the relevant reversal potential or driving voltage.

For an ideal open ion channel with constant electrical conductance $g$, membrane potential $V$ and reversal potential $E_{\mathrm{rev}}$, define outward electric current as positive:This is an Ohm's law model of the open state, not a universal channel law or a rule for opening probability. In the synthetic figure, $g=10\mathrm{pS}$ and $E_{\mathrm{rev}}=0\mathrm{mV}$. Because $1\mathrm{pS}\cdot 1\mathrm{mV}=10^{-3}\mathrm{pA}$, the open-state electric currents at $-80$ and $-40\mathrm{mV}$ are $-0.8$ and $-0.4\mathrm{pA}$. Zero electric current at the reversal potential does not imply a closed ion channel. Real ion channels can have nonlinear responses, so the straight line is an explicit idealization.

Consider a logistic choice rule $P(\mathrm{accept})=1/(1+\exp (-U))$ with $U=\alpha R-\kappa E$ and fixed noise scale. At one positive offer $(R_0,E_0)$, changes satisfying $\delta \alpha R_0=\delta \kappa E_0$ leave utility and acceptance probability unchanged. One offer therefore cannot identify reward gain and effort cost separately.

For example, $(\alpha ,\kappa )=(0.6,0.2)$ and $(1,1)$ both give $U=1$ at $(R,E)=(2,1)$. Their predictions differ at other offers. Independently varying reward and effort can identify the two coefficients within this model when the design has rank two and choice data are informative. If noise scale is also unknown, its tradeoff with the coefficients requires additional constraints.

If $|b⟩$ lies in the kernel of $A$, then $A|b⟩=0$ has no normalized quantum state. A request to produce that normalized vector with positive probability is consequently undefined. For example, $A=\mathrm{diag}(0,1/2)$ on one qubit has distinct dyadic eigenvalues, but annihilates $|0⟩$. The corrected filtering assertion requires $A|b⟩\ne 0$; it cannot be repaired by assigning a positive success probability to this input.

In independent site percolation with parameter $p$, each vertex of a graph is open with probability $p$, independently of the others. Retain those vertices and every original edge whose two endpoints are retained. This gives an induced subgraph, whose connected components of a graph are the percolation clusters. A closed vertex is absent from the usable network, whereas bond percolation keeps all vertices and instead removes individual edges.

A specified simple graph path of $m$ edges has $m+1$ distinct vertices. Its probability of being entirely open is therefore $p^{m+1}$, including the starting vertex. Conditional on that vertex being open, it is $p^m$ by independence. For a two-edge graph path at $p=1/2$, the unconditional probability is $1/8$, compared with $1/4$ in bond percolation. This comparison concerns a fixed route; percolation critical probabilities require analysis of all possible routes on the specified infinite graph.

Under the studentized normal sample mean assumptions, let $t_{q,\nu }$ denote the quantile at cumulative probability $q$ of the Student's t-distribution. For $0<\alpha <1$, rearranging the event $|T|\le t_{1-\alpha /2,n-1}$ gives a Student t confidence interval for the unknown mean $\mu$ with exact coverage probability $1-\alpha$. ^[ref]

Choose the list $8,9,10,11,12$ as an original arithmetic example, not as reported measurements or a claim that these five numbers establish a normal distribution. If such values were obtained under the assumed independent and identically distributed normal distribution model with unknown $\sigma$, thenThere are four statistical degrees of freedom. For a 95% confidence interval, $t_{0.975,4}=2.776445105\dots$, soUsing the standard normal quantile $z_{0.975}=1.959963985\dots$ with the same estimated standard error would give only $10\pm 1.385903824\dots$. That underestimates the correct margin by about 29.4%. Indeed its actual coverage probability under this normal distribution model is $\mathbb{P}(|t_4|\le z_{0.975})\simeq 0.8784$, rather than 0.95.

The observations, the sample mean and the confidence interval all have different meanings. The bar estimates the population mean; it is not an interval intended to contain 95% of individual observations. In repeated independent samples of size five from the assumed normal distribution, 95% of intervals constructed using the exact $t$ quantile cover the same fixed $\mu$. This is not a 95% posterior probability that the fixed statistical parameter lies in this particular realized interval. Such a probability requires a separate Bayesian inference model.

For this four-degree case, the cumulative distribution function has an especially simple form:One can check the formula by differentiation: $F_4^{\prime }(t)=\frac{3}{8}(1+t^2/4{)}^{-5/2}=f_4(t)$, with limits zero and one at the two ends of the real line. Solving the increasing equation $F_4(t)=0.975$ by the bisection method gives the quantile used above.

A smoothing bandwidth $h>0$ determines the scale over which a regression kernel averages data, or a kernel for density estimation spreads each observation. It has the same units as the observations. In kernel density estimation, changing $h$ changes the visible detail while preserving the total integral one. For a Gaussian density kernel, each bump has standard deviation $h$ and its individual contribution has peak height $1/(nh\sqrt{2\pi })$.

If the smoothing bandwidth is too small relative to the spacings in a statistical sample, the kernel density estimator shows narrow spikes tied to individual observations: apparent peaks may be sampling accidents. If it is too large relative to features of the unknown probability density function, it blurs distinct groups together and spreads fitted probability far from the observations. An intermediate smoothing bandwidth can reveal broad structure, but visual smoothness alone cannot establish accuracy.

The first three panels use exactly $(-1.8,-1.4,-1.1,-0.7,0.7,1.1,1.4,1.8)$ with smoothing bandwidths $0.08$, $0.35$ and $1.2$, respectively. Every curve has area one; the two groups in the observed statistical sample become less distinct as $h$ increases. No true probability density function is specified in this illustration, so the middle panel is illustrative, not a certified optimal fit. The fourth panel uses $(0.1,0.4,0.9)$ and $h=0.5$ to illustrate boundary bias of a kernel density estimator.

For a fixed nonnegative kernel for density estimation $K$ with finite squared integral, a probability density function $f$ that is bounded and continuous at an interior point $x$ with $f(x)>0$, and a deterministic sequence $h=h_n\downarrow 0$, the pointwise sampling variance obeysThus decreasing the smoothing bandwidth increases the leading sampling variance while reducing interior smoothing bias of an estimator: this is a bias-variance tradeoff. The conditions $h_n\to 0$ and $n{h}_n\to \infty$ express two goals, increasingly fine resolution and enough observations contributing locally. Practical methods such as plug-in bandwidth selection use the data to choose $h$; pointwise optimal kernel bandwidth depends on local features of the unknown probability density function. The smoothing bandwidth remains a tuning choice in nonparametric statistics, and knowing the support of a probability distribution can require more than a width adjustment near a boundary.

A prescribed-velocity motor can attach only over a narrow displacement window, remain attached over a protected interval, then detach with constant rate. Converting time to displacement gives a rising attachment probability, a plateau $p_b$, and an exponential tail with spatial length $v/k_{off}$. Independent successive-site encounters require detachment before the next window, not merely a large geometric site separation. Earlier attachments can be retained in a periodic availability correction.

Where a nonzero wavefunction can be written as $\psi =\sqrt{\rho }{e}^{i\theta }$, and there is no vector potential, substituting into the Schrodinger probability current gives $\mathbf{j}=\rho \hslash \nabla \theta /m$. The direction of probability flow is determined by the phase gradient rather than the slope of the real part. A plane wave and its complex conjugate have the same real part and density but opposite currents. In a vector potential the gauge-invariant kinetic momentum replaces $\hslash \nabla \theta$.

For spin one-half, choose the normalized spin eigenstates $|+z⟩,|-z⟩$ with $S_z|\pm z⟩=\pm (\hslash /2)|\pm z⟩$. The spin one-half along an axis pure state in the $xz$ plane isIts norm is one because the two eigenstates are orthogonal and ${\cos }^2(\theta /2)+{\sin }^2(\theta /2)=1$. Applying the Born rule to a $z$ projective measurement givesConsequently the expectation value, second moment, and variance areAt $\theta =\pi /3$, the outcome probabilities are $3/4$ and $1/4$, the mean is $\hslash /4$, and the variance is $3{\hslash }^2/16$. Every individual outcome remains $\pm \hslash /2$. At $\theta =\pi /2$, the mean is zero and the variance is maximal, ${\hslash }^2/4$. At either pole, one outcome has probability one and the variance vanishes.

The diagonal mixed state ${\rho }_{\mathrm{diag}}=p_{+}|+z⟩⟨+z|+p_{-}|-z⟩⟨-z|$ has the same $S_z$ probabilities, expectation value, and variance as $|{\psi }_{\theta }⟩$. It is nevertheless a different quantum state for $0<\theta <\pi$: the displayed pure state has $⟨S_x⟩=\hslash \sin \theta /2$, while ${\rho }_{\mathrm{diag}}$ has $⟨S_x⟩=0$. A single component's mean does not reveal all relative quantum phases or establish purity of a density operator.

For a spin one-half system, an ideal projective measurement along a unit vector $\mathbf{n}$ measures ${\sigma }_{\mathbf{n}}=\mathbf{n}\cdot \mathbf{\sigma }$. Its eigenvalues are $\pm 1$, corresponding to physical spin angular momentum components $\pm \hslash /2$, and its orthogonal projections are ${\Pi }_{\pm }=(I\pm {\sigma }_{\mathbf{n}})/2$. For density matrix $\rho =(I+\mathbf{r}\cdot \mathbf{\sigma })/2$, the Born rule and the Pauli matrix multiplication law give the displayed probabilities. Thus a spin measurement along one direction tests only that component of the Bloch vector; measurements along three independent directions determine the entire density matrix.

For a state initially $z$-up and precessing about $y$, measurement along $\mathbf{n}=(0,\sin \theta ,\cos \theta )$ has the displayed probabilities. eigenvectors can be chosen as $(\cos (\theta /2),i\sin (\theta /2){)}^T$ and $(i\sin (\theta /2),\cos (\theta /2){)}^T$. Their phases must be retained when taking inner products. Full $2\pi$ rotation changes their signs but leaves isolated-state probabilities unchanged.

In the ordered z-projection basis $|1,1⟩,|1,0⟩,|1,-1⟩$, the normalized eigenvector of $J_x\sin \theta +J_z\cos \theta$ with eigenvalue $+\hslash$ is the displayed vector. It follows by a y-axis rotation of the maximum-weight state, or direct solution of the tridiagonal eigenvalue equation. Squaring its entries gives the three z-projection measurement probabilities.

Prepare the $J_1=+\hslash$ spin coherent state and evolve with Hamiltonian operator $H=\omega J_3$ for time $t$. In the descending-$J_3$ basis of the spin-one angular momentum matrices, the initial vector is $(1,\sqrt{2},1{)}^{\mathsf{T}}/2$ and the evolved vector is $(e^{-i\theta },\sqrt{2},e^{i\theta }{)}^{\mathsf{T}}/2$, where $\theta =\omega t$. The $J_1$ orthonormal basis has vectors $(1,\sqrt{2},1{)}^{\mathsf{T}}/2$, $(1,0,-1{)}^{\mathsf{T}}/\sqrt{2}$ and $(1,-\sqrt{2},1{)}^{\mathsf{T}}/2$. Taking their inner products with the evolved vector gives probability amplitudes $(1+\cos \theta )/2$, $-i\sin \theta /\sqrt{2}$ and $(\cos \theta -1)/2$. Their squared moduli give the Born rule probabilities in the title. At integer half-turns $\theta =\pi k$, the outcome is deterministically $(-1{)}^k\hslash$.