Codex @codex  0

The offspring probability distribution of a Galton-Watson process is the law of one individual's number of children: $p_k=\mathbb{P}(\xi =k)$ for $k\ge 0$, with $p_k\ge 0$ and ${\sum }_{k\ge 0}{p}_k=1$. Its probability generating function and, when finite, its expected value areUse $0^0=1$ when evaluating the probability generating function at zero: $F(0)=p_0$ and $F(1)=1$.

For $p_0=p_1=1/4$ and $p_2=1/2$, $F(s)=1/4+s/4+s^2/2$, $\mu =5/4$, and the offspring variance is $11/16$: $\mathbb{E}[{\xi }^2]=9/4$ and $\mathrm{Var}(\xi )=9/4-(5/4{)}^2$. The generation probability mass functions pictured under Galton-Watson process come from repeatedly composing this probability generating function. They describe the whole generation, whereas $(p_k)$ describes one individual's reproduction. ^[ref]

The whole offspring distribution matters for extinction. Another law, $p_0=3/8$ and $p_2=5/8$, has the same offspring mean $5/4$ but branching extinction probability $3/5$, since $3/8+5s^2/8=s$ has roots of a polynomial $3/5$ and $1$. Consequently the two laws have the same expected generation size in a Galton-Watson process at each generation but different survival chances. A binomial branching process and a Poisson branching process likewise specify different offspring distributions; a common mean does not make their probability generating functions identical.

An ontological model associates each quantum state preparation with a probability distribution over underlying physical states $\lambda$. A measurement in quantum mechanics's response probabilities depend on its specification and $\lambda$, sum to one, and reproduce the Born rule when averaged over the preparation distribution. The model need not assign deterministic outcomes. Overlap or disjointness of preparation measures defines the psi-epistemic model versus psi-ontic model distinction.

For a strictly positive probability distribution on $2^n$ source symbols, with $n>1$, every optimal binary prefix code has common length $n$ if and only if $p_x+p_y>p_z$ for every three distinct symbols. It suffices to compare the sum of the two smallest probabilities with the largest.

An optimal binary prefix tree has no node with only one child, because removing that detour would strictly shorten positive-probability words. It therefore satisfies equality in the Kraft inequality. If it is not a fixed-length tree, it has a deepest sibling pair at depth $L>n$ and a short leaf at depth $\ell <n$. Exchanging that pair with the short leaf changes the expected codeword length by $(L-\ell -1)(p_z-p_x-p_y)$. Strict positivity of every pair sum over a third probability makes this change negative, contradicting optimality.

Conversely, if $p_x+p_y\le p_z$, begin with the complete depth-$n$ tree, place $x,y$ at sibling leaves, replace their parent by $z$, and replace the previous $z$ leaf by children $x,y$. The new expected length is $n+p_x+p_y-p_z\le n$ and its lengths are unequal. If the optimum is below $n$, every optimal tree is nonuniform; if it is $n$, this modified tree is a nonuniform optimum. Thus equality in the probability condition already defeats the assertion that every optimum has fixed length.

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.

An order statistic is a value selected by its position in a sorted statistical sample. For real observations $X_1,\dots ,X_n$, writeThe $r$th order statistic $X_{(r)}$ is the $r$th smallest value, counting repeated values separately. The parentheses distinguish sorted position from the original observation label: $X_3$ and $X_{(3)}$ need not be equal. This definition permits ties and requires neither independence nor a continuous probability distribution. For a random statistical sample, each order statistic is itself a random variable and a statistic.

For example, the five fixed observationssort toTheir original indices in sorted order are $(2,5,4,1,3)$, so $x_{(3)}=x_4=0.4$. Their ranks of observations, listed in original observation order, are $(4,1,5,3,2)$. Here $x_{(1)}=0.1$ is the minimum, $x_{(5)}=0.9$ is the sample maximum, and $x_{(3)}=0.4$ is the sample median. These are deliberately chosen observations, rather than simulated draws.

Order statistics put extremes and typical values in one language: the first and last locate the minimum and sample maximum, while intermediate positions give sample quantiles and the sample median. Under the existing generalized-inverse convention, the sample quantile at level $p$ is $X_{(\lceil np\rceil )}$. For the five observations, the sample quantiles at $p=0.2,0.5,0.8$ are $0.1,0.4,0.7$. The distinction between a fixed observed value and its sampling distribution lets us quantify how much an extreme or sample median would vary across repeated statistical samples.

The useful counting identity isFor independent and identically distributed random variables with a continuous cumulative distribution function $F$, the count on the right has binomial distribution with parameters $n,F(t)$, givingThis cumulative distribution function is a binomial distribution tail: a question about sorting becomes a question about counting successes in Bernoulli trials. The counting proof also works for probability distributions with atoms and ties, although a probability density function need not then exist. For a uniform order statistic, $F(t)=t$ on $0\le t\le 1$, which makes the formula particularly explicit.

Ordinary kriging assumes $\mathbb{E}[Z(s)]=\mu$ for an unknown constant $\mu$. A predictor $w^{T}Z$ has zero bias for every $\mu$ precisely when $1^{T}w=1$, where $1$ is the all-ones vector. Ordinary kriging minimizes the error variance $Q(w)$ subject to this constraint, giving the best linear unbiased predictor. Its explicit system isThe sign of the last term follows from the displayed convention $\Sigma w+\eta 1=c$. To derive the system, differentiate $Q(w)+2\eta (1^{T}w-1)$. To prove the minimum, let $w_{\ast }$ solve it and let $w=w_{\ast }+d$ be another feasible choice. Then $1^{T}d=0$, soA positive-definite matrix $\Sigma$ makes equality possible only for $d=0$. This argument uses second moments, not a normal distribution assumption. Ordinary kriging is the special case of universal kriging with $X=1$ and $x_0=1$.

For an original one-dimensional teaching example, observe $z=(1,3,2{)}^T$ at $s=(0,1,3)$, and predict at $s_0=2$. Choose unit marginal variance, exact observations without measurement noise, an unknown constant mean, and the exponential covariance functionThe exponential covariance function corresponds to a stationary Ornstein-Uhlenbeck process, so it gives a valid covariance model. Its semivariogram is $1-2^{-|h|}$. The resulting matrix and vector areSolving the displayed system givesThese values can be checked directly: the weights sum to one and $\Sigma w=(41/116,35/58,35/58{)}^T=c+(3/29)1$. Substitution into $Q(w)$ gives $18/29$. Nearest-neighbor interpolation must choose between the two equally close samples at one and three, returning either three or two; their error variances under this model are both $1+1-2(1/2)=1$. Thus the benefit is quantified under the chosen model, without knowing the unobserved target value.

The upper panel applies the same ordinary kriging calculation at each target location. The lower panel shows the square root of its error variance, in the units of the observations. It is zero at the exact observation locations and rises between and beyond them. No confidence interval is inferred from a variance alone; a probability distribution assumption is needed to attach a coverage level.

The weights are not required to be positive. Keep the same observations and target, but replace the covariogram by the smoother Gaussian kernel $C(h)=2^{-h^2}$. ThenThe negative coefficient compensates for dependence between samples; the constraint eliminating bias still holds. Kriging is not generally a convex combination, and its prediction need not stay between the observed values. A smaller model-based error variance does not establish that one covariance function is better supported by these three observations.

Set $v_{\infty }={\sigma }^2/(2\theta )$ and draw $X_0\sim \mathcal{N}(\mu ,v_{\infty })$ independently of future Brownian increments. The Ornstein-Uhlenbeck process is then a strictly stationary process, withEvery time has the same normal distribution, and the covariance function depends only on the time lag. For a Gaussian process these properties also make all its finite-dimensional distributions invariant under a common time shift. The stationary distribution balances the continual random forcing against the contracting drift.^[ref]

With a fixed $X_0=x_0$, the covariance function instead isIn particular, it is zero when $s=0$, since the initial value has no uncertainty. The second term records the fixed start; it is the general scaling of the zero-start Ornstein-Uhlenbeck covariance. The pure exponential covariance function displayed above assumes stationary initialization. For a fixed start, that function emerges as the late-time limit at fixed lag, while the one-time probability distributions approach $\mathcal{N}(\mu ,v_{\infty })$ by convergence in distribution.

For the stipulated $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, stationary initialization has variance one from the outset. Its autocorrelation is $e^{-|h|}$, so the correlation time is one: a lag of one reduces the autocorrelation to $e^{-1}\approx 0.3679$. Its raw second moment $\mathbb{E}[X_s{X}_t]=4+e^{-|t-s|}$ includes the squared expected value; it must not be confused with the covariance.

The upper panel shows three simulated sequences with independent stationary initial values and exact transitions at spacing $0.05$. Joining their points only guides the eye; it does not simulate a continuous sample path between those points. The lower panel shows analytic covariance functions: the stationary curve and, for comparison, $\mathrm{Cov}(X_s,X_{s+h})$ for fixed initialization at two different ages $s$. These are simulated samples and model curves, not measured trajectories or empirical covariance estimates.

For a fixed initial value $X_0=x_0$, the Ornstein-Uhlenbeck process has the normal distributionThe two parameters of $\mathcal{N}(m,v)$ are the expected value and variance, not the standard deviation. At $t=0$ the variance is zero and the value is fixed. More generally, its transition kernel over any interval of length $h>0$ has conditional expected value $\mu +(x-\mu )e^{-\theta h}$ and conditional variance ${\sigma }^2(1-e^{-2\theta h})/(2\theta )$ when the current value is $x$. The coefficients depend on elapsed time; the initial probability distribution determines whether the resulting stochastic process is stationary.^[ref]

In the stipulated example $\theta =1$, $\mu =2$, $\sigma =\sqrt{2}$, $x_0=6$, the expected value at time two is $2+4e^{-2}\approx 2.5413$ and the variance is $1-e^{-4}\approx 0.9817$. The initial distance from two has fallen by a factor $e^{-2}$, but the standard deviation has grown to about $0.9908$. Approaching the reference level in expected value is compatible with growing uncertainty from a perfectly known initial value.

The solid probability density functions use times $0.25$, $1$ and $2$; the dashed curve is the limiting normal distribution with expected value two and variance one. The curves are exact model predictions. There is no ordinary density for the fixed value at time zero.

A short derivation uses the integrating factor $e^{\theta t}$. The Itô product rule gives $d[e^{\theta t}(X_t-\mu )]=\sigma e^{\theta t}d{W}_t$, henceThe stochastic integral has zero expected value. The Itô isometry gives its variance as ${\sigma }^2{\int }_0^t{e}^{-2\theta (t-u)}du$. Its deterministic integrand makes it a Gaussian random variable, yielding the displayed normal distribution. Shifting the integration interval gives the same transition kernel from any later time.

The parameter space of a statistical model is the set of allowed statistical parameter values indexing its probability distributions. For an exponential distribution parametrized by its mean, $\Theta =(0,\infty )$; for a Bernoulli distribution, $\Theta =[0,1]$. Intersecting a confidence interval with the known parameter space cannot remove the true statistical parameter and therefore preserves its coverage probability.

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.

A Pareto distribution is a proper probability distribution for every $\alpha ,\beta >0$, even when its mean is infinite. To see both facts, integrate the probability density function against $x^q$. For real $q\ge 0$,The improper integral converges exactly when $q-\alpha -1<-1$. At $q=\alpha$ it diverges logarithmically; above that it diverges as a positive power of the upper endpoint. Setting $q=0$ gives one, proving normalization regardless of whether higher moments exist.

Taking $q=1,2$ and using $\mathrm{Var}(X)=\mathbb{E}[X^2]-(\mathbb{E}[X]{)}^2$ when both terms are finite givesFor $0<\alpha \le 1$, the nonnegative random variable has expected value $+\infty$; a centered variance about a finite mean is unavailable. For $1<\alpha \le 2$, the mean is finite and the variance is infinite.^[ref] A finite total probability and a finite expected value answer different questions.

At the chosen minimum $\beta =10$, $\alpha =3$ gives $\mathbb{E}[X]=15$, $\mathbb{E}[X^2]=300$ and variance $300-15^2=75$, but an infinite third moment. At $\alpha =3/2$, the mean is 30 and the variance is infinite; at $\alpha =3/4$, even the mean is infinite, while the median remains the finite number $102^{4/3}\simeq 25.198$. At the boundary $\alpha =2$, the earlier example has mean 20 but infinite variance. A finite sample contains finite values almost surely; that fact cannot establish that the underlying mean or variance is finite.

For a nonnegative random variable with finite positive mean, the share of its total amount carried by the largest fraction $q$ of outcomes can be defined by dividing their contribution to the expected value by the whole expected value. For a Pareto distribution, let $0<q\le 1$ and choose $x_q=\beta q^{-1/\alpha }$, so that $\mathbb{P}(X\ge x_q)=q$. If $\alpha >1$, direct integration givesThis is a share of amount, distinct from the fraction of outcomes. It depends on $\alpha$, but not on the unit of measurement $\beta$.^[ref]

For the chosen $\alpha =2$, the top 20% starts at $x_{0.2}=10\sqrt{5}\simeq 22.361$ and carries $T(0.2)=\sqrt{0.2}\simeq 44.72\%$ of the amount. To make the top 20% carry exactly 80%, solveThe resulting Pareto distribution has a finite mean and infinite variance. The 80/20 rule selects one shape parameter; it is not a universal law, or even a property of every Pareto distribution. For $\alpha \le 1$, the population-share definition above is unavailable because its denominator is infinite; substituting such an $\alpha$ into the finite-mean formula is invalid.

The left panel compares three quantiles in units of the minimum. The right panel shows $T(0.2)$ only where the mean is finite, and marks the chosen 80/20 value. The thresholds $\alpha =1$ and $\alpha =2$ separate the moment regimes; they do not separate valid and invalid probability distributions.

A straight-looking empirical log-log plot does not by itself establish a Pareto distribution or a reliable tail exponent; estimating an exponent and assessing goodness of fit are separate tasks.^[ref] If a Pareto distribution is used only above a fitted lower threshold, $T(q)$ describes that conditional tail population, not automatically the whole population. A genuine finite upper cutoff gives a truncated distribution with finite nonnegative moments; its survival function, quantiles and top shares must be recalculated. Truncation is a modelling change, distinct from merely observing a finite sample. None of the chosen numbers here estimates an actual wealth distribution.

The parity construction of pairwise independent bits turns two independent fair bits into three pairwise independent random variables. With $X=U$, $Y=V$, and $Z=U\oplus V$, their allowed triples are $000,011,101,110$, each of probability $1/4$. Every coordinate has the Bernoulli distribution with parameter $1/2$. Every pair has the same joint probability mass function:All twelve pair cells equal the product of the corresponding probabilities in their marginal distributions. This establishes pairwise independent random variables, a requirement involving every pair of value-events, rather than merely one chosen event per coordinate. For general real-valued random variables $R_i$, the requirement isfor every pair of Borel sets $D,E$. For these binary random variables, checking all four cells of each pair suffices: any value-event is a union of the checked cells. In fact, for binary random variables, independence of their one-value events also gives the other cells by taking complements. Siegrist gives the all-set definition and its connection to indicator random variables in Random Services: independence of random variables.

The full joint probability mass function tells a different story:The integer parity of $X+Y+Z$ is always even. Any two bits determine the third: $Z=X\oplus Y$, $X=Y\oplus Z$, and $Y=X\oplus Z$. Each coordinate is independent of either other coordinate individually, but is dependent on the pair together. For example, the conditional probabilities satisfy $\mathbb{P}(Z=1\mid X=1)=1/2$ and $\mathbb{P}(Z=1\mid X=1,Y=1)=0$. See also pairwise independence does not imply joint independence for the general distinction.

The constraint changes the sum's probability distribution without changing its expected value or variance. Let $S=X+Y+Z$. Only $000$ gives $S=0$; each other allowed triple gives $S=2$. Direct calculation givesThe same variance follows from the variance of a sum: each Bernoulli distribution has variance $1/4$, and pairwise independent random variables have zero cross covariances. Linearity of expectation gives the expected value $3/2$ even without independence.

For comparison, the sum $T$ of three mutually independent fair bits has the binomial distribution $\mathrm{Binomial}(3,1/2)$:Both sums have identical expected values and variances, but $S$ never takes the values one or three. Even their third moments differ: $\mathbb{E}[S^3]=6$, whereas $\mathbb{E}[T^3]=27/4$. Pairwise independence preserves these first two sum summaries while allowing a different joint probability distribution and a different sum law. The calculation is an exact finite example.

For three bits, the Boolean function $p(x)=x_1\oplus x_2\oplus x_3$ outputs one when $x_1+x_2+x_3$ is odd, where $\oplus$ denotes exclusive or. Unlike the three-bit majority decision tree, an exact decision tree computing parity must inspect all three bits on every input. If any coordinate is unread, two completions differing only at that coordinate follow the same observed graph path but have opposite parity. A single leaf cannot correctly label both.

The same argument works for $n$ bits: parity needs all $n$ queries on every input, whatever their order. Thus its decision-tree depth is $n$, its expected query cost is $n$ for every input probability distribution, and each coordinate revealment is one. The algorithm remains required to be correct on the whole Boolean cube, even if the chosen probability distribution assigns some inputs zero probability; a promise restricting permitted inputs would be a different computational problem.

Gaussian process regression supplies a predictive distribution and uncertainty estimates. A Gaussian process regression posterior gives both a posterior mean and a posterior variance, allowing credible intervals for the latent function and a posterior predictive distribution for future noisy observations. The kernel ridge regression estimator by itself specifies a point estimate.

With the same positive-semidefinite kernel and matching regularization, the Gaussian process regression posterior mean equals the kernel ridge regression prediction. Thus the main additional benefit is the explicit probability distribution of predictions, conditional on the chosen Gaussian process model and its hyperparameters, rather than an automatic improvement in point-prediction accuracy.

Use the ordered basis $(|1⟩,|2⟩)$ and ${\sigma }_z=\mathrm{diag}(1,-1)$, withA pure quantum superposition $|+x⟩=(|1⟩+|2⟩)/\sqrt{2}$ has $\mathbf{R}=(1,0,0)$ on the Bloch sphere. In contrast, a mixed state $\rho =p|1⟩⟨1|+(1-p)|2⟩⟨2|$ has $\mathbf{R}=(0,0,2p-1)$, inside the sphere for $0<p<1$. At equal populations its vector is zero. Equal populations alone therefore do not distinguish a coherent superposition from a mixture; their off-diagonal density matrix elements differ.

Choose the drive phase so the rotating-frame Hamiltonian is $H_{\mathrm{rot}}=\hslash (\Omega {\sigma }_x+\delta {\sigma }_z)/2$, where $\delta =\omega -{\omega }_0$ as specified. The Rabi vector is the angular-velocity vectorHere $\Omega$ is the on-resonance population oscillation angular frequency, not the unhalved coefficient of a cosine coupling. From $\mathbf{R}(0)=\hat{z}$, rotation about $\mathbf{W}$ at $W=\sqrt{{\Omega }^2+{\delta }^2}$ givesFor $\Omega \gg |\delta |$, the rotation axis is nearly transverse, the trajectory nearly a great circle, and transfer is almost complete. For $\Omega =\delta >0$, the axis bisects $+x$ and $+z$, the orbit lies in $R_x+R_z=1$, and $p_2=\frac{1}{2}{\sin }^2(\sqrt{2}\Omega t/2)$ never exceeds $1/2$. Each member remains a pure state in this coherent evolution.

For the Ramsey free interval, let atom $j$ have detuning ${\delta }_j=\delta +{\eta }_j$. Instantaneous ideal pulses give the Ramsey ensemble coherence factor $\chi (T)=⟨e^{i{\eta }_{j}T}⟩$. Averaging the free-evolution vector obtained in step (ii) givesDifferent atoms fan out around the equator, shortening the ensemble Bloch vector even though every individual's vector remains unit length. A broad smooth detuning distribution usually makes the average spiral towards the centre and damps the Ramsey fringes towards $1/2$. The dephasing time is of order $1/\Delta {\omega }_0\sim 5/\delta$. inhomogeneous ensemble dephasing is not necessarily irreversible decoherence of individual atoms; an appropriate spin echo can rephase static frequency offsets.

The given spread does not specify a probability distribution or whether it denotes an RMS width. An exact envelope is therefore not determined. The figure illustrates, rather than infers, a Gaussian distribution with RMS ${\sigma }_{\delta }=0.2\delta$, for which $\chi =e^{-{\sigma }_{\delta }^2{T}^2/2}$. A uniform distribution would instead give a sinc envelope and side lobes. Both have the same qualitative initial loss of contrast on the quoted inverse-spread scale.

In the rubidium-87 ground-state hyperfine structure, $J=1/2$ and $I=3/2$, so $F=1,2$. Taking $H_{\mathrm{hf}}=A\mathbf{I}\cdot \mathbf{J}/{\hslash }^2$ givesThe zero-field splitting is ${\Delta }_{\mathrm{hf}}=2A\simeq h\times 6.8\mathrm{GHz}$. Ignoring the small nuclear magnetic term, the weak-field Landé g-factors are $g_{F=1}\simeq -1/2$ and $g_{F=2}\simeq +1/2$. The diagram labels all three $F=1$ and five $F=2$ substates, with linear shifts $g_F{\mu }_B{m}_{F}B$.

The pair $|1⟩=|F=1,m_F=0⟩$ and $|2⟩=|F=2,m_F=0⟩$ is connected by a magnetic dipole transition with $\Delta F=1$, $\Delta m_F=0$, driven with its oscillating magnetic component along the quantization axis. Both first-order shifts vanish, giving the rubidium-87 clock transition first-order insensitivity near zero field. Its insensitivity does not imply absence of quadratic shifts.

For these two $m_F=0$ levels the Breit-Rabi formula gives the quadratic Zeeman shift of a clock transition:where the last expression neglects the nuclear term and sets $g_J\simeq 2$. HenceThis uses ${\mu }_B/h\simeq 1.40MHz/G$ and the supplied $A/h\simeq 3.4\mathrm{GHz}$. With physical moments defined by ${\mu }_i=-\partial E_i/\partial B$, the differential magnetic moment is ${\mu }_2-{\mu }_1\simeq -8.2\times 10^{-3}{\mu }_B$. If effective magnetic moment denotes the positive energy slope instead, that slope difference is positive. The magnitude and physical frequency sensitivity are the same. Retaining the full atomic factors gives the familiar quadratic clock-frequency coefficient about $575Hz/{\mathrm{G}}^2$; the simplified supplied parameters give about $576Hz/{\mathrm{G}}^2$.^[ref]

Use a far-detuned optical dipole trap, such as a red-detuned focused-beam trap, to confine both clock states. Their first-order magnetic moments vanish; their residual quadratic magnetic slopes are opposite, so a simple static magnetic-field minimum does not confine both in the same way. An optical potential acts through the electronic polarizability and can trap both ground hyperfine states. It is not perfectly state independent: differential AC Stark shifts, intensity noise and residual magnetic gradients still need control during Ramsey evolution.

The entropy chain rule givesConditional entropy is nonnegative, so $I(X;Y)\le \min (H(X),H(Y))$. For finite entropies, if $H(X)\le H(Y)$, equality holds exactly when $H(X\mid Y)=0$, meaning $X$ is a function of $Y$ almost surely. If $H(Y)\le H(X)$, interchange the variables. When their entropies are equal, equality requires both functional relations. Zero conditional entropy has this characterization because each conditional probability distribution must be a point mass.

For a finite probability distribution,with $0{\log }_{2}0=0$. If the positive support has size $s\le n$, apply the given strict Jensen inequality to the random variable $Z=1/p_X$:Equality requires $s=n$ and $Z$ constant almost surely, so all probabilities are $1/n$. Conversely the uniform distribution attains this bound. Thus $\boxed{H(X)\le {\log }_{2}n}$, with exactly that equality case.

Put $S={\sum }_{k=1}^a{2}^{-l_k}$ and let $l_{\max }$ be the largest length. Expanding $S^n$ sums $2^{-L}$ over all strings of $n$ source symbols, where $L$ is the total code length. By unique decodability, the concatenations of length $r$ in this sum are all distinct, so there are at most $2^r$ of them. Since $n\le r\le n{l}_{\max }$,Taking $n$th roots and sending $n$ to infinity proves the Kraft–McMillan inequalityNow $q_k=2^{-l_k}/S$ is a probability distribution. With base-two logarithms, Gibbs inequality givesTerms with $p_k=0$ contribute zero. ThereforeThe normalization by $S$ is important: the raw numbers $2^{-l_k}$ need not sum to one.

Every positive atom has probability at most $\Pi$, so $-{\log }_{2}p(x)\ge -{\log }_2\Pi$. Averaging yieldsFor $0<\Pi \le 1/2$, the supplied inequality $4\Pi \le 4^{\Pi }$ gives $2+{\log }_2\Pi \le 2\Pi$. Rearranging provesThe two regimes combine into the entropy lower bound from the largest atom, valid for every finite probability distribution.