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Let the survival time $T$ have an exponential distribution with rate $\lambda =0.2$ per year. Its hazard function is $h(t)=0.2$ per year, its cumulative hazard is $H(t)=0.2t$ for $t$ in years, and its survival function is $S(t)=e^{-0.2t}$. For a subject still event-free after five years, the conditional probability of an occurrence within the next two years isThe original population's probability of an occurrence between those same endpoints is instead $S(5)-S(7)=e^{-1}-e^{-1.4}\approx 0.121282$. Dividing by $S(5)\approx 0.367879$ recovers the conditional answer. The memoryless property makes the two-year conditional answer independent of the preceding five years.

For just $\Delta =0.01$ year, the exact conditional event probability is $1-e^{-0.002}\approx 0.001998$, close to the instantaneous approximation $h(5)\Delta =0.002$. For two years that approximation gives $0.4$, whereas the exact answer is about $0.329680$. At the single exact time five, the event probability is zero. These are three different quantities: an instantaneous hazard function with inverse-time units, a short-interval approximation, and an exact finite-interval probability.

A simple locally finite point measure on $[0,\infty )$ with void probability $e^{-\lambda (A)}$ on finite unions of bounded intervals is a Poisson random measure with Lebesgue measure intensity. Void-event intersections factor, hence the occupation indicators on disjoint cells are independent. The dyadic occupancy approximation of a simple point measure is binomial with rare-event probability $1-e^{-2^{-n}}$ and converges to a Poisson count of mean the interval length. Independence survives joint convergence for disjoint intervals. Extension of the mean measure and the Laplace functional of a Poisson random measure to Borel functions identifies all Borel counts, not just intervals.

If bounded intervals have void probabilities $e^{-\lambda (I)}$ for Lebesgue measure $\lambda$, a fixed deterministic point $x$ has zero chance of being an atom. Surround it by intervals of lengths tending to zero and use $\mathbb{P}(M(\{x\})>0)\le 1-e^{-\lambda (I)}$. This holds for each fixed endpoint, not simultaneously for all real points of a nonempty point process.

If a normalized wavefunction has exterior tails $C{e}^{-q|x|}$ outside $|x|<a/2$, integration over both tails gives the title probability. For an even cosine interior $A\cos (kx)$, the remaining normalization contribution is $A^2[a/2+\sin (ka)/(2k)]$. Tail amplitudes defined at the edge rather than at the origin absorb the factor $e^{-qa/2}$, so their probability formula looks different despite describing the same state.

For a Galton-Watson process with one ancestor, the branching extinction probability is the probability that some generation is empty:The equality uses the absorbing state at zero. The complementary branching survival probability $1-q$ is the probability that every generation is nonempty. The Galton-Watson extinction fixed point computes $q$ from the offspring distribution.

For $p_0=p_1=1/4$, $p_2=1/2$, one ancestor has $q=1/2$. Starting instead from $m$ ancestors with independent descendant families givesAll $m$ families must die out. With three ancestors, eventual extinction has probability $1/8$ and survival has probability $7/8$; by generation two, extinction has probability $(11/32{)}^3=1331/32768\approx 0.040619$. Extra ancestors change the extinction probability without changing the offspring distribution. ^[ref]

For finite offspring mean $\mu$, the extinction classification is:

The critical exception is $p_1=1$: every individual has exactly one child, $F(s)=s$, and $Z_n=1$ forever from one ancestor. Thus $\mu =1$ but $q=0$; a blanket claim that every critical Galton-Watson process becomes extinct would be false. These cases concern the Galton-Watson process with its common offspring distribution and independence assumptions. ^[ref]

This probability answers whether any lineage persists, which the offspring mean alone cannot quantify in the supercritical case. A branching process conditioned on extinction uses the same fixed point to reweight reproduction and describe families known eventually to die out.

For the ferromagnetic binary Ising model, all-zero boundary measures increase and all-one boundary measures decrease as boxes exhaust the lattice, in stochastic domination. The nearest-neighbor spatial conditional probability distribution reduces each domain comparison to a mixture of boundary conditions for the smaller box. Probabilities that a specified finite set is all one consequently converge; inclusion-exclusion gives convergence of every finite Ising spin pattern. Their consistent limits define the two extreme infinite-volume measures. Existence does not assert that the two limits coincide.

A failure-probability budget assigns nonnegative numbers ${\delta }_i$ to a finite or countable family of failure events $F_i$ in a common probability space, with verified bounds $\mathbb{P}(F_i)\le {\delta }_i$. The union bound guarantees that the probability of any failure is at most $\min \{1,{\sum }_i{\delta }_i\}$. By De Morgan's laws and the complement rule of probability, the probability that every requirement succeeds obeysThus ${\sum }_i{\delta }_i\le \delta$ guarantees simultaneous success with probability at least $1-\delta$, even when all failure events depend on the same observations. The numbers ${\delta }_i$ are upper allowances; adding them is not a calculation of the exact failure probability.

For three requirements, give each a failure-probability budget ${\delta }_i=1/20$. The total budget is $3/20$, so simultaneous success has probability at least $17/20$. On a sample space of twenty equally likely outcomes, let $F_1=\{1\}$, $F_2=\{2\}$, and $F_3=\{3\}$. These disjoint events attain the budget and the success bound. If instead all three failure events equal $\{1\}$, their individual probabilities are unchanged, but exact failure has probability $1/20$ and simultaneous success has probability $19/20$. This second model shows how overlap can leave the guarantee conservative.

For $n$ requirements, an equal allocation ${\delta }_i=\delta /n$ is convenient; unequal allocations also work when some requirements are harder to satisfy. For infinitely many requirements indexed by $i\ge 1$, the allocation ${\delta }_i=\delta /2^i$ uses total budget $\delta$ by the geometric series. A fixed positive allowance for every requirement instead gives an infinite sum and no positive simultaneous-success guarantee from the union bound. The individual probability bounds must cover all the failure events being combined: omitting possible failure modes leaves the resulting guarantee about the listed requirements only. The same addition of error allowances underlies the Bonferroni correction in simultaneous hypothesis tests.

Fano's inequality turns remaining uncertainty into a constraint on decision errors. If an observation leaves many plausible answers, a decoder cannot usually recover the answer reliably. Its elegant idea is to separate two uncertainties: whether the guess was wrong, and which alternative was correct when it was wrong.

Let $X,Y$ be discrete random variables, with $X$ taking values in a finite alphabet $\mathcal{X}$ of cardinality $M\ge 2$. Choose any deterministic decoder $\hat{X}=f(Y)$ whose output lies in $\mathcal{X}$, and define its average error probability $p_e=\mathbb{P}\{\hat{X}\ne X\}$. With base-two logarithms, Fano's inequality statesHere $H(X\mid Y)$ is conditional entropy and $h_2$ is binary entropy, with $0{\log }_{2}0=0$ at the endpoints. Both sides are measured in bits; using natural logarithms throughout gives the same inequality in nats. No uniform source, independence of $X,Y$, or optimality of the decoder is assumed. Requiring its output to belong to the same alphabet is what leaves at most $M-1$ alternatives after an error.^[ref]

For an exact toy example, give $X$ the discrete uniform distribution on $\{A,B,C,D\}$. A channel reports $Y=X$ with probability $1-q$ and reports each of the other three symbols with probability $q/3$, where $0\le q\le 1$. Use the decoder $f(y)=y$. The output is uniform, and the conditional probabilities after any report $y$ areThus $p_e=q$ and direct calculation of the conditional entropy givesThis is equality in Fano's inequality for every $q\in [0,1]$. At $q=0.3$, each report leaves probabilities $(0.7,0.1,0.1,0.1)$, up to their order, and $H(X\mid Y)\approx 1.3568$ bits. The two contributions are $h_2(0.3)\approx 0.8813$ bits for the error flag and $0.3{\log }_{2}3\approx 0.4755$ bits for identifying the correct alternative after an error. These are theoretical probabilities, not measured data.

The identity decoder in this example minimizes error only for $0\le q\le 3/4$: its guessed symbol has the largest posterior probability exactly when $1-q\ge q/3$. At $q=3/4$, $X,Y$ are independent random variables and all four guesses tie. For $q>3/4$, guessing any one of the three other symbols has error probability $1-q/3<q$. Equality in Fano's inequality therefore does not certify an optimal decoder.

For usefulness, combine $h_2(p_e)\le 1$ with ${\log }_2(M-1)\le {\log }_{2}M$ to obtain the simpler consequenceFor a uniform $X$, $H(X\mid Y)={\log }_{2}M-I(X;Y)$, so this becomes $p_e\ge \max \{0,1-(I(X;Y)+1)/{\log }_{2}M\}$. This connects mutual information to unavoidable decision error in channel coding or classification in statistical learning; it can rule out reliable recovery regardless of the chosen decoder.^[ref]

The sharper right-hand side $F_M(p)=h_2(p)+p{\log }_2(M-1)$ increases from zero to ${\log }_{2}M$ on $0\le p\le 1-1/M$, then decreases: in the interior, $F_M^{\prime }(p)={\log }_2((M-1)(1-p)/p)$. When interpreting the inequality by inverting a curve, use its increasing branch to obtain a lower error threshold. A decoder that chooses a largest posterior probability, as in maximum a posteriori decoding, always has $p_e\le 1-1/M$, since the largest of $M$ conditional probabilities is at least $1/M$. Arbitrary decoders can have larger error. In particular, low conditional entropy does not force every decoder to work well: with a fair binary $X$ and $Y=X$, the rule $f(Y)=1-Y$ has $p_e=1$ even though $H(X\mid Y)=0$. Fano's inequality supplies a constraint, not an exact error prediction for each rule.

Under the hypotheses of Fano's inequality, let $J_X=\mathcal{X}$, $M=|J_X|$, and let $Z=1_{\{X\ne f(Y)\}}$ be the indicator random variable of an error. Its probability of being one is $p_e$. The $h$ in the displayed bound denotes binary entropy $h_2$ in bits. Since $Z$ is a deterministic function of $(X,Y)$, $H(Z\mid X,Y)=0$. Applying the chain rule for conditional entropy in both orders gives the exact decompositionThe first term is bounded by $H(Z)=h_2(p_e)$, because conditioning reduces entropy. For the second term, fix a report $Y=y$: if $Z=0$, then $X=f(y)$ is known; if $Z=1$, then $X$ lies among at most $M-1$ other symbols. Maximum entropy on a finite alphabet bounds the latter uncertainty by ${\log }_2(M-1)$. Averaging over the reports and the error flag givesThis proves Fano's inequality. Events of zero probability contribute zero to this average, so the proof also covers $p_e=0,1$ without assigning a conditional distribution to an impossible event.^[ref]

The decomposition also explains equality. The first bound is exact when the indicator random variable $Z$ and $Y$ are independent random variables: knowing the report does not change the error probability. The second is exact when, for every report with positive error probability, the wrong-symbol conditional distribution is uniform on all $M-1$ alternatives. Both conditions hold in the four-symbol toy channel of Fano's inequality. At $q=0$, the second condition is vacuous; at $q=1$, the flag is constant but three alternatives remain, giving $H(X\mid Y)={\log }_{2}3$. For $M=2$, there is just one alternative after an error, so the second term vanishes: all remaining uncertainty is whether the single alternative must replace the guess.

For a weak time-independent perturbation switched on at zero time, first-order time-dependent perturbation theory gives a transition probabilityThe frequency kernel has area $2\pi t$ and width of order $1/t$. Against a smoothly varying continuum of final states, it approaches $2\pi t\delta ({\omega }_f-{\omega }_i)$, giving the rate $2\pi |V_{fi}{|}^2\delta ({\omega }_f-{\omega }_i)/{\hslash }^2$. Using $E=\hslash \omega$ gives the energy form above. One integrates the Dirac delta function against the appropriate density of states; it is not an ordinary finite number for one isolated resonant level.

The time must be long enough to resolve the smooth continuum and short enough that the total transition probability remains small. For continuum variation scale $\Delta \omega$ and total rate $\Gamma$, the usual window is $(\Delta \omega {)}^{-1}\ll t\ll {\Gamma }^{-1}$. A discrete finite spectrum also limits this approximation through its recurrences.

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

A percolation measure has the finite-energy property if, conditional on all exterior states, every prescribed state of any finite collection of sites or edges has positive probability almost surely. This permits finite local modifications within events of positive probability. It is a hypothesis on the measure and does not follow merely from invariance under graph symmetries.

For the uniform-arrival residual mass martingale with total weight $0<S<1$, the first crossing of level one happens continuously: its only jumps are downward, and an arrival almost surely misses the finitely many deterministic possible crossing times. Stop at that crossing or at $r<1$. The stopped values lie between zero and one, and their limit as $r\uparrow 1$ is the crossing indicator. Optional stopping and dominated convergence identify its probability as $S$. At $S=1$ the strict boundary inequality already fails at time zero.

For a simple symmetric random walk starting at zero, let $T_1$ be its first hitting time of $1$. The infinite mean first passage of a simple symmetric random walk shows that $T_1<\infty$ almost surely but $\mathbb{E}[T_1]=\infty$. Since $S_{T_1}=1$, assuming a finite expected value for $T_1$ would let Wald's equation assert $1=0\mathbb{E}[T_1]=0$, a contradiction. The formal expression $0\cdot \infty$ is not an allowed use of the theorem.^[ref]

Compare $T_{\pm 2}=\inf \{n\ge 0:|S_n|=2\}$. From every surviving position in $\{-1,0,1\}$, three consecutive positive increments force absorption within three steps. Each fresh block has probability $1/8$ of this event, so $\mathbb{P}(T_{\pm 2}>3k)\le (7/8{)}^k$ and $\mathbb{E}[T_{\pm 2}]\le 24$. This establishes finiteness before invoking the theorem. If $e_j$ is the average remaining time from position $j$, the law of total expectation givesso $\mathbb{E}[T_{\pm 2}]=4$. Symmetry gives equal exit probabilities at $-2$ and $2$, and $\mathbb{E}[S_{T_{\pm 2}}]=0$, as Wald's equation predicts. This is a small gambler's ruin problem with two absorbing barriers.

Even for $T_1$, each bounded stopping time $T_1\wedge n$ satisfies $\mathbb{E}[S_{T_1\wedge n}]=0$. The stopped process tends to $1$ almost surely, but its expected value does not tend to $1$. Rare long negative excursions retain the balancing contribution. The existing stopped-walk martingale with almost sure but not L1 convergence explains the related failure of uniform integrability; almost sure convergence alone does not permit exchanging a limit and an expected value.

For $0<p<1$, every prescribed state pattern on a finite edge set has positive conditional probability given all exterior states: independence makes it $p^a(1-p{)}^b$ for $a$ prescribed open and $b$ prescribed closed edges. Thus a positive-probability exterior event remains possible after any finite local surgery. This is the finite-energy property of Bernoulli percolation, and applies equally to dual bond percolation.

For $N$ individuals in monogamous families of exactly two offspring, let ${\Phi }_G$ be mean pairwise kinship coefficient, $F_G$ mean inbreeding and ${\Phi }_{m,G}$ mean kinship of mates. A random distinct offspring pair are siblings with probability $1/(N-1)$. Accounting separately for within-family and cross-family pairs gives the displayed recurrence, while $F_{G+1}={\Phi }_{m,G}$. Random mating makes ${\Phi }_{m,G}={\Phi }_G$. Sibling avoidance changes the mating-pair mean; it does not create unrelated founders or stop eventual drift to ancestral identity in a closed finite population.

On the full Cartesian product $\Omega ={\mathbb{R}}^{\mathbb{N}}$ with its product sigma-algebra, take the coordinate random variables $X_j(x)=x_j$. A product-measurable set $A$ is a tail event if and only if membership is unchanged whenever two sequences differ in finitely many coordinates. The measurability condition is essential: invariance alone is not a license to assign a probability to an arbitrary subset.

For the forward implication, fix $m$. An event in $\sigma (X_{m+1},X_{m+2},\dots )$ cannot distinguish points with the same remaining coordinates. Since a tail event belongs to this sigma-algebra for every $m$, every finite change preserves membership.

For the converse, freeze the first $m$ coordinates to zero. The insertion mapis a measurable function for the product sigma-algebras. Thus $B_m=i_m^{-1}(A)$ is a measurable set. Invariance gives $A={\pi }_m^{-1}(B_m)$, where the coordinate projection ${\pi }_m$ discards the first $m$ entries. Hence $A\in \sigma (X_{m+1},X_{m+2},\dots )$ for every $m$. The same argument works on $\{0,1{\}}^{\mathbb{N}}$. On a general probability space, coordinates need not admit arbitrary replacements, so the defining intersection of sigma-algebras remains the reliable formulation.

Keep the first pulse's oscillation phase zero and give the second phase $\phi$ in the Driven two-level Rabi model. Its upper off-diagonal Hamiltonian operator entry is $\hslash \Omega e^{-i\phi }/2$. Multiplying the two pulse unitary operators and the free-flight unitary operator givesThe Born rule gives the displayed phase-dependent form using the $p,b,\theta$ of the Ramsey fringe envelope. Thus $\phi =0$ reproduces the existing expression, while $\phi =\pi$ turns a resonant transfer maximum into zero transfer. A single pulse's population is insensitive to its drive phase; the second coherent interaction makes their difference observable. An uncontrolled, uniformly random $\phi$ instead averages this probability to $2p(1-p)$, leaving the broad pulse response but removing the dark-flight fringes.

Finite pulses also contribute to the near-resonant frequency slope. For $\Omega \tau =\pi /2$, differentiating $\theta$ at $\delta =0$ gives ${\theta }^{\prime }(0)=1/\Omega$. Consequently the phase near resonance isAt $\phi =\pi /2$ the exact resonant slope is $\partial P/\partial \nu =-\pi T_{\mathrm{eff}}$. For the plotted parameters $T_{\mathrm{eff}}\simeq 5.637\mathrm{ms}$; the near-resonant estimate $1/(2T_{\mathrm{eff}})\simeq 88.71\mathrm{Hz}$ explains why the exact finite-pulse width below differs from $100\mathrm{Hz}$. The slope is exact at resonance under these assumptions, while this width estimate uses the local phase expansion. A velocity distribution requires averaging the full response, not substituting one effective flight time.

Here is an original reproducible finite set with 40 points and 512 coordinates. Let $e_1,\dots ,e_4$ be the first four vectors of the standard basis of ${\mathbb{R}}^{512}$. With point indices $i=0,\dots ,39$, generateThe four groups have different centers, and the standard normal random variables perturb all 512 coordinates. Generate the coordinates in row order using NumPy's PCG64 pseudorandom number generator with seed 20261013. Then keep this finite set fixed. A second PCG64 stream with seed 424242 generates a $32\times 512$ matrix of independent random variables with the standard normal distribution; divide it by $\sqrt{32}$ to obtain the Gaussian Johnson–Lindenstrauss transform.

For all 780 unordered pairs, compute the distance ratio $r_{ij}=\Vert T{x}_i-T{x}_j{\Vert }_2/\Vert x_i-x_j{\Vert }_2$. In this draw, its minimum is approximately $0.733010$ and its maximum $1.478785$, giving a maximum relative error in Euclidean distance of about $0.478785$. The squared-distance ratios $r_{ij}^2$ instead range from approximately $0.537304$ to $2.186806$: their maximum relative error is about $1.186806$. Thus this particular draw fails a 20% distance tolerance. The Johnson–Lindenstrauss lemma does not assert that every draw into 32 dimensions must succeed.

The two bands of original Euclidean distances distinguish pairs within a group from pairs between groups. Their spread after the linear map shows why checking an average distance is insufficient: the Johnson–Lindenstrauss lemma controls every pair. The dashed 20% distance band is an illustrative tolerance, not a confidence interval or a squared-distance band. The affine hull of 40 points has dimension at most 39; 512 is the ambient coordinate count, and no full-dimensionality assumption is required. For this draw the bi-Lipschitz distortion is approximately $1.478785/0.733010=2.017414$.

Now keep the same points and draw 48 fresh Gaussian Johnson–Lindenstrauss transforms for each target dimension $k\in \{8,16,32,64,128\}$. NumPy's SeedSequence(314159).spawn(240) supplies distinct PCG64 streams, ordered first by target dimension, then by draw. Each stream generates a new $k\times 512$ random matrix, normalized by $1/\sqrt{k}$; no map is a row prefix of another map. Each plotted value is

The observed median maximum relative errors in Euclidean distance are approximately $0.735$, $0.520$, $0.381$, $0.267$, and $0.184$, respectively. All pairs fall inside the 20% distance tolerance in $0$, $0$, $0$, $0$, and $35$ of the 48 draws, respectively. These are computations on one synthetic finite set, rather than optimal dimension bounds or measured physical data. The displayed quantiles describe the sampled maps; they do not certify a success probability for other sets. Even at 128 dimensions, 13 of the sampled maps fail this tolerance.

The practical benefit is that each point acquires fewer coordinates while its pairwise Euclidean distances can remain useful. The choice of target dimension balances compression, accuracy, and failure probability. For example, a conservative sufficient squared-distance tolerance $\epsilon =0.2$ with $n=40$ and $\delta =0.05$ gives $k\ge 1195$ by the explicit bound above. Here that exceeds the ambient dimension, so the identity map is better for a certified guarantee. The much smaller dimensions in the plots illustrate observed behavior without claiming that this sufficient bound is sharp.

Consider a finite-duration Markov process with discrete, time-even states coupled to one thermal reservoir at positive temperature $T$. Let $P_F[\gamma ]$ include the initial probability ${\rho }_0(x_0)$ of the sample trajectory $\gamma$. Define $P_R$ by reversing the control protocol and initializing it with the actual forward final probability distribution ${\rho }_{\tau }$. Write $\Theta \gamma$ for the same states visited in reverse time order. Require normalized path laws with matching probability support: a path has positive forward weight exactly when its reverse has positive reverse weight. This excludes irreversible forbidden transitions and missing initial states.

Under local detailed balance, with positive heat $Q_{\mathrm{bath}}$ delivered to the thermal reservoir, the full path-weight logarithm isThe endpoint term is the change in stochastic system entropy; the conditional path ratio alone accounts for the thermal reservoir. Here $k_B$ is the Boltzmann constant. Both terms matter even for a short observation. ^[ref]

Define the conjugate reverse functional by ${\Sigma }_R[\eta ]=\ln (P_R[\eta ]/P_F[\Theta \eta ])$, so ${\Sigma }_R[\Theta \gamma ]=-{\Sigma }_F[\gamma ]$. Summing the path identity over paths with a given value gives the exact finite fluctuation theoremfor nonzero denominator, using probability masses or corresponding probability density functions. This ${\Sigma }_R$ is defined by the paired laws: it need not be the total entropy production obtained by independently restarting the thermodynamic construction for the reverse experiment. A generic driven experiment therefore does not imply $p_F(s)/p_F(-s)=e^s$.