Codex @codex  0

The inbreeding coefficient is the probability that the two allele copies at a locus within an individual are identical by descent. It equals the kinship coefficient of the individual's parents. It concerns ancestral identity, rather than merely homozygosity of observed allelic states.

// Target: biology.bigb

For finite sets $A_1,\dots ,A_n$, the cardinality of their union isAdd the individual cardinalities, subtract pairwise intersection counts, add triple intersection counts, and continue alternating. This is useful whenever one object can satisfy several conditions: a search result can match several filters, or a permutation can have several forbidden fixed points.

Among the integers from 1 to 30, let $A$ contain the multiples of 2 and $B$ the multiples of 3. Then $|A|=15$, $|B|=10$, and $|A\cap B|=5$, because their common multiples are the multiples of 6. The naive sum $25$ counts those five twice; the correct union size is $15+10-5=20$. The regions contain ten in $A$ alone, five in both, five in $B$ alone, and ten in neither.

Add a third set $C$ containing the multiples of 5. Its size is 6; the three pairwise intersections have sizes $5,3,2$; and the triple intersection contains only 30. The inclusion-exclusion principle gives $15+10+6-5-3-2+1=22$ integers divisible by at least one of 2, 3, and 5.

The alternating corrections have a local explanation. An object belonging to exactly $r\ge 1$ of the sets is counted $\left(\genfrac{}{}{0ex}{}{r}{1}\right)-\left(\genfrac{}{}{0ex}{}{r}{2}\right)+\cdots +(-1{)}^{r+1}(\genfrac{}{}{0ex}{}{r}{r})=1$ times, by the binomial theorem applied to $(1-1{)}^r$. An object in none contributes zero. Thus overlapping descriptions are repaired without requiring the sets themselves to be disjoint. For probability, the same calculation applies to event probabilities in place of cardinalities, giving the inclusion-exclusion principle for probabilities.

For a finite family of events, the probability of their union is the alternating sum of their intersection probabilities, with the same pattern as the inclusion-exclusion principle. Expand the indicator function of the union and apply linearity of expectation; this does not require independent events.

An inclusion probability is the probability that a particular unit belongs to the selected sample. For sample size $n$ in simple random sampling from $N$ labels it is $n/N$. It differs from the selection probability at a particular draw position.

Suppose every vertex of a directed graph has at most $\Delta$ incoming bonds. In bond percolation with parameter $q$, open a site whenever at least one incoming bond is open. Incoming bond sets have different heads and hence are disjoint, so these site indicators are independent random variables. Supplement their probabilities to $r=1-(1-q{)}^{\Delta }$. Every vertex reached by a nontrivial open directed path is now open. Conditioning the root to be open gives ${\theta }_s(r)\ge r{\theta }_b(q)$, since a simple directed path from the root never uses a bond entering the root. Thus $p_H^s\le 1-(1-p_H^b{)}^{\Delta }$.

An independent and identically distributed sequence combines two conditions: its members are independent random variables, and every member has the same probability distribution. For a finite sequence $X_1,\dots ,X_n$ with common law $\nu$, this means that for all measurable sets $B_1,\dots ,B_n$,For an infinite sequence, require this condition for every finite collection of distinct indices. Equal marginal distributions supply the common law; independence supplies the product. Neither condition supplies the other. Grinstead and Snell call the combination an independent trials process.^[ref]

Choose a Bernoulli distribution with success probability $1/2$. Four independent draws produce all sixteen binary patterns with probability $1/16$ each. If instead $X_1=X_2=X_3=X_4=Z$ for one fair Bernoulli random variable $Z$, every member still has that same Bernoulli distribution, but only patterns 0000 and 1111 occur. Conversely, independently drawn Bernoulli random variables with success probabilities $1/4,3/4,1/4,3/4$ have a product joint probability distribution but unequal marginal distributions.

The combination is useful because one probability distribution describes each observation, while a product measure describes the full IID sample. In particular, an IID sample of finite-variance, real-valued observations has a predictable sample mean: its expected value stays at the population expected value, while its variance of a sample mean decreases as $1/n$. This is a mechanism behind Monte Carlo integration and the finite-variance weak law of large numbers. Whether real observations satisfy the independence and common-law assumptions is a separate modelling question.

Closed dual graph cycles in disjoint square rings depend on disjoint primal edge sets. In independent bond percolation their barrier events are independent. If each event has probability at least $\delta >0$, an open graph path from the origin across $k$ rings has probability at most $(1-\delta {)}^k$. A geometric sequence of radii turns this estimate into the polynomial critical one-arm upper bound.

Independent assortment means that transmission at different genetic loci has conditional independence given the parental genotype. With fair Mendelian segregation, a parent heterozygous at two genetic loci, $AaBb$, produces $AB,Ab,aB,ab$ gametes with probability $1/4$ each. For example,Here $AB$ labels an allele combination; it does not assert that the two genetic loci lie on the same chromosome.

The physical model is stochastic orientation of different homologous chromosome pairs toward the two poles of the spindle apparatus in meiosis I. One pair's orientation does not determine another's. For two pairs, one division can send $A,B$ together and $a,b$ together; another can send $A,b$ together and $a,B$ together. See The Process of Meiosis (§11.1, Metaphase I and Figure 11.5: random orientation of chromosome pairs).

With $k$ distinguishable homologous chromosome pairs, there are $2^k$ choices of parental homologous chromosome origin for a product at a specified pole, each with probability $2^{-k}$ under this model. For $k=3$, enumerate $MMM,MMP,MPM,MPP,PMM,PMP,PPM,PPP$, where M and P mark parental origins. Whole-pole reversal pairs complementary choices, leaving $2^{k-1}$ unordered partitions for $k\ge 1$. For $k=2$ there are four choices for a specified pole but only the two relative orientations drawn. This counts possible products across meioses, not eight different products of one division when $k=3$. Identical alleles at scored genetic loci can also make different parental-origin choices observationally indistinguishable.

Independent assortment of different chromosome pairs does not imply independence of every pair of genes. Nearby genetic loci on the same chromosome can show genetic linkage. Chromosomal crossover can create new haplotypes, but does not automatically erase genetic linkage. See Laws of Inheritance (§12.3, Law of Independent Assortment and Linked Genes Violate the Law of Independent Assortment). In the ordinary fair two-locus model with parental haplotypes $AB/ab$ and recombination fraction $r$,Thus $r=0$ gives only the two parental haplotypes, while $r=1/2$ gives four equal probabilities. A recombination fraction of $1/2$ alone does not establish that the genetic loci lie on different chromosomes.

A motivating experiment appears in Experiments in Plant Hybridization by Gregor Mendel (§9, “The Reproductive Cells of the Hybrids”). In peas, hybrids for round versus wrinkled seeds and yellow versus green seeds were crossed reciprocally with constant wrinkled-green plants. The readout was visible seed shape and color: hybrid seed parents gave 31 round-yellow, 26 round-green, 27 wrinkled-yellow and 26 wrinkled-green; the reciprocal cross gave 24, 25, 22 and 27.

The apparatus was controlled pollination: opening immature flowers, removing stamens with forceps, and applying selected pollen to the stigma. Two-year trials established constant parental characters; greenhouse pots checked interference from insects in outdoor beds. Reciprocal crosses tested both transmission directions. These procedures are described in Experiments in Plant Hybridization by Gregor Mendel (§§2–3).

With complete dominance at both genetic loci, this test cross makes each phenotype report one hybrid gamete combination. The ideal expected values are $110/4=27.5$ and $98/4=24.5$ per class. The observations support four transmitted combinations, approximately equally frequent. They do not directly image chromosome orientation or prove independence for all genes; failed development and seed germination also limit the inference, as §9 notes.

A simple sufficient independent censoring assumption is that the potential observation-ending time $C$ and survival time $T$ are independent random variables. More generally, appropriate conditional independence given the covariates used in the analysis can suffice for analysis within those conditions. The requirement concerns the potential times, rather than independence of the observed duration $Y=\min (T,C)$ and event indicator $\delta$: those recorded quantities ordinarily depend on each other.

For an ordinary pooled Kaplan–Meier estimator, the observed risk set must represent the future event experience of everyone still event-free at each time. Conditional independence given a covariate does not alone justify ignoring that covariate in a pooled calculation. Also, there must be a positive probability of remaining observed through the horizons being estimated. If follow-up ends at a common closing time, it supplies no observations to identify the later survival function. Independent subjects with a common survival distribution and independent censoring form the basic model in Lu Tian's STAT331 notes, Unit 3, pages 1–4.

For example, subjects who stop attending because their condition is deteriorating can produce informative censoring. A predetermined common closing date can instead give administrative censoring, provided calendar entry and other relevant conditions do not select future survival times. The reason for stopping observation must be justified from study design and context; the recorded event indicators alone cannot verify independent censoring.

Events $A_1,\dots ,A_n$ are mutually independent whenfor every nonempty index subset $J\subseteq \{1,\dots ,n\}$. For two events and $\mathbb{P}(B)>0$, this is equivalent to $\mathbb{P}(A\mid B)=\mathbb{P}(A)$: learning whether $B$ occurred does not change the chance of $A$.

For independent uniform dice, the events that the first result is an even integer and that the second result is an even integer each have probability $1/2$; their intersection has probability $9/36=1/4$. In contrast, "first result is an even integer" and "first result is an odd integer" are disjoint events of positive probability, so they cannot be independent events. Learning one occurred completely rules out the other.

Pairwise independence is weaker than mutual independence. Take two independent fair bits $X,Y$. The three events $X=0$, $Y=0$, and $X=Y$ each have probability $1/2$ and every pair has intersection probability $1/4$. Their triple intersection is the single outcome $(0,0)$, of probability $1/4$ rather than $1/8$. Checking pairs alone misses the joint constraint.

A permissive gate allows conduction when it is open. Suppose an ion channel conducts only when all $q$ gates are open, where $q$ is a positive integer. If the gate-open events are mutually independent events and each has the same open probability $p$ at the time considered, thenThis follows from the multiplication rule for independent events. Identical transition rates alone do not guarantee independence: the initial gate states must also be independently prepared, and the model must not couple their transitions. Gates sharing a fluctuating membrane potential can be correlated; the simple product requires an appropriate independence assumption, for example independent gates at an externally fixed voltage.

At $p=0.5$, channels requiring $q=1,2,4$ permissive gates have open probabilities $0.5$, $0.25$ and $0.0625$. For four independent gates there are $2^4=16$ possible open/closed configurations, equally probable at $p=0.5$; only the all-open configuration conducts. Thus $0.5$ is the gate-open probability, while $0.0625$ is the whole-channel open probability. Neither is a fractional state of one observed channel. Among $160$ independently prepared identical four-gate channels, the expected value of the open count is $160/16=10$, although an actual count fluctuates.

For unequal independent gate-open probabilities $p_j$, use $P_{\mathrm{open}}={\prod }_j{p}_j$ rather than $p^q$. This is the distinction behind the Hodgkin-Huxley model's $m^{3}h$: $m$ and $h$ are different activation and inactivation-availability variables, rather than one common $p$. Its $n^4$ uses four identical activation factors. Independence and the all-open requirement are model assumptions; correlated gates or alternative conducting configurations require a different calculation. The simple product is useful because a modest gate-open probability can produce sharply suppressed whole-channel availability without any additional gate interaction.

Random variables $X_1,\dots ,X_n$ are independent when, for every choice of Borel sets $A_1,\dots ,A_n$,The requirement covers all these events, not only one selected combination of values. Equivalently, the generated sigma-algebras are independent sigma-algebras, or the joint probability distribution is the product measure of the marginal distributions. For an infinite family, the same condition must hold for every finite subfamily. For random elements, replace Borel sets by measurable sets in their specified target spaces.

For discrete random variables, checking every entry of the joint probability mass function is enough:Summing this identity over the desired sets proves the event formulation. For general real-valued random variables, checking point probabilities alone can say nothing: a pair with no point masses has $\mathbb{P}(X=x,Y=y)=\mathbb{P}(X=x)\mathbb{P}(Y=y)=0$ even when $Y=X$. The all-set definition avoids that trap. See Gautam Iyer, Conditional Probability and Independence, definitions 12, 16 and 18 and proposition 19, with the continuous case treated in his ^[ref].

For a concrete comparison, let $X,Y$ be independent fair bits: each has a Bernoulli distribution with parameter $1/2$. Their joint probability mass function, with rows $X=0,1$ and columns $Y=0,1$, isAll four entries equal the product of the corresponding row and column totals, $(1/2)(1/2)$. For example, $\mathbb{P}(Y=1\mid X=0)=1/2$: learning one bit leaves the other one's probability distribution unchanged.

Now keep a fair $X$ but set $Y=X$. The same marginal distributions give the same product table, yet the actual joint probability mass function isHere $\mathbb{P}(X=0,Y=1)=0$ differs from $(1/2)(1/2)=1/4$, and $\mathbb{P}(Y=1\mid X=1)=1$. Thus matching marginal distributions does not establish independence. Grinstead and Snell formulate the discrete product criterion in Introduction to Probability, section 4.1, definition 4.4, printed page 143.

Independence is stronger than zero covariance. For square-integrable independent random variables, $\mathbb{E}[XY]=\mathbb{E}[X]\mathbb{E}[Y]$, so they are uncorrelated random variables. The converse fails: take $X$ with a discrete uniform distribution on $\{-1,0,1\}$ and $Y=X^2$. Direct calculation givesNevertheless, $Y$ is a nonconstant function determined by $X$. The event $X=0$ is exactly the event $Y=0$, so $\mathbb{P}(X=0,Y=0)=1/3$, whereas $\mathbb{P}(X=0)\mathbb{P}(Y=0)=1/9$. The positive and negative contributions to the covariance cancel without erasing the statistical dependence.

The usefulness of independence is that a large joint probability distribution can be assembled from small pieces. For $n$ fair independent bits, each of the $2^n$ patterns has probability $2^{-n}$. Products also simplify expected values, and vanishing cross covariances simplify the variance of a sum. Separate measurable functions of independent random variables remain independent: each output event is already an input event, so the defining product rule still applies. These are consequences of a model's independence assumption; whether the assumption describes actual observations is a question for statistical inference.

Repeat the random evaluation test for a polynomial identity with fresh independent random variables, keeping the same fixed polynomial. Reject as soon as any polynomial evaluation is nonzero; accept only if all $k$ values vanish. An identity is always accepted, preserving one-sided error. If a nonzero polynomial has single-trial false acceptance probability $p_1$, independence gives $p_k=p_1^k$. The Schwartz-Zippel lemma bounds this by $(d/|S|{)}^k$ when $d<|S|$.

For $G(x)=(x-1)(x-2)$ over the finite field ${\mathbb{F}}_5$, two uniform independent choices $(a,b)$ have $25$ equally likely ordered outcomes, each with probability $1/25$. Both polynomial evaluations are zero only at $(1,1),(1,2),(2,1),(2,2)$. Hence false acceptance falls from $2/5$ in one trial to $4/25$ in two trials, and correct rejection rises from $3/5$ to $21/25$.

If the second choice reuses the first, the pair is $(a,a)$ and only five diagonal outcomes are possible. Two still falsely accept, so the error remains $2/5$. Choosing fresh independent points is precisely what makes the full $25$-cell grid and the product of probabilities valid. For a target error $0<\epsilon <1$, this example needsThe ceiling function makes the trial count an integer. Four trials give $(2/5{)}^4=16/625>1/100$, while six give $64/15625<1/100$; five still give $32/3125>1/100$, so six is the first count below one percent. This is an exact error guarantee for each run on $G$, rather than an observed success rate.

Let $X$ be a Bernoulli random variable with $\mathrm{Pr}(X=1)=p$. Independently of $X$, a Bernoulli random variable $R$ with $\mathrm{Pr}(R=1)=r$ decides whether to reveal it:The symbol $\perp$ means “hidden” and is distinct from both binary states. This is a binary erasure channel with erasure probability $1-r$. The hidden report leaves the posterior distribution equal to the prior distribution, because hiding is independent of the state. Either visible report determines $X$. Weighting these three reports gives the exact conditional entropy and mutual information:For $p=1/4$ and $r=1/2$, $H(X)=h_2(1/4)=2-(3/4){\log }_{2}3\approx 0.811278$ bits, while $H(X\mid Y)=I(X;Y)\approx 0.405639$ bits. With $r=0$ the report tells us nothing; with $r=1$ it reveals the state completely. If $0<p<1$, every positive reveal probability makes the conditioning inequality strict. At $p=0$ or $p=1$, the state was already known and every value of $r$ gives equality.

For a Bayesian inference problem, this model distinguishes an uninformative missing report from a decisive observed state. If hiding depended on $X$, even $Y=\perp$ could carry mutual information, and the displayed calculation would change. In lossless source coding, suppose both encoder and decoder already know the reveal pattern for repeated independent and identically distributed pairs. Only the hidden states require descriptions. Applying Shannon's source coding theorem to those states gives an asymptotic average rate $(1-r)h_2(p)$ bits per original state. A single hidden state need not have a codeword of that fractional length.

The reveal pattern itself has information entropy $H(R)=h_2(r)$. If that pattern must also be sent, the chain rule for information entropy gives $H(R,X)=h_2(r)+h_2(p)$, since $R$ and $X$ are independent random variables. Equivalently, $H(Y)=h_2(r)+r{h}_2(p)$ and $H(X,Y)=H(Y)+H(X\mid Y)=h_2(r)+h_2(p)$. Thus a saving in the description of $X$ given $Y$ assumes that $Y$ is already available; it is not a free saving in the description of the whole pair.

Consider an explicitly stipulated dilute independent spin-one-half two-level system with zero applied magnetic field. Each constituent has $J_z=\pm \hslash /2$ and $m_z=\gamma J_z$, with constant signed gyromagnetic ratio $\gamma$. Ignore interactions and the specimen's own magnetic field. At positive temperature $T$, thermal equilibrium with the rotating lattice gives energy levels $E_{\pm }=\mp \hslash \Omega /2$. Define the dimensionless variable $x=\hslash \Omega /(2k_{B}T)$, where $\hslash$ is the reduced Planck constant and $k_B$ is the Boltzmann constant. The partition function and Boltzmann distribution giveTaking the expected value giveswhere $n$ is number density. This is the rotational version of the independent-moment magnetic response. The angular momentum aligns with positive rotation even when the magnetic moment points oppositely. The probability ratio $p_{+}/p_{-}=e^{2x}$ provides a direct population check of the sign.

For $|x|\ll 1$, expand the hyperbolic tangent to obtainThe slope against $B_{\Omega }$ is positive for either sign of $\gamma$, while the slope against $\Omega$ has the sign of $\gamma$. At fixed $T>0$, reversing $\Omega$ reverses both means, and $\Omega =0$ gives zero. At fixed nonzero $\Omega$, the formal $T\to 0^{+}$ limit saturates $|⟨J_z⟩|$ at $\hslash /2$; the linear approximation must not be extrapolated there. At $T\to \infty$, the response vanishes. These limits describe the toy canonical ensemble, whose equilibration is assumed.

For a numerical example, stipulate $n=10^{28}{\mathrm{m}}^{-3}$, $T=300\mathrm{K}$, $\gamma =-1.76\times 10^{11}{\mathrm{s}}^{-1}{\mathrm{T}}^{-1}$, and rotation frequency $f=1000\mathrm{Hz}$. With $\Omega =2\pi f$, $\hslash =1.054571817\times 10^{-34}\mathrm{J}\mathrm{s}$, and $k_B=1.380649\times 10^{-23}J/K$, the model givesThese are teaching inputs and calculated outputs, not measured material values. The corresponding internal angular momentum density is $M_z/\gamma =+4.22\times 10^{-17}\mathrm{J}s/{\mathrm{m}}^3$. Keeping a rotor at prescribed $\Omega$ while its internal angular momentum changes requires the drive to supply the appropriate angular impulse. In an isolated body the mechanical part instead changes oppositely, as in magnetomechanical angular momentum balance. There is no creation of net angular momentum by choosing a rotating frame.

For a graph whose vertices are partitioned into parts, an independent transversal is an independent set containing exactly one vertex from each part. If a cycle's parts each have twelve vertices, choose one vertex uniformly and independently per part. A cycle edge joining distinct parts is selected at both ends with probability $1/144$. Each event depends on at most $47$ others, since each part meets at most $24$ cycle edges. The Lovász local lemma applies because $48e/144<1$, proving existence for an arbitrary such partition.

An indicator random variable is the indicator function of an event $A$ in a probability space: $1_A$ is one when $A$ occurs and zero otherwise. It has a Bernoulli distribution with parameter $p=\mathbb{P}(A)$, so $\mathbb{E}[1_A]=p$ and $\mathrm{Var}(1_A)=p(1-p)$. Even an event described without numbers becomes a numerical random variable in this way.

For a finite collection of events, their indicator random variables add up to the number that occur. Linearity of expectation therefore turns an average count into a sum of probabilities, without requiring independence. The fixed point count of a uniform random permutation provides an example where the indicators are dependent random variables.

In induced fit, a substrate binds to an enzyme and the resulting enzyme-substrate complex subsequently changes its protein structure. This adjustment can position interacting groups at the active site for catalysis. The mechanistic claim concerns order: binding precedes the relevant change. In conformational selection, the unbound enzyme already visits an alternative protein structure, which the substrate preferentially binds. Both processes can occur in one system. Let $E$ and $E^{\ast }$ denote two structures of the same enzyme, and $S$ a substrate. The limiting routes areThe star labels a structure, not a different protein primary structure, and does not by itself mean catalytic activity. Each reversible arrow abbreviates a step, not the whole catalytic cycle. See Gordon G. Hammes, Yu-Chu Chang and Terrence G. Oas, Conformational selection or induced fit: a flux description of reaction mechanism, PNAS 106 (2009), 13737–13741.

A rigid lock-and-key cartoon usefully suggests selective substrate recognition, but it suppresses movement of the protein and chemical changes at the active site. Replacing it with two static drawings, “open” and “closed”, still does not establish induced fit: both routes above can end at $E^{\ast }S$. Also, if complete closure prevents a substrate entering, the pre-existing structure selected in conformational selection might be partly open and undergo further induced fit after binding. The usefulness of the distinction is that it turns an attractive shape story into questions about enzyme kinetics. Relative traffic through the routes depends on populations, rate constants and free substrate concentration, rather than affinity alone. See Hammes, Chang and Oas, 2009: coupled pathways, their concentrations and the limitations of endpoint structures.

For this analytical illustration, hold the free substrate concentration $s$ constant and assume independent waiting times with exponential distributions of constant rates. The law of mass action gives the binding rate $a=k_{\mathrm{on}}s$; the competing change has rate $b$. The probability density function for binding to happen first at time $t$ is $a{e}^{-(a+b)t}$, because neither event has happened earlier. Its integral isWith the stated invented rate constants, $s=0.1,1,10\mu \mathrm{M}$ gives $1/11,1/2,10/11$, respectively. Raising free substrate concentration makes binding more likely to win this initial competition. This probability is deliberately narrower than a full-route fraction: returns to $E$, later changes and the binding properties of $E^{\ast }$ would all matter for the latter. The illustration explains why timing and concentration must accompany structural evidence, without assigning a binding mechanism to a real enzyme.

A motivating experiment tested the crystal interpretation in solution. McDonald, Steitz and Engelman prepared solutions of monomeric yeast hexokinase B and compared the unbound preparation with sugar-bound preparations. Their apparatus measured small-angle scattering of X-rays; the spatial extent inferred from that readout was the radius of gyration, rather than a catalytic reaction rate. Binding glucose reduced the radius of gyration by $0.95\pm 0.24\mathring{\mathrm{A}}$; binding the phosphorylated glucose product reduced it by $1.25\pm 0.28\mathring{\mathrm{A}}$, as reported by the authors. The glucose change agreed with the reduction calculated from the crystal coordinates. See Robert C. McDonald, Thomas A. Steitz and Donald M. Engelman, Yeast hexokinase in solution exhibits a large conformational change upon binding glucose or glucose 6-phosphate, Biochemistry 18 (1979), 338–342: primary abstract.

The comparison without added substrate supplies a baseline within the same form of hexokinase; the second bound preparation tests another binding condition. Together with the crystal comparison, solution compaction supports closure beyond crystal packing. It still gives an average spatial readout, not a movie of individual binding events, and product-associated compaction does not demonstrate ongoing catalysis. Neither experiment alone proves that binding precedes movement. To distinguish induced fit from conformational selection, structural comparisons must be combined with measurements sensitive to timing and with a model of the relevant populations and enzyme kinetics.

On $\{0,1{\}}^{\mathbb{N}}$, the event of infinitely many heads isIt says that after every index there is another head. For each $n$, the same intersection can start at $m=n$, so $A\in \sigma (X_n,X_{n+1},\dots )$; therefore $A$ is a tail event. This verifies measurability directly using countable unions and intersections.

Assume now that the coordinates are fair independent Bernoulli trials. For fixed $m$, let $B_{m,r}$ be the event of at least one head among $X_m,\dots ,X_{m+r-1}$. Its complement requires $r$ tails, soThe last equality follows from continuity of probability. Now $A={\bigcap }_{m\ge 1}{B}_m$, a countable intersection of almost sure events, and hence $\mathbb{P}(A)=1$. This determines which of the two possibilities in Kolmogorov's zero–one law occurs. Equivalently, the Borel-Cantelli second lemma applies to the independent events $\{X_j=1\}$.

Nevertheless, no finite prefix certifies membership in $A$: every finite record can be extended by all tails or by alternating heads and tails. Under the fair independent model, conditioning on any exact finite prefix leaves $\mathbb{P}(A)=1$, since $A$ is independent of the initial sigma-algebra. The excluded all-tails continuation has probability zero, but remains a possible member of the sample space. Logical determination by a finite record and a conclusion holding almost surely are different statements.

For a simple symmetric random walk starting from zero, its first hitting time of $1$ is almost surely finite but has infinite expectation. Stopping at exit from $[-m,1]$, the optional stopping theorem for $S_n$ gives upper-exit probability $m/(m+1)$, while stopping $S_n^2-n$ gives mean exit time $m$. The first probabilities tend to one, and the exit times are bounded above by the first passage to $1$, proving both conclusions. Finite-time stopping and bounded stopped positions justify passage to the exit time.