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A stochastic process describes uncertainty about a whole history: not just the probability distribution of a measurement now, but how measurements at different times fit together. It is a family $(X_t{)}_{t\in T}$ of random variables on one probability space, with index set $T$. Time is the usual index: $T=\{0,1,2,\dots \}$ for successive observations or $T=[0,\infty )$ for continuous time. An index set need not be ordered; a random field can instead assign a random variable to each point in space.

There are two complementary views. Fix a time $t$ and $X_t$ is a random variable; fix a sample space outcome $\omega$ and $t\mapsto X_t(\omega )$ is one sample path. A model specifies the probability distribution of these histories. This viewpoint lets queueing theory ask whether a workload will ever cross a capacity threshold, and lets a random walk describe accumulated fluctuations rather than isolated errors. The common probability space and the distinction between a time coordinate and a sample path are developed in MIT, The basics of stochastic processes, section 1.

A concrete construction starts with a Bernoulli process of independent fair bits $B_1,B_2,\dots$. Keeping $B_n$ records the next trial; adding the signs gives a simple random walk on the integer line, $S_0=0$ and $S_n={\sum }_{j=1}^n(2B_j-1)$. Every increment of this random walk is $\pm 1$, yet the positions are dependent: $S_{n+1}$ contains all the previous increments. Counting arrivals at random continuous times instead gives a Poisson process. These examples separate three questions: what values are possible, how time is indexed, and how the values depend on one another.

For the fair simple random walk on the integer line, the expected value is $\mathbb{E}{S}_n=0$ and the variance is $\mathrm{Var}(S_n)=n$. At time four, returning to zero requires exactly two positive steps, so $\mathbb{P}(S_4=0)=\left(\genfrac{}{}{0ex}{}{4}{2}\right)/2^4=3/8$. A rate-two Poisson process instead has $\mathbb{P}(N_1=0)=e^{-2}\approx 0.1353$. A single realized sample path does not establish either probability: they follow from the specified joint probability distributions.

The most revealing feature is often dependence. Two stochastic processes can have identical marginal distributions at every time while one forgets everything at the next step and the other repeats its first value forever. Their finite-dimensional distributions and covariance functions distinguish them. That extra structure is what makes a stochastic process useful for prediction.

Stratified sampling partitions a population into strata and samples separately within each. It guarantees specified representation and can improve precision when the strata describe population variation. Equal sampling fractions give equal unit inclusion probabilities; unequal fractions require appropriate weighting for population summaries.

For continuous semimartingales, the Stratonovich integral is defined by ${\int }_0^t{Y}_s\circ d{X}_s={\int }_0^t{Y}_{s}d{X}_s+\frac{1}{2}[Y,X{]}_t$, where the first term is an Itô integral and the correction is quadratic covariation. It obeys the ordinary chain rule and is the limit in probability of symmetric endpoint sums.

For $c,r>0$, an intersection inside a disk requires at least two lines to meet that disk. The latter count is Poisson with mean $2\pi cr$, giving the non-strict bound first. It is strict because two lines with nearby normal angles but separated perpendicular distances can both hit the disk while intersecting outside. Choose two positive-measure parameter boxes forcing this configuration; the event of exactly one line in each box and no other disk-hitting line has positive probability. This also yields a void probability too large to equal that of a Poisson point process with the intersection process's intensity measure.

Suppose $B$ is a nonzero separable Banach space, the Cameron-Martin space of a Gaussian random variable in a Banach space is a dense subset of $B$, and every centered ball of positive radius has positive probability. Then $t\mapsto \mathbb{P}(\Vert X{\Vert }_B\le t)$ is strictly increasing on $(0,\infty )$. For $0<a<b$, choose $h$ in that Cameron-Martin space of a Gaussian random variable in a Banach space with $\Vert h{\Vert }_B=(a+b)/2$ and $0<\delta <(b-a)/2$. The closed ball of radius $\delta$ centered at $h$ lies in the annulus $a<\Vert x{\Vert }_B<b$ and has positive probability by the symmetric Gaussian translation lower bound. The nonzero assumption is necessary: on $B=\{0\}$ the distribution function equals one everywhere.

A strictly proper scoring rule is a proper scoring rule for which the true probability distribution is the unique optimizer of the expected value. Equivalently, any different admissible probability distribution gives strictly larger expected loss. For probability density functions, two densities that agree almost everywhere represent the same probability distribution. Ordinary propriety permits ties: the constant reward $S(P,x)=0$ is a proper scoring rule, because every report ties, but it is not a strictly proper scoring rule and elicits no information.

Strict propriety concerns expectation, not guaranteed success on each observation or each realized dataset. With the stipulated Bernoulli distribution $q=0.7$, suppose the outcome is $Y=0$. The honest report $p=0.7$ then has Brier score loss $0.49$, while $p=0.2$ has loss $0.04$. The logarithmic loss similarly gives losses $-\ln (0.3)\approx 1.203973$ and $-\ln (0.8)\approx 0.223144$. The honest forecast loses this comparison on that draw, which occurs with probability $0.3$ in the stipulated statistical model.

The $Y=0$ loss curves favor $p=0$, and the $Y=1$ loss curves favor $p=1$. Their weighted average, with weights $0.3$ and $0.7$, favors $p=0.7$. Choosing a report after observing $Y$ would use information unavailable at forecast time. Even an average loss on a finite dataset can rank an honest forecast below a competing forecast through outcome variation; strict propriety guarantees the ranking in expected value under the specified information and true predictive distribution.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$, and let $\mu =\mathbb{E}[X_1]$ be their expected value. Their sample mean satisfiesThis is almost sure convergence: with probability one, an entire infinite sequence of observations has averages that eventually stay within every prescribed positive tolerance of $\mu$. The variance need not be finite. The strong law of large numbers implies the weak law of large numbers, which asserts convergence in probability at each increasing sample size.

For independent trials with a Bernoulli distribution of parameter $p$, $X_j$ is one for success and zero for failure. Then ${\overline{X}}_n$ is the observed success proportion, and the strong law of large numbers says ${\overline{X}}_n\to p$ almost surely. This turns a theoretical probability into a quantity one can estimate by repetition. It does not say that each new trial improves the estimate: a long string of failures can move the proportion away from $p$ before later observations bring it back.

The strong law of large numbers gives the limiting location of the sample mean, while the central limit theorem describes its fluctuations when the variance is finite and positive. The two assertions can hold together:The first error shrinks to zero; the second has been enlarged enough to retain a nontrivial probability distribution. The second assertion is convergence in distribution, not a claim that the enlarged error along one sequence approaches a particular normal random variable.

The assumptions make a practical difference. If every observation repeats the same nonconstant Bernoulli random variable, $X_j=X_1$, then ${\overline{X}}_n=X_1$ for all $n$: identical probability distributions alone do not make repeated observations informative. For independent observations with a Standard Cauchy distribution, the Cauchy sample mean has an unchanged distribution, so it does not concentrate around a constant. Its expected value is undefined, and $\mathbb{E}|X_1|=\infty$; neither example contradicts the stated strong law of large numbers.

A brief proof outline in the case of finite variance uses the Kolmogorov maximal inequality for the centered partial sums. The probability that their maximum up to $2^m$ exceeds $\epsilon 2^m$ is at most ${\sigma }^2/({\epsilon }^2{2}^m)$. These bounds are summable, so the Borel-Cantelli lemma controls all sufficiently large such blocks and hence every intermediate sample size. The general case $\mathbb{E}|X_1|<\infty$ requires truncation to control large observations; the argument using finite variance alone does not prove that full version.

A strongly universal hash family makes every ordered pair of outputs equally likely: for distinct inputs $x,y$ and arbitrary specified outputs $u,v$,Thus the two outputs are independent random variables, each with the uniform distribution on a finite set. Adding the $q$ equal ordered pairs gives collision probability $q/q^2=1/q$, so the family is also a universal hash family.

Here is a completely enumerable example. In the finite field ${\mathbb{F}}_5$, take the affine functions $h_{a,b}(x)=ax+b$, using modular arithmetic modulo $5$. Choose $a$ and $b$ as independent random variables, each with the uniform distribution on a finite set of all five finite field elements $0,1,2,3,4$, including $a=0$. There are $25$ equally likely hash functions. For $x=0,y=1$, the ordered pair is $(b,a+b)$. For instance $(a,b)=(2,3)$ gives $(h(0),h(1))=(3,0)$; $(0,3)$ gives $(3,3)$. Every prescribed $(u,v)$ occurs exactly once, at $b=u,a=v-u$.

Each row contains five choices, so each individual output has probability $5/25=1/5$. Each cell has probability $1/25$, and the five diagonal cells give collision probability $5/25=1/5$. For any distinct $x,y$, the same conclusion follows from $a=(v-u)/(y-x)$ and $b=u-ax$: a nonzero finite field element has a unique inverse. This is the beauty of the construction: two small random coefficients provide exact pair balance without choosing five unrelated values.

The example uses equal-sized input and output sets to make every choice visible; it illustrates the family definitions rather than compression. In a larger input set, a hash function cannot avoid all collisions into five buckets. What a universal hash family controls is the fraction of its members that collide on each fixed pair, rather than the existence of any collision anywhere.

Restricting to $a\ne 0$ leaves $20$ permutations with zero collision probability for distinct inputs. Individual outputs remain uniform, but given $h(0)=u$, the four values other than $u$ each have conditional probability $1/4$ and $u$ has zero. Hence these outputs are not independent random variables: this is a universal hash family but not a strongly universal hash family. Finally, the full $25$-member example contains five constant functions, with immediately visible collisions. Its pair guarantee concerns a random choice of hash function, and makes no claim of cryptographic collision resistance.

A plan or policy that reaches a goal after finitely many steps under every permitted outcome sequence. In a finite state space with finite outcome sets, a terminating finitely branching plan tree has bounded depth. Strong planning winning-set iteration constructs exactly the states from which such finite guaranteed-success plans exist. A repeated state cannot be accepted merely because repeated attempts might eventually succeed: without a fairness or probability assumption, an unfavorable outcome can recur forever.

A centered random vector is sub-Gaussian with variance proxy ${\tau }^2$ when the displayed inequality holds for every deterministic vector $a$. Equivalently every scalar inner product is a sub-Gaussian random variable with proxy ${\tau }^2\Vert a{\Vert }_2^2$. Independence of coordinates is not required. For vectors $a_j$ with Euclidean norms at most one, the union bound gives ${\max }_j|⟨a_j,g⟩|\le \tau \sqrt{2\log (2m/\delta )}$ with probability at least $1-\delta$. This assumption alone is weaker than concentration for all Lipschitz functions.

Let a constant input lie a fraction $f$ of the way between two consecutive quantizer levels separated by $\Delta$. With independent uniform additive dither spanning one step and nearest-level rounding, the two output codes occur with probabilities $1-f$ and $f$. Their mean equals the input and their sample-average variance is the displayed expression. This simple model shows both the sub-step information and the remaining noise cost.

Apply the independence axiom for lottery preferences in both weak directions of the indifference relation. At $p=0$ the two mixtures are identical. Repeating this substitution and using transitivity lets one replace any finite collection of indifferent component lotteries, after which reduction of compound lotteries combines their outcome probabilities.

In independent Bernoulli trials with success probability $p$ and failure probability $q=1-p$, let $T_n$ be the trial of success $n$ and $U_m$ the trial of failure $m$, for positive integers $n,m$. If success $n$ occurs after $k<m$ failures, its final trial is a success and the preceding trials have $n-1$ successes and $k$ failures. The negative binomial stopping argument therefore gives the title sum.

Put $F=n+m$. By trial $F-1$, at least one target must have been reached, so the same event is that a binomial distribution with $F-1$ trials has at least $n$ successes. Conditioning instead on the final trial of a block of $F$ trials givesIndeed, more than $n$ successes in the full block guarantee the success target won; with exactly $n$ successes it won only when the final trial was a failure. These disjoint cases explain the extra boundary term. A zero success target wins immediately; a zero failure target has already been reached. To include $n=0$ in the displayed finite sum requires the explicit generalized convention $\left(\genfrac{}{}{0ex}{}{-1}{0}\right)=1$; an upper summation limit of $-1$ denotes an empty sum.

A jump from positive frequency $\omega$ to $\Omega$ keeps an oscillator mode and its first derivative continuous. A normalized positive-frequency incident solution has outgoing coefficients $\alpha =(\Omega +\omega )/(2\sqrt{\Omega \omega })$ and $\beta =(\Omega -\omega )/(2\sqrt{\Omega \omega })$, satisfying ${\alpha }^2-{\beta }^2=1$. This temporal matching is not a sum-one spatial-scattering probability rule.

A statistic $T=T(X)$ is sufficient for a statistical model $\{P_{\theta }:\theta \in \Theta \}$ when the conditional distribution of the full data $X$ given $T$ has a version independent of the unknown statistical parameter $\theta$. Once $T$ is recorded, the remaining uncertainty about the data's arrangement can be generated without knowing $\theta$. ^[ref]

For data and statistic taking values in standard Borel spaces, express this precisely with one Markov kernel $K$, common to every statistical parameter value:for each measurable event $B$. Thus $K(t,\cdot )$ is a regular conditional distribution of $X$ given $T=t$. Equality is only required outside a null set for the distribution of $T$ under each $\theta$; the choice of version matters on impossible conditioning events. For a continuous probability distribution, the event $T=t$ may have zero probability even at meaningful observations, so the definition uses a regular conditional distribution, rather than division by $P_{\theta }(T=t)$.

A sufficient statistic is a property of a specified statistical model, rather than of a formula for summarizing data in isolation. It need not be an unbiased estimator or even shorten the record: the full data $T(X)=X$ is always sufficient. An estimator instead targets a quantity such as a mean. An ancillary statistic has a parameter-independent marginal probability distribution, whereas a sufficient statistic has a parameter-independent conditional full-data law. A minimal sufficient statistic permits the greatest reduction among sufficient statistics. A complete statistic satisfies a separate condition about zero expectations; a complete sufficient statistic must satisfy both conditions.

Take independent random variables $U_n$, each with the continuous uniform distribution on $[0,1]$, and set $A_n=\{U_n\le p_n\}$. Then $\mathbb{P}(A_n)=p_n$, and the events are mutually independent events. The product probability space supplies a model for the entire infinite sequence; a finite list of simulated values is only an illustration.

For $p_n=1/(n+1{)}^2$, the p-series is summable:The Basel problem supplies the total sum, and the integral test for convergence gives the displayed tail bound. The First Borel-Cantelli lemma makes the limsup of events have probability zero. In fact, the tail bound implies that the probability of even one alarm at or after $m$ is at most $1/m$; for $m=100$, it is at most $0.01$. This bound remains valid for dependent events with the same individual probabilities.

For $p_n=1/(n+1)$, the harmonic series diverges, so the Second Borel-Cantelli lemma makes the limsup of events have probability one. There is also an exact finite-horizon calculation using independent events:For any fixed $m$, this finite product of numbers tends to zero as $M\to \infty$. Every infinite tail therefore contains an alarm almost surely, even though the individual probabilities tend to zero.

Start traps at a unit-intensity Poisson random measure in Euclidean space and attach mutually independent Brownian motions, independent of the initial measure. For a deterministic target path $f$, independently thin initial positions according to whether their marked path comes within radius one of the target by time $t$. The retention probability $p_t(x)$ integrates, by the Tonelli theorem, to the expected volume of the sausage along $f-B$. Symmetry of Brownian motion makes this the expected volume of the sausage along $f+B$. The Poisson zero-count probability proves the displayed survival formula. This derives the formula directly from independent thinning, without assuming a separate marking or displacement theorem.

For a nonnegative survival time $T$, the survival function iswhere $F$ is the cumulative distribution function. Thus $S(t)$ is the probability that the chosen occurrence is still in the future at time $t$. We consistently use the strict inequality $T>t$: a probability mass at $t$ has already left the survival function at $t$ itself. With this convention $S$ is a right-continuous function and a nonincreasing function taking values in $[0,1]$. If $\mathbb{P}(T=0)=0$, then $S(0)=1$; if $T$ is finite with probability one, then $S(t)\to 0$ as $t\to \infty$.

For $S(a)>0$ and $b>a$, conditional probability givesFor example, $S(2)=0.8$ and $S(5)=0.6$ mean that $0.6/0.8=0.75$ of those still event-free at year two remain event-free through year five. Their intervening event probability is $0.25$, whereas the original population's probability of an occurrence in $(2,5]$ is $0.8-0.6=0.2$. Conditioning changes the denominator.

The area under the survival function up to a horizon gives the restricted mean survival time. A positive terminal plateau in a plotted Kaplan–Meier estimator does not establish that the occurrence will never happen: terminal censoring and survival-mean identifiability explains why the unobserved tail remains undetermined.

Calibration compares predicted survival probabilities with observed survival experience at relevant horizons or over time. Censoring must be handled, for example by suitable survival estimates or inverse-probability weighting. Calibration should be assessed on validated predictions, because the fitting sample can hide optimism from variable selection and tuning. It is distinct from the risk-ranking information measured by a concordance index.

For a Galton-Watson process started from one ancestor, this is the probability that every generation is nonempty. It is the complement of the branching extinction probability. The binomial branching survival correction distinguishes finite-binomial reproduction from the Poisson branching process near mean one.