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If $Y_n$ are random variables bounded above by an integrable function $Z$, with all relevant expectations well defined, Fatou's lemma applied to the nonnegative variables $Z-Y_n$ gives ${\mathrm{limsup}}_n\mathbb{E}{Y}_n\le \mathbb{E}{\mathrm{limsup}}_n{Y}_n$. In particular, bounded indicator random variables that tend almost surely to zero have probabilities tending to zero.

For a stochastic process $X_t$ and a fixed sample space outcome $\omega$, its sample path is the function $t\mapsto X_t(\omega )$. One row of observed fair bits, one trajectory of a random walk, or one staircase of Poisson process arrival counts is a sample path. Fixing $t$ instead gives a random variable ranging across possible histories. The same two-variable object $X_t(\omega )$ therefore describes both observations across outcomes and evolution within an outcome.

Statements about continuous, càdlàg or Hölder continuous function sample paths concern these functions simultaneously in time. They differ from assertions made at each fixed time separately. For example, take $U$ with a uniform distribution on $[0,1]$ and set $X_t=1_{\{t=U\}}$. For every fixed $t$, $X_t=0$ almost surely, but every sample path has a single spike. Thus this process is a modification of a stochastic process that is identically zero, while the two are not indistinguishable stochastic processes. Continuity of both sets of sample paths rules out this discrepancy: equality at a countable dense set then implies equality everywhere. This is why path regularity is an additional part of constructing a continuous-time model.

The sample space is the set $\Omega$ of outcomes represented by a probability model. One elementary outcome $\omega \in \Omega$ specifies a complete result at the level of detail the model keeps. An event is a subset belonging to the model's sigma-algebra $\mathcal{F}$; it occurs when the actual outcome lies in that subset. Together with a probability measure $\mathbb{P}$, these form a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, with $\mathbb{P}(\Omega )=1$.^[ref]

Choosing a sample space means deciding which distinctions matter. For two tosses, recording the ordered head-or-tail sequence distinguishes $HT$ from $TH$; recording only the number of heads merges them. Neither representation is automatically wrong, but the second cannot answer a question about which toss was heads. Keeping a detailed sample space and defining a random variable for the desired observation lets several questions share one model.

A sample space alone assigns no probabilities. Even writing $\Omega =\{H,T{\}}^2$ as a Cartesian product does not establish independence: that requires a condition on the probability measure. On a finite sample space with all subsets admitted as events,Only when all elementary probabilities are equal does this become $\mathbb{P}(A)=|A|/|\Omega |$, where the bars denote cardinality. This is the uniform distribution on a finite set, an additional modelling choice.

Sample spaces can also be infinite. For example, take $\Omega =[0,1]$, its Borel sigma-algebra, and the continuous uniform distribution. Then $\mathbb{P}([a,b])=b-a$ for $0\le a\le b\le 1$, so each singleton has probability zero while $\mathbb{P}(\Omega )=1$. Thus an elementary outcome need not carry positive probability, and a continuous model cannot be understood by counting equally likely points. The distinction is developed in zero-probability event and countable additivity.

For finite alphabets, $H(X_1,\dots ,X_n)\ge {\max }_{i}H(X_i)$, soFor $n$ identical copies of one fair bit, the upper bound is attained and total correlation is $n-1$ bits. Adding another copy increases the result without introducing a new underlying random choice. The choice of coordinates and their grouping therefore matters when comparing systems of different sizes.

For more general random variables, the relative entropy against the product of the marginals supplies a definition of total correlation that may be infinite. Subtracting differential entropies is justified only when those quantities are finite and the entropy identity is valid; an undefined difference of infinities is not a definition. The finite-alphabet examples above avoid that problem entirely.

A scoring rule turns a probability distribution forecast and an observed outcome into a number. For example, a forecast of rain with probability $0.7$ should express uncertainty more usefully than a bare prediction of rain. A scoring rule lets us compare such forecasts after observations arrive, and its choice determines whether exaggerating confidence pays. A proper scoring rule rewards an honest account of uncertainty in expected value.

There are two sign conventions. A reward $S(P,x)$ is maximized; a loss function $L(P,x)$ is minimized. The conversion $L=-S$ reverses every comparison. Here the linear probability score is a reward, while the binary Brier score and the negative logarithmic scoring rule are losses. Calling all three a "score" without specifying the direction would hide the central difference.

Throughout the binary examples, the random variable $Y$ has a Bernoulli distribution with actual success probability $q=0.7$, and the report $p\in [0,1]$ assigns probability $p$ to $Y=1$ and $1-p$ to $Y=0$. The value $q=0.7$ is stipulated by this statistical model; it is not inferred from data or a claim that a forecasting system is calibrated. In practice the true predictive distribution is unknown, and the relevant $q$ is the conditional probability of success given the information available when the forecast is issued. Propriety evaluates incentives for reporting that predictive distribution; it does not supply the unknown probabilities.

The second moment of a real random variable is $\mathbb{E}[X^2]$. Its variance is $\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$ when these quantities are finite.

For a nonnegative random variable $N$ with positive expected value and finite second moment, write $N=N{1}_{\{N>0\}}$. The Cauchy-Schwarz inequality giveswhich proves the bound in second moment method. It is the zero-threshold case of the Paley-Zygmund inequality. Equality occurs when $N$ is a fixed positive value on the event $N>0$ and zero elsewhere, almost surely. For $N=10B$ with $B$ a Bernoulli random variable of parameter $1/10$, $(\mathbb{E}N{)}^2/\mathbb{E}{N}^2=1/10$ equals the exact positivity probability.

The variance identity $\mathbb{E}{N}^2=(\mathbb{E}N{)}^2+\mathrm{Var}(N)$ turns the lower bound into a statement about variance relative to the squared mean. A small ratio forces positivity to be likely. A positive expected value alone does not do so: keeping $\mathbb{E}N=1$ while taking $N=1/q$ with probability $q$ and zero otherwise makes the positivity probability as small as any chosen $q>0$.

The second moment method uses a random variable's expected value and second moment to establish that a counted or measured quantity is positive with substantial probability. It is useful when a positive average alone could be caused by extremely large values on very rare outcomes. Controlling the variance rules out too much of that concentration in rare outcomes.

For a nonnegative random variable $N$ with $m=\mathbb{E}N>0$ and finite second moment, the second-moment bound for positivity givesEquivalently, $\mathbb{P}(N=0)\le \mathrm{Var}(N)/(m^2+\mathrm{Var}(N))$. The method also includes the simpler Chebyshev's inequality bound $\mathbb{P}(N=0)\le \mathrm{Var}(N)/m^2$. Thus a sequence of such random variables with $\mathrm{Var}(N)/(\mathbb{E}N{)}^2\to 0$ has $N>0$ with probability tending to one. These statements require no independence assumption. The variance argument and its use for random counts are developed in Zhao's lecture notes. ^[ref].

For an original toy count, let $N$ count successes in ten independent Bernoulli trials with success probability $p=1/10$. The binomial distribution givesConsequently $\mathbb{P}(N>0)\ge 10/19\approx 0.526316$. The exact probability is $1-(9/10{)}^{10}\approx 0.651322$. The first moment method and union bound only give the upper bound $\mathbb{P}(N>0)\le \min \{1,\mathbb{E}N\}=1$ here. First and second moments therefore answer complementary existence questions.

For a sum of indicator random variables, the variance of a sum isIndependence makes the covariances vanish, but the formula itself also covers dependent counts. For the same ten success probabilities, let every trial share a single Bernoulli random variable $B$: then $N=10B$, with mean one, variance nine, and $\mathbb{P}(N>0)=1/10$. The second-moment lower bound is exactly $1/10$. Ignoring the positive covariances would produce the false independent-count conclusion. Zhao's discussion of sums of indicators explains why overlapping counted structures require these pairwise terms. ^[ref].

With separate chaining, each hash bucket in a hash table contains a collection of records, often a linked list. Lookup computes the bucket index, then follows that chain until a key equality comparison succeeds or the chain ends. A hash collision therefore affects the search work without merging unequal keys. Deletion searches the same chain and unlinks the matching record; in a singly linked list, remembering the preceding record during the search permits the unlink.^[ref]

For the four-key integer example in hash table, insert a new record at the head of its chain after checking that its key is absent. Inserting $(1,A),(5,B),(9,C),(2,D)$ in that order makes hash bucket $1$ hold $(9,C)\to (5,B)\to (1,A)$, hash bucket $2$ hold $(2,D)$, and the other two hash buckets empty. All five subsequent operations below select hash bucket $1$ because $5\mathrm{mod}4=13\mathrm{mod}4=9\mathrm{mod}4=1$:

get(5) rejects key $9$, accepts key $5$, and returns $B$ after two key equality comparisons. get(13) rejects $9,5,1$ and reports absence after three comparisons. put(5,B*) examines $9,5$ and replaces the value without changing $n=4$. delete(9) examines just the head record, removes it, and gives $n=3$. Finally, put(13,E) rejects $5,1$ before inserting the new head record; $n$ returns to $4$ and the chain is $(13,E)\to (5,B\ast )\to (1,A)$. The successive hash table load factors are $1,1,1,3/4,1$. These counts follow from the stipulated chain order; a different insertion policy changes the successful-search counts.

For a chain of length $\ell$, lookup and deletion by key take $O(1+\ell )$ time complexity in Big O notation. The same applies to put before any resizing, since replacement semantics require checking for an existing equal key. Head insertion alone is $O(1)$ if absence is already known; that is a different precondition. These bounds assume $O(1)$ bucket-index hashing, array access, key equality comparisons, allocation and link updates. A long string or large integer can make hashing or comparison depend on its encoded length, so counting record visits alone does not establish constant running time.

Under simple uniform hashing, each distinct key's bucket index is an independent random variable with a uniform distribution on a finite set of $m$ indices. The keys and query are fixed independently of this random assignment. Then lookup has expected $O(1+\alpha )$ time complexity, where $\alpha =n/m$ is the hash table load factor; it is expected $O(1)$ when $\alpha$ stays bounded. These are model assumptions, not an automatic consequence of using a deterministic hash function.^[ref]

Here is the short calculation. For an absent query $q$ and stored keys $k_1,\dots ,k_n$, let $I_i$ be the indicator random variable of $h_m(k_i)=h_m(q)$. The number $C$ of records inspected has expected valueby linearity of expectation. For a present fixed query, scanning its entire chain inspects at most itself plus all other colliding records, with expected value $1+(n-1)/m$; stopping at the match only decreases this upper bound. Adding constant hashing and bucket access yields expected $O(1+\alpha )$ for either query. The argument needs pairwise collision bounds; it does not require all the indicator random variables to be independent. A universal hash family can provide a related bound by random choice of a hash function for a fixed key set, as explained in that article.

The worst case remains $O(n)$, and an absent query takes $n$ comparisons if all $n$ records occupy its hash bucket. For instance, $1,1+m,1+2m,\dots ,1+(n-1)m$ all collide under $h_m(k)=k\mathrm{mod}m$. Separate chaining remains correct, but a sequence of insertions that each checks the existing chain can take quadratic total work. Amortized analysis of resizing does not remove this search cost, and an adaptive choice of keys based on a known hash function is outside the stated simple uniform hashing model.

A shape parameter changes the form of a probability distribution without merely translating it or rescaling its random variable.

Draw one random variable $R$, taking values zero and two with equal probabilities. Conditional on $R$, generate a Poisson point process on the real line with constant rate $R$. This is a Cox process directed by the random measure $Rdx$: the same rate is shared by every window. It is a locally finite simple point process and a stationary point process. The law of total expectation gives $\mu (A)=\mathbb{E}[R]|A|=|A|$; the law of total variance and law of total covariance give, for bounded Borel sets $A,B$ with $A\cap B=\varnothing$,The conditional covariance vanishes by independence, but the shared conditional means fluctuate together. This distinguishes conditional from unconditional independence.^[ref]

For a window of length $L$, the exact probability mass function isIn particular, its void probability functional is $(1+e^{-2L})/2$, compared with $e^{-L}$ for the unit-rate Poisson point process. Whole realizations can be empty or comparatively crowded, while their mean density remains one. This example demonstrates shared-rate statistical dependence, rather than a model of short-range offspring clusters.

Choose one random variable $U$ with the continuous uniform distribution on $[0,1]$, and reuse it in every event:Each probability is $1/(n+1)$, so their harmonic series diverges, just as in summable and harmonic event probabilities. But the events are decreasing: $A_{n+1}\subseteq A_n$. For every outcome $U=u>0$, all indices with $n+1>1/u$ fail. Only finitely many events occur on that outcome. The limsup of events is exactly $\{U=0\}$, a zero-probability event. This is a concrete closed-interval version of the shrinking-interval dependence example.^[ref]

The missing assumption is visible already in two events. For $i<j$, their intersection is $A_j$, henceMoreover, the whole tail union is just $A_m$, with probability $1/(m+1)\to 0$. This calculation explains directly why divergence alone fails: overlapping events keep spending their individual probabilities on the same outcomes.

Let $N={\sum }_{n\ge 1}{1}_{A_n}$ be the total number of occurrences, a sum of indicator random variables. Although $N<\infty$ almost surely, Tonelli theorem gives $\mathbb{E}[N]={\sum }_{n\ge 1}1/(n+1)=\infty$. More explicitly, $\mathbb{P}(N\ge k)=1/(k+1)$ for $k\ge 1$: very small positive values of $U$ give very long, but finite, initial runs. Infinite expected value does not imply a positive probability of an infinite count.

For a real random variable $X$ on a probability space, $\sigma (X)$ is the smallest sigma-algebra making $X$ a measurable function into the Borel sigma-algebra on $\mathbb{R}$. The collection of all inverse images of Borel sets is already a sigma-algebra, since inverse images preserve complements and countable unions; this proves the displayed formula without completing any measure.

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

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

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

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

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

For a positive integer-valued random variable with finite mean $\mu$, the size-biased distribution has probabilities $k\mathbb{P}(Y=k)/\mu$. They sum to one by the definition of expected value. Multiplying the differentiated probability generating function by $s/\mu$ gives the displayed identity for $0\le s<1$, with the endpoint interpreted by its limit.

For a random variable $X$ with probability density function $e^{-|x|}/2$, the event $X^2\le y$ is $-\sqrt{y}\le X\le \sqrt{y}$ for $y\ge 0$. Integrating the symmetric density gives the displayed cumulative distribution function; it is zero for negative $y$. The density on $y>0$ is $e^{-\sqrt{y}}/(2\sqrt{y})$. Its integrable singularity at zero does not create an atom, because the cumulative distribution function is continuous there.

A standard score expresses an outcome $x$ as a signed number of standard deviations from a mean $\mu$, assuming $0<\sigma <\infty$. It is a dimensionless quantity. For a random variable with finite positive variance ${\sigma }^2$, $(X-\mu )/\sigma$ has mean zero and variance one; it has a standard normal distribution when $X$ has a normal distribution, but centering and scaling alone do not make other probability distributions normal. ^[ref].

Two random variables are statistically dependent when their joint probability distribution differs from the product of their marginal probability distributions. Equivalently, some events determined by the two variables violate the product rule that defines independence. Nonzero covariance implies statistical dependence when the moments exist, but zero covariance does not in general imply independence.

A statistical estimate is the realized numerical value of an estimator after observing a statistical sample. Before observation, the estimator is a random variable or random vector; its realized value is the estimate. Both are commonly written $\hat{\theta }$, so context distinguishes them.

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.