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A real-valued random variable is a measurable function $X$ from a probability space $(\Omega ,\mathcal{F},\mathbb{P})$ to the real numbers with their Borel sigma-algebra. More general targets give random elements. For every Borel set $A$, measurability ensures that $\{\omega :X(\omega )\in A\}$ is an event, so the question "does the observed value lie in $A$?" has a defined probability.

The function $X$ is fixed; its input outcome is uncertain. This separates what we measure from how the uncertainty is modeled. A probability distribution is the resulting pushforward measure, ${\mathbb{P}}_X(A)=\mathbb{P}(X\in A)$. A cumulative distribution function, probability mass function, or probability density function describes that measure in an appropriate setting. The random variable and its probability distribution are different objects: the former also retains its relationship to other measurements on the same sample space.

For two independent uniform dice, let $U,V$ record their results and define $S=U+V$ and $M=\max (U,V)$. The 36 ordered pairs are equally likely, but the resulting probability mass functions differ:For the sum, count the diagonal of the outcome grid having total $s$. For the maximum, subtract the $(m-1{)}^2$ pairs with both results below $m$ from the $m^2$ pairs with both results at most $m$. These examples show how a deterministic function transforms a simple input probability distribution into a nonuniform output one.

The corresponding expected values are $\mathbb{E}S=7$ and $\mathbb{E}M=161/36$. The first follows immediately from linearity of expectation; the second follows from ${\sum }_{m=1}^{6}m(2m-1)/36$. Neither number must be an attainable result. An expected value describes an average under the probability measure, not a prediction that the next observation will equal it.

The distinction between a random variable and its probability distribution matters for dependence. $U$ and $V$ have the same discrete uniform distribution and are independent random variables. $U$ and a second copy of $U$ also have that same marginal probability distribution, but they always agree. Their joint probability distribution is concentrated on the diagonal; the marginals alone cannot say which relationship holds. This is why a model for several measurements must specify their joint law or an appropriate independence assumption.

Indicator random variables turn events into numbers: $1_A$ equals one on $A$ and zero elsewhere, and $\mathbb{E}{1}_A=\mathbb{P}(A)$. This connects event counting with expected value and variance, and lets tools for random variables answer questions about probabilities. More complicated functions of observations can describe totals, waiting times, or estimation errors using the same framework.

A regular conditional distribution of a random variable $X$ given $Y$ is a Markov kernel $K$ such that, for all measurable sets $A,B$,For each $y$, $K(y,\cdot )$ is a probability measure; for each $B$, $K(\cdot ,B)$ is measurable. Thus it assigns an actual probability distribution to a specified observation even when $P(Y=y)=0$. Such kernels exist for standard Borel state spaces, including real-valued variables, and their uniqueness is only up to $P_Y$-null sets. With a joint density and positive marginal density, the ratio of the two densities defines the usual conditional density kernel. Values outside the marginal support or on a null set are not uniquely determined by the joint law.

For a countable discrete group, the Reiter condition requires a sequence of probability measures in ${\ell }^1(G)$ asymptotically invariant under every left translation. It is equivalent to the existence of an invariant mean and a Følner sequence. Separation by bounded functions shows that amenability gives approximate invariance simultaneously on each finite set of translations; countability makes the successive finite tests into a sequence.

A risk-neutral measure $Q$ is a probability measure equivalent to the physical probability measure $P$, meaning they agree on which events have zero probability. For assets without intermediate payments, prices expressed in the chosen numéraire must be martingales under $Q$. For a positive bank account $B$ and a stock without dividends, this requires $S/B$ to be a martingale; with dividends, use discounted dividend gains. In an unrestricted finite-state one-period financial market, existence of such a measure is equivalent to absence of arbitrage, by the fundamental theorem of asset pricing.

Physical probability $p=P(\text{up})$ describes the modeled frequency or belief about an up move. The risk-neutral probability $q=Q(\text{up})$ is constrained by traded prices. In the worked binomial market, $q=5/8$ even if $p=1/2$ or $p=4/5$. Calling $Q$ risk-neutral does not assume that investors have no risk aversion; it changes the weights used for risk-neutral pricing.

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.

A finite signed measure with total mass one under which discounted asset gains have zero integrals. Its density can be negative, so it need not be a probability measure or give arbitrage-free prices to every nonnegative contingent claim. A square-integrable signed density is useful in one-period least-squares hedging, even when it is not an equivalent martingale measure.

A statistical functional assigns a target value to each probability measure in a statistical model. Examples include an expected value, a quantile, and an integral of a power of a probability density function. Its behavior along statistical paths determines whether first-order influence-function representers are available.

A statistical path is a one-dimensional family of probability measures inside a statistical model, passing through a specified baseline $P$. A differentiability in quadratic mean condition gives its score function, the first-order direction of the family in square-root-density coordinates. The path must respect all restrictions of the statistical model, including normalization and any fixed moments.

A function into a Banach space is strongly measurable if, outside a null set, it is a pointwise norm limit of measurable simple functions. A measurable function with values in a separable Banach space is strongly measurable: approximate values by a countable collection of balls of shrinking radii and then truncate the resulting countably valued approximants to simple functions. A strongly measurable function is Bochner integral integrable exactly when its norm is integrable. Finite second moments under a probability measure therefore imply existence of its Bochner integral expected value by the Cauchy-Schwarz inequality.

For a probability measure on a topological space, its support consists of the points whose every open neighborhood has positive probability. For a continuous effort distribution supported on an interval, every open subinterval of that interval has positive probability.

Surface area is the two-dimensional measure induced on the unit sphere in three dimensions. It is invariant under orthogonal rotations. It is not the restriction of ambient three-dimensional Lebesgue measure, under which the sphere has measure zero. Normalizing by the sphere area gives a probability measure.

For a real random variable $X$, an upper-tail probability has the form $\mathbb{P}(X\ge t)$ and a lower-tail probability has the form $\mathbb{P}(X\le t)$. Each is the probability of an event lying beyond a chosen threshold in one direction. A tail probability need not be small: the threshold determines how much of the probability distribution it includes. A concentration inequality bounds such probabilities, often for deviations from an expected value.

Endpoint conventions matter. Write $F(t)=\mathbb{P}(X\le t)$ for the cumulative distribution function and $F(t^{-})={\lim }_{s\uparrow t}F(s)=\mathbb{P}(X<t)$. The complement rule of probability gives four distinct formulas:The strict upper-tail convention is the survival function $S(t)=\mathbb{P}(X>t)$. Its value differs from the inclusive upper tail probability by exactly the probability mass at the threshold:A singleton with positive probability mass is an atom of a measure. Thus $1-F(t)$ excludes an atom of a measure at $t$, whereas $1-F(t^{-})$ includes it. These endpoint identities follow directly from the cumulative distribution function properties in Kyle Siegrist, Distribution and Quantile Functions: endpoint probabilities and right tails.

As the threshold increases, both upper tail probabilities are nonincreasing functions. The strict survival function is a right-continuous function; the inclusive upper tail probability takes the upper endpoint value at a downward jump, equal to the left limit of the survival function. Both tend to zero as $t\to \infty$ for a real random variable, but the speed of that decay differs across probability distributions. The cumulative distribution function properties and their proof from continuity of probability measures are given in MIT 6.436J, Lecture 4: random variables and cumulative distribution function properties.

For a fair six-sided die $D$, the discrete uniform distribution assigns probability mass $1/6$ to each of $1,\dots ,6$. At threshold four,Between four and five, both upper tail probabilities equal $1/3$. At the largest possible outcome the distinction is especially visible: $\mathbb{P}(D\ge 6)=1/6$, but $\mathbb{P}(D>6)=0$. The diagram shows exact probabilities for this mathematical model; its dots specify which endpoints belong to each step function.

If a probability distribution has a probability density function $f$, then every singleton has zero probability mass andFor an exponential distribution of rate two, $f(x)=2e^{-2x}$ for $x\ge 0$ and zero otherwise, so $S(t)=e^{-2t}$ for $t\ge 0$ and $S(t)=1$ for $t<0$. Hence $\mathbb{P}(X\ge 1)=e^{-2}\approx 0.135335$. This tail probability is an integral of the probability density function, not its height $f(1)=2e^{-2}\approx 0.270671$. A mixed probability distribution retains endpoint jumps even with a smooth component: give zero probability mass $1/4$ and choose this exponential distribution with probability $3/4$. Then $\mathbb{P}(X\ge 0)=1$, $\mathbb{P}(X>0)=3/4$, and $S(t)=(3/4)e^{-2t}$ for $t\ge 0$.

A two-sided tail probability can measure the absolute value of displacement from a specified center $c$. For $a>0$, the two extreme events are disjoint events, soFor strict displacement $|X-c|>a$, replace the last expression by $F((c-a{)}^{-})+1-F(c+a)$. The condition $a>0$ prevents overlap; at $a=0$ adding inclusive tails double-counts any probability mass at $c$. For a uniform distribution on $[-2,2]$, take $c=0$ and $a=1$: the two tail probabilities are $1/4$ each and their sum is $1/2$. Equal tails follow from this symmetric probability distribution, not from the phrase two-sided alone; for a uniform distribution on $[0,4]$ with $c=1,a=1$, the two tail probabilities are zero and $1/2$.

The tail-sum formula for expectation turns a whole curve of tail probabilities into an expected value. For any nonnegative random variable $Y$, without assuming a probability density function or a finite expected value,For a proof outline, write $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise and exchange the expected value and Lebesgue integral using Tonelli theorem; $1$ is an indicator function. The inclusive version has the same Lebesgue integral: positive probability masses occur at only a countable set of thresholds, which has Lebesgue measure zero. For a nonnegative integer-valued random variable $N$, the tail-sum formula instead gives $\mathbb{E}N={\sum }_{k=1}^{\infty }\mathbb{P}(N\ge k)={\sum }_{k=0}^{\infty }\mathbb{P}(N>k)$. The integer hypothesis is essential for these sums. The integral identity, its indicator function proof, and the integer version appear in Kyle Siegrist, Additional Properties of Expected Value: integration and summation of right tails.

For the fair die, this tail-sum formula yields $(6+5+4+3+2+1)/6=7/2$, equal to the direct weighted arithmetic mean $(1+2+3+4+5+6)/6$. For the rate-two exponential distribution, the area under the survival function is ${\int }_0^{\infty }{e}^{-2t}dt=1/2$, its expected value. The first panel below shades an integral of the probability density function for a single threshold; the second shades an integral of the survival function across all nonnegative thresholds. These are analytical illustrations, with the displayed curves truncated at three; their exact total areas use an infinite integration range.

Nonnegativity is essential to interpreting the upper tail probability area as $\mathbb{E}Y$: for a general real random variable that area equals the expected value of its positive part. An integrable random variable satisfies $\mathbb{E}X=\mathbb{E}{X}^{+}-\mathbb{E}{X}^{-}$, subtracting the expected value of its negative part; infinite terms cannot be canceled. Even when $Y$ is finite almost surely, its expected value can be infinite. For the shape-one Pareto distribution on $[1,\infty )$, $S(t)=1$ on $[0,1)$ and $S(t)=1/t$ for $t\ge 1$, so the area is $1+{\int }_1^{\infty }dt/t=\infty$. The finite versus infinite expected value distinction is illustrated in Kyle Siegrist, Definitions and Basic Properties: exponential and Pareto expected values.

Why are tail probabilities useful? They answer threshold questions directly, and a concentration inequality can bound them without knowing the full probability distribution. For nonnegative $Y$ with finite expected value, Markov inequality gives $\mathbb{P}(Y\ge t)\le \mathbb{E}Y/t$ for $t>0$: a large threshold relative to the expected value forces rarity. The quantile function reverses the question by choosing a threshold for a prescribed cumulative level. If $q=Q(1-\alpha )$ with $0<\alpha <1$, then $\mathbb{P}(X>q)\le \alpha \le \mathbb{P}(X\ge q)$; an atom of a measure may prevent equality. For the die, $Q(0.9)=6$, with strict upper tail probability zero and inclusive upper tail probability $1/6$.

A tail probability describes rarity under a specified probability distribution; it is not automatically a conditional probability after extra information. For $u\ge t$ with $\mathbb{P}(X\ge t)>0$,because the threshold events are nested. Thus the die has $\mathbb{P}(D\ge 5)=1/3$ but $\mathbb{P}(D\ge 5\mid D\ge 4)=2/3$. Conditioning changes the sample space and renormalizes its probability measure; the same extreme outcomes can become common in that restricted population.

A probability measure is tight if compact sets carry arbitrarily nearly all its mass. Every Borel probability measure on a Polish space is tight. The stronger family condition is uniform tightness.

Total variation distance asks how much two probability distributions can disagree about the same event. For probability measures $P,Q$ on one measurable space $(S,\mathcal{F})$, its definition isThe supremum ranges over all measurable sets, so $d_{\mathrm{TV}}(P,Q)\le \epsilon$ guarantees that every yes-or-no prediction has probability error at most $\epsilon$. This metric takes values between zero and one; it is zero exactly when the two probability measures agree. This makes it useful for comparing a sampling algorithm's output with its target probability distribution, and for defining the mixing time of a Markov chain.

On a finite set, add the excess probability mass wherever $P$ exceeds $Q$. That excess is also half the sum of all absolute differences:The last expression reveals the common probability mass that the two probability distributions can share.^[ref] The coupling characterization of total variation distance turns this overlap into an actual joint construction; minimum sum of errors in a simple hypothesis test gives it a statistical interpretation. One quantity connects probability bookkeeping, comparison of sampling methods, and the limits of a statistical test.

A transport plan is a coupling of probability distributions $\mu ,\nu$, that is, a probability measure on the product space with those marginal distributions. Unlike a transport map, it need not be concentrated on the graph of a function.

Model two fair independent coin tosses, recording their order. Here $H$ denotes heads and $T$ tails, andThe power set $\mathcal{P}(\Omega )$ includes every question expressible as a subset of these four outcomes. The mass $1/4=(1/2)(1/2)$ comes from the fairness and independence assumptions. These are exact model probabilities, rather than measured toss frequencies.

Let $A$ be the event that the first toss is heads and $B$ the event that the second toss is heads. ThenThe union means “at least one head,” while the intersection means “both heads.” Their probabilities are $3/4$ and $1/4$. Counting the union through the inclusion-exclusion principle gives $1/2+1/2-1/4=3/4$; alternatively its complement of an event is $\{TT\}$, of probability $1/4$.

On the same sample space, assigning masses $3/8,1/8,1/8,3/8$ to $HH,HT,TH,TT$ would still make each toss fair: $\mathbb{P}(A)=\mathbb{P}(B)=1/2$. But $\mathbb{P}(A\cap B)=3/8\ne 1/4$, so the tosses would be dependent. The event of at least one head would have probability $5/8$. Changing the probability measure changes the answer even when the sample space and the single-toss probabilities stay the same.