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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.

Let $n\ge 2$, and let $s_{n-1}$ denote the total surface area of the unit sphere $S^{n-1}$. A spherical cap $D_{\rho }$ of angular radius $0<\rho <\pi /2$ satisfiesIn particular, the condition holds for $\rho \ge \pi /4$ in every such dimension. The normalization matters: $s_{n-1}$ is the area of the entire sphere, rather than the enclosed volume or a unit-normalized probability measure.

Here is a proof from spherical slices. The normalized cap area isSince $\cos t\ge \cos \rho$ throughout the integration interval,This also shows by differentiation that $F(\rho )/{\sin }^{n-1}\rho$ is increasing. Its endpoint value at $\pi /2$ is $1/2$, giving a second bound $F(\rho )\le {\sin }^{n-1}\rho /2$.

The Cauchy-Schwarz inequality applied to the gamma function integral gives $\Gamma (x+1/2{)}^2\le \Gamma (x)\Gamma (x+1)=x\Gamma (x{)}^2$, so $C_n\le \sqrt{n-1}/2$. On $\arcsin (n^{-1/2})\le \rho \le \pi /4$, the first bound yieldsFor $\rho \ge \pi /4$, the second bound gives the same conclusion because $2\sin \rho \ge 1$.

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 $\mu$ on a topological space, its support isA point belongs to the support when observations can fall arbitrarily close to it with positive probability. This is a neighbourhood condition, not a requirement that the point itself have positive probability mass or that a chosen density be positive there.

For the uniform distribution on $(0,1)$ viewed as a law on the real line, the support is $[0,1]$, including both endpoints. A density $f(x)=3x^2/2$ on $[-1,1]$ vanishes at zero, but zero still lies in its support because every open neighbourhood of zero has a positive integral of $f$. A discrete distribution with mass $2^{-n}$ at $1/n$, $n\ge 1$, likewise has zero in its support despite assigning it probability mass zero.

If a distribution's support equals an interval, every open subinterval of that interval has positive probability. The weaker condition $\mu (I)=1$ does not imply that the support equals $I$: a uniform law on $(0,1)$ also assigns probability one to $[0,2]$, but assigns zero to its subinterval $(3/2,7/4)$. This distinction matters when using support to describe feasible observations or randomized effort in a contest.

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.

If each probability measure ${\mu }_i$ has Poincaré constant $C_i$, then the product measure ${⨂}_i{\mu }_i$ has Poincaré constant at most ${\max }_i{C}_i$. This follows by iterating the law of total variance and applying Jensen inequality to derivatives of conditional expectations.

A precise sum-cost form of tensorization of a transport-entropy inequality is as follows. Let ${\mu }_i$ be probability measures on Polish spaces and $c_i$ nonnegative continuous transport cost functions. Suppose for every probability measure ${\nu }_i$ thatThen for the product measure $\mu ={⨂}_i{\mu }_i$ and sum cost $c(x,y)={\sum }_i{c}_i(x_i,y_i)$,To understand the mechanism first for finite spaces, reveal the coordinates under $\nu$ in order. Given the previous coordinates $x_{<i}$, couple the conditional law ${\nu }_i(\cdot \mid x_{<i})$ to ${\mu }_i$. Choosing these couplings successively preserves the product measure as the second marginal. The total expected cost is at most $C{\sum }_i{\mathbb{E}}_{\nu }D({\nu }_i(\cdot \mid X_{<i})\Vert {\mu }_i)$, which equals $CD(\nu \Vert \mu )$ by the chain rule for relative entropy. Approximate optimal couplings give the infimum. This statement spells out the cost and constant; it does not identify every different distance-based normalization with the same formula.^[ref]

Work with the real Lp space $L^{\infty }((0,1))$ for Lebesgue measure, so its predual $L^1((0,1))$ is separable. Choose a measurable representative $f$ of a unit-ball element with $-1\le f\le 1$ and putFor every $t$, $H_t$ has modulus one almost everywhere, so it is an extreme point of the real $L^{\infty }$ closed unit ball. The map $t\mapsto H_t$ is Borel for the weak-star topology: pairing with any $g\in L^1$ is measurable, and separability makes the weak-star unit ball metrizable. Push forward uniform probability measure on $[0,1]$ through this map.

For almost every $s$,Fubini's theorem then gives ${\int }_0^1\int g(s)H_t(s)dsdt=\int g(s)f(s)ds$ for every $g\in L^1$. Thus the resulting Borel probability measure is concentrated on the extreme points and has barycenter $f$.

For $f(s)=2s-1$, the profiles are the threshold steps $H_t(s)=21_{\{t\le s\}}-1$. Averaging these sign-valued extreme points produces the linear profile. This is an explicit Choquet theorem representation in a setting where the required metrizability is actually available.

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-entropy inequality compares the cost of moving one probability measure to another with their relative entropy. For a nonnegative transport cost function $c$ and reference probability measure $\mu$, definewhere $\Pi (\nu ,\mu )$ consists of couplings with those marginals. One common convention is ${\mathcal{T}}_c(\nu ,\mu )\le CD(\nu \Vert \mu )$ for every probability measure $\nu$; the orientation of the relative entropy matters.

For a metric $d$, with the measures having finite first moments, the $T_1(C)$ convention is Wasserstein distance $W_1(\nu ,\mu {)}^2\le 2CD(\nu \Vert \mu )$, while $T_2(C)$ replaces $W_1$ by the second Wasserstein distance $W_2$. The first convention bounds fluctuations of all 1-Lipschitz functions: $\log {\mathbb{E}}_{\mu }{e}^{\lambda (f-{\mathbb{E}}_{\mu }f)}\le C{\lambda }^2/2$. The choice of a squared distance or an unsquared transport cost function is part of the statement, rather than a universal normalization.^[ref]

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.

A Tukey median maximizes halfspace depth. For a probability measure $P$ on ${\mathbb{R}}^d$, the depth of a point $x$ isFor a sample, $P$ is replaced by its empirical distribution, so $P(H)$ is the fraction of observations in the closed half-space $H$. Every direction is considered; choosing a single coordinate direction would give a different procedure.

In one dimension, the depth is the smaller of the masses in the closed rays to the left and right of $x$. For the sample $(0,1,2,3,100)$, point $2$ has depth $3/5$, while $100$ has depth $1/5$. The maximizing point is the sample median. In higher dimensions this gives a geometric extension of a central location that depends on how many observations lie on each side, rather than on their squared distances. Maximizers need not be unique, so a single-valued estimator needs a selection rule.

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.

A Markov kernel $P$ has Wasserstein contraction factor $c<1$ for a ground metric $\rho$ whenfor all probability measures for which the distances are finite. Writing $c=e^{-\alpha }$ gives a positive one-step contraction rate $\alpha$. Iteration yields $W_{\rho }(\mu P^n,\nu P^n)\le c^n{W}_{\rho }(\mu ,\nu )$. If $\pi$ is a stationary distribution with finite first moment, this proves geometric convergence in Wasserstein distance to $\pi$ and uniqueness within that moment class. A continuous-time version instead uses factor $e^{-\alpha t}$ for its transition semigroup.

For the random affine recursion $X_{n+1}=a{X}_n+{\xi }_{n+1}$ on the real line with $|a|<1$ and independent finite-first-moment noise, Synchronous coupling uses the same fresh ${\xi }_{n+1}$ for both chains. It gives $|X_{n+1}-Y_{n+1}|=|a||X_n-Y_n|$, hence the contraction factor $|a|$ for $W_1$. The stationary random series ${\sum }_{j\ge 0}{a}^j{\xi }_{-j}$ converges absolutely almost surely because its expected absolute sum is finite. Thus this example also supplies existence, which a uniqueness argument alone would not establish.

The first Wasserstein distance measures the least expected separation achievable when two random variables are coupled with specified marginal probability measures. For a metric space $(E,\rho )$ and measures $\mu ,\nu$ with finite first moments,A transport plan $\pi$ says how much probability mass moves from each starting location to each destination; the distance pays the corresponding metric cost. For point masses on the real line, $W_1({\delta }_a,{\delta }_b)=|a-b|$, because there is only one coupling. In contrast, their Total variation distance is one whenever $a\ne b$. Wasserstein distance therefore records that nearby displacements are small even when the two laws give mass to disjoint sets.

Every coupling gives an upper bound by its expected separation. Conversely, the Kantorovich–Rubinstein theorem gives lower bounds from differences of expectations of Lipschitz functions. The p-Wasserstein distance generalizes the construction by using the $p$th power of the ground metric and then taking a $p$th root.

probability measures ${\mu }_n$ on a metric space converge weakly to $\mu$ when $\int fd{\mu }_n\to \int fd\mu$ for every bounded continuous real function $f$.