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A centered random variable $X$ is sub-Gamma in the right tail with variance parameter $v$ and scale parameter $c$ whenfor $0\le \lambda <c^{-1}$.

A centered random variable $X$ is sub-Poisson in the right tail with variance parameter ${\sigma }^2$ whenfor every $\lambda \ge 0$.

If a finite group $\Gamma$ acts by measure-preserving maps on a probability space, let $\mathcal{I}$ be its invariant sigma-algebra. For an integrable random variable $Z$, the displayed group average is $\mathcal{I}$-measurable. For $A\in \mathcal{I}$, changing variables under each group element gives ${\int }_{A}Z\circ T_{g}d\mathbb{P}={\int }_{A}Zd\mathbb{P}$. The average therefore satisfies the defining integral identity for conditional expectation. For permutations of $N$ exchangeable coordinates, averaging a product attached to $d$ distinct indices yields the sum over all ordered distinct $d$-tuples divided by the falling factorial $(N{)}_d=N!/(N-d)!$.

A tail event is an event $A\in \mathcal{T}$, where $\mathcal{T}$ is the tail sigma-algebra of the chosen sequence of random variables. It concerns information available arbitrarily far along the sequence. In the measurable coordinate models described in finite-prefix invariance of tail events, any finite edit preserves whether it occurs.

This is stronger than merely needing a long waiting time. On $\{0,1{\}}^{\mathbb{N}}$, with $1$ denoting heads, let $K=\inf \{j\ge 1:X_j=1\}$, taking $K=\infty$ if there is no head. The eventis not a tail event: changing only $X_1$ from zero to one changes membership. Under fair independent Bernoulli trials its probability is $1/8$. Even the event $\{K<\infty \}$, of probability one in this model, is not a tail event as a literal set: the all-zero sequence and the sequence with just one initial head differ in one coordinate. A probability of zero or one does not characterize tail events.

For a nonnegative random variable, $Y={\int }_0^{\infty }{1}_{\{Y>t\}}dt$ pointwise, including when $Y=\infty$. Tonelli's theorem permits taking expectation under the integral and proves the formula in the extended nonnegative reals. This relates finiteness of a Brownian exit time expectation to integrability of its survival probability.

For $q>0$ and a random variable $X$ with finite absolute $q$th moment, Tonelli theorem gives

Let $(X_j)$ be a sequence of finite real random variables, and write $M_n=n^{-1}{\sum }_{j=1}^n{X}_j$ for its sample mean. For fixed $m$ and $n>m$, put $M_n^{(m)}=n^{-1}{\sum }_{j=m+1}^n{X}_j$. Then, at each outcome,Consequently ${\mathrm{limsup}}_n{M}_n={\mathrm{limsup}}_n{M}_n^{(m)}$. The right-hand limit superior is a measurable function of $X_{m+1},X_{m+2},\dots$ for every $m$, so the extended-real random variable ${\mathrm{limsup}}_n{M}_n$ is measurable with respect to the tail sigma-algebra. The same holds for the limit inferior. Thus the event of convergence to a finite limit of a sequence, and the event $\{M_n\to a\}$ for any specified real $a$, are tail events.

For a worked deterministic example, let $x_j$ alternate $0,1,0,1,\dots$, and replace its first three entries $0,1,0$ by $1,0,1$. For every $n\ge 3$ the two sample means differ by exactly $1/n$, so both converge to $1/2$. Finite edits can change small-sample answers while preserving the infinite limit of a sequence.

Under independence, Kolmogorov's zero–one law makes each specified convergence event have probability zero or one. It does not select which value occurs. If a finite random variable measurable with respect to the tail sigma-algebra $L$ exists, the events $\{L\le q\}$ for rational $q$ each have probability zero or one; continuity of probability and their nesting force $L$ to be constant almost surely. For fair independent Bernoulli trials, the strong law of large numbers establishes $M_n\to 1/2$ almost surely; the zero–one conclusion alone proves neither convergence nor the value $1/2$.

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.

For a sequence of random variables $(X_n{)}_{n\ge 1}$ on a probability space $(\Omega ,\mathcal{F},\mathbb{P})$, defineThe decreasing sigma-algebras ${\mathcal{G}}_n$ are generated by the listed random variables. Their intersection consists of events expressible after discarding every finite prefix. No independence is required.^[ref]

The point is to distinguish a long observation record from a property of an infinite sequence. The event of seeing infinitely many heads and the event that sample means converge survive any finite change, while an exact first-head time does not. Kolmogorov's zero–one law adds the striking conclusion that under independence each tail event has probability zero or one; the tail sigma-algebra itself can still contain many nonempty sets.

The tail-sum formula expresses an expected value as the area under a survival function. For every nonnegative random variable $Y$,This Lebesgue integral and the expected value may both be $+\infty$; a probability density function is unnecessary. For a random variable $N$ taking values in the nonnegative integers, the corresponding sum isThe events $\{N>k\}$ and $\{N\ge k+1\}$ coincide because $N$ takes integer values. Thus the strict sum starts at zero, while the inclusive sum starts at one.^[ref]

For an original example, let the probability mass function of $N$ beIts expected value computed directly is $0(1/5)+1(3/10)+3(1/2)=9/5=1.8$. The strict tail probabilities at thresholds $0,1,2,3$ are $4/5,1/2,1/2,0$: their sum is again $1.8$. There is no probability mass at two, yet the event $\{N>2\}$ still has probability $1/2$. The survival function has heights $4/5,1/2,1/2$ on the successive intervals $[0,1),[1,2),[2,3)$; each interval has length one, so its integral is the same sum.

The indexing matters numerically. Starting the inclusive sum at zero adds $\mathbb{P}(N\ge 0)=1$, giving $2.8$ for this example. Starting the strict sum at one omits $\mathbb{P}(N>0)=4/5$, giving $1$. For a general nonnegative random variable, replacing $>$ with $\ge$ inside the integral is harmless: the difference is $\mathbb{P}(Y=t)$, nonzero at only a countable set of thresholds. For each positive integer $m$, at most $m$ thresholds can have probability mass at least $1/m$; their countable union contains all thresholds with positive probability mass and has Lebesgue measure zero. This reasoning concerns an integral, not equality of the two survival function conventions at every point.

For a continuous original example, take the uniform distribution on the interval $[0,4]$. Its survival function is $S(t)=1-t/4$ for $0\le t\le 4$ and zero for $t>4$. The area is a triangle of base four and height one:Directly integrating against the probability density function gives the same expected value, ${\int }_0^{4}y(1/4)dy=2$. The integral of the probability density function alone is one; it is the survival function whose area is the expected value.

The proof counts layers of each outcome. For a realized value $y\ge 0$, the indicator function $1_{\{y>t\}}$ equals one over an interval of length $y$, soTonelli theorem permits the interchange because the indicator function is nonnegative; no finite expected value is assumed. Similarly, an outcome $N=n$ contributes one indicator random variable for each of the $n$ events $\{N\ge k\}$ that occur:Again Tonelli theorem applies to the nonnegative sum, or equivalently the monotone convergence theorem applies to its partial sums. This exchanges counting layers within each outcome with averaging across outcomes. The existing integer-valued hypothesis in the tail-sum formula explains what changes when the values are not integers; the tail integral formula for moments weights the layers to obtain higher moments.

For an integrable random variable $X$ that can have either sign, apply the nonnegative formula separately to its positive part $X_{+}=\max (X,0)$ and negative part $X_{-}=\max (-X,0)$:Integrability makes both integrals finite. More generally the difference defines an extended expected value when at least one is finite; if both are infinite, subtraction does not define an expected value.^[ref]

Tail probabilities are often easier to measure or bound than a whole probability density function. For a device's survival time, the survival function records the probability of still operating at each threshold; its integral gives the mean operating time. A finite observation window naturally gives the expected value of a capped nonnegative random variable. For a mathematical bound, $\mathbb{P}(Y>t)\le e^{-t/2}$ for all $t\ge 0$ implies $\mathbb{E}[Y]\le {\int }_0^{\infty }{e}^{-t/2}dt=2$. The tail-sum formula turns control of a whole family of events into control of an average.

For a nonnegative random variable $X$, Tonelli theorem givesConsequently, for each $a>0$, $\mathbb{E}[X]=\infty$ exactly when ${\sum }_{n\ge 1}\mathbb{P}(X>an)=\infty$; compare the integral on successive intervals of length $a$.

A terminal portfolio wealth floor is a prescribed lower bound, possibly a random variable, on terminal portfolio portfolio wealth. In a complete market, its state-price density cost cannot exceed initial portfolio wealth. Nonnegative admissibility replaces a possibly negative floor by its positive part.

Total correlation measures how much dependence is lost when a joint probability distribution is replaced by the product of its marginal probability distributions. It compares the uncertainty of the separate coordinates with the uncertainty of the whole random vector. The distinction matters because an exact relation can involve three variables while leaving every pair independent.

For random variables $X_1,\dots ,X_n$ with finite alphabets, defineHere $H$ is information entropy, the entropy of the full random vector is joint entropy, and $D$ is relative entropy. Base-two logarithms give bits; natural logarithms give nats. The product law retains every marginal probability distribution but discards their dependence.^[ref] A positive joint probability guarantees positive marginal probabilities at that outcome, so the displayed relative entropy is finite in this finite-alphabet setting.

For two variables, total correlation is exactly mutual information. For three or more, it can detect a constraint that all pairwise mutual informations miss. Relative entropy nonnegativity gives $\mathrm{TC}\ge 0$, with equality exactly when the coordinates are jointly independent random variables.

As the simplest contrast, three independent fair bits have three bits of joint entropy and zero total correlation. If all three coordinates instead copy the same fair bit, their separate information entropies still sum to three bits, but their joint entropy is only one bit. Their total correlation is therefore two bits. The separate uncertainties count the same underlying choice three times.

Suppose one observation comes from either $P$ or $Q$, each with prior probability $1/2$. A decision rule that chooses $P$ on an event $E$ and $Q$ otherwise has success probabilityOn a finite set, choosing $E=\{x:P(x)>Q(x)\}$ therefore gives $\frac{1}{2}(1+d_{\mathrm{TV}}(P,Q))$. This is the equal-prior version of minimum sum of errors in a simple hypothesis test; allowing a statistical test to use an auxiliary random variable cannot improve it, because for each outcome the better of the two decisions already maximizes its contribution.

For the four-outcome example, choose $P$ after $A$ and $Q$ after $B$ or $C$, with either decision allowed after $D$. Choosing $Q$ after $D$ yields $\frac{1}{2}(0.4+0.9)=0.65$. The best success probability is 65%, with a 35% error probability. A smaller total variation distance means the two sources are harder to distinguish from one observation. Unequal prior probabilities, unequal costs, or multiple observations change this particular decision problem; the $65\%$ figure is not a universal bound for every experiment.

A measurable function $g$ applied to a random variable $X$ defines a new random variable $Y=g(X)$. Its probability distribution is the pushforward measure under $g$.

For a nonnegative random variable with finite mean $\mu$ and $0\le \theta \le 1$, the part of the expectation on $X\le \theta \mu$ is at most $\theta \mu$. Subtracting gives $\mathbb{E}[X{1}_{\{X>\theta \mu \}}]\ge (1-\theta )\mu$. If the second moment is positive and finite, applying the Cauchy-Schwarz inequality to $X$ and the indicator variable gives the Paley-Zygmund inequality $\mathbb{P}(X>\theta \mu )\ge (1-\theta {)}^2{\mu }^2/\mathbb{E}[X^2]$.

Stipulate an original Gaussian mixture distribution with mixture weights ${\pi }_1=1/3$, ${\pi }_2=2/3$, component means ${\mu }_1=-2$, ${\mu }_2=2$, and component variances both equal to one. These are chosen statistical parameters, with no observed sample or fitting. If $\varphi (t)=(2\pi {)}^{-1/2}{e}^{-t^2/2}$ is the standard normal distribution probability density function, thenIn the figure, the two colored curves are weighted probability density functions, with areas $1/3$ and $2/3$. Adding their heights at every $x$ gives the black curve. Neither colored curve stops at a dividing line.

The law of total expectation and law of total variance give the total mean and variance:All of the extra variance $32/9$ comes from variation between component means. Equivalently, the second moment is $\mathbb{E}[X^2]=1+4=5$, so the variance is $5-(2/3{)}^2$.

A mixture distribution is different from a deterministic linear combination of random variables. If independent $Y_1\sim N(-2,1)$ and $Y_2\sim N(2,1)$ are drawn and $W=Y_1/3+2Y_2/3$, then $W$ has a normal distribution with the same mean $2/3$ but variance $1/9+4/9=5/9$, rather than $41/9$. Every draw of $W$ uses both random variables. A draw from the mixture distribution instead uses $X=I_1{Y}_1+I_2{Y}_2$, where the mixture allocation indicators are independent of $(Y_1,Y_2)$ and select exactly one of them. The expression for $f$ adds densities, not simultaneous random outcomes.

Two square-integrable random variables $X,Y$ are uncorrelated random variables when $\mathrm{Cov}(X,Y)=0$. Independent random variables with finite second moments are uncorrelated random variables, but zero covariance does not generally imply independence.

For an explicit counterexample, let $X$ take $-1$, $0$, $1$ with equal probabilities, and let $Y=X^2$. Then $\mathbb{E}X=\mathbb{E}{X}^3=0$, so $\mathrm{Cov}(X,Y)=\mathbb{E}{X}^3-\mathbb{E}X\mathbb{E}{X}^2=0$. Nevertheless, $Y$ is completely determined by $X$: $\mathbb{P}(Y=0\mid X=0)=1$, while $\mathbb{P}(Y=0)=1/3$. Thus these are uncorrelated random variables with statistical dependence. In this example, opposite contributions from the odd function $x^3$ cancel; the even function $x^2$ still retains information about $|X|$. The last panel of the covariance diagram shows the relationship.

There are useful special cases. Uncorrelated jointly normal variables are independent, and the Bernoulli independence from zero covariance result supplies another setting where the converse holds. Their distribution assumptions do the extra work; zero covariance alone does not.

The variance measures the mean squared distance of a random variable from its expected value. For a real square-integrable random variable $X$,The square prevents positive and negative deviations from cancelling. It also gives the variance squared units: if $X$ measures a length in metres, its variance has units of square metres. Its nonnegative square root is the standard deviation, which has the same units as $X$.

For example, let $X$ take $-1$, $0$ and $1$ with equal probabilities. Its expected value is zero, but its variance is $[(-1{)}^2+0^2+1^2]/3=2/3$, and its standard deviation is $\sqrt{2/3}$. The constant random variable zero has the same expected value and zero variance: a mean alone does not describe the spread. The variance is zero exactly when $X=\mathbb{E}X$ almost surely.

The variance is useful because squared deviations interact neatly with linear combinations. The covariance supplies the cross terms when combining different random variables, and a covariance matrix organizes those terms for many variables. The characterization of variance as the minimum mean squared error of a constant also explains why the expected value is the best constant prediction under squared error.

Let $X_1,\dots ,X_n$ be an IID sample of real-valued random variables with finite expected value $\mu$ and finite variance ${\sigma }^2$, where $n\ge 1$. Its sample mean $\overline{X}=n^{-1}{\sum }_i{X}_i$ satisfiesLinearity of expectation gives the first identity. The variance of a sum and vanishing cross covariances give $\mathrm{Var}({\sum }_i{X}_i)=n{\sigma }^2$; scaling by $1/n$ scales the variance by $1/n^2$. Pairwise independent random variables with the same expected value and variance suffice for these identities, even without full independence. Grinstead and Snell state the corresponding finite-variance independent-trials result.^[ref]

The standard deviation of the sample mean is $\sigma /\sqrt{n}$. Four times as many observations halve this scale; they do not divide it by four. In the fair Bernoulli distribution example, $\sigma =1/2$, so sample sizes $4$ and $64$ give standard deviations $1/4$ and $1/16$, respectively. The common-coin averaging example shows why merely repeating observations with a common marginal law does not justify this calculation.

Chebyshev's inequality turns the variance identity into a probability bound: for $\epsilon >0$,This proves convergence in probability of the sample mean to $\mu$ in this finite-variance setting, a form of the weak law of large numbers. It is an upper bound, so values above one convey no improvement over the universal probability bound one. Cauchy sample mean stability lies outside these hypotheses.