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Normalization of the probability density function requiresHence $\boxed{c=t-1}$. The distribution is a Pareto distribution with minimum one and shape parameter $t-1$.

The probability density function is nonnegative for $\alpha >0$. Its integral over the support isso it is normalized for every $\alpha >0$. This is the unit-threshold Pareto distribution.

The support is $x\ge 1$, and its normalized Pareto distribution density integrates to

Let $X$ have the Pareto distribution with probability density function $f_X(x)=3x^{-4}$ for $x\ge 1$, and zero otherwise. Its normalization of a probability distribution and first two moments follow fromSet $Y=X-3/2$. ThenThe probability distribution of $Y$ is not a symmetric distribution: $Y\ge -1/2$ almost surely, whereas $\mathbb{P}(Y>1/2)=\mathbb{P}(X>2)=1/8>0$ and $\mathbb{P}(Y<-1/2)=0$. For $X\ge 3$ we have $|Y|\ge X/2$, soThus $\boxed{Y=X-3/2}$ has every property required.

Let $y_{(n)}={\max }_i{y}_i$ and $c=\max (\beta ,y_{(n)})$. For the observed positive data, the uniform distribution likelihood function isMultiplying by the Pareto distribution prior density gives the posterior density kernel ${\theta }^{-(\alpha +n+1)}{1}_{\{\theta >c\}}$. Sincethe normalized posterior distribution isThis is uniform-Pareto conjugacy. The posterior density is strictly decreasing over its support, so the highest posterior density interval starts at $c$, where the admissible density is largest in the limiting sense. Its upper endpoint $u$ satisfiesHence a 95% highest posterior density interval isInclusion of endpoints does not change its posterior probability. Every point inside has higher posterior density than every admissible point to its right, and points below $c$ have density zero. This is the Pareto highest density interval; equal endpoint densities are unnecessary because one endpoint is the support boundary.

Let $M=\max \{b,y_1,\dots ,y_n\}$. The continuous uniform distribution likelihood contributes ${\varphi }^{-n}{1}_{\{\varphi \ge {\max }_i{y}_i\}}$, while the Pareto distribution prior contributes ${\varphi }^{-a-1}{1}_{\{\varphi \ge b\}}$. Their product proves Uniform-Pareto conjugacy:Thus the posterior distribution is Pareto with shape $a+n$ and lower bound $M$. Its density is strictly decreasing on its support, so the highest posterior density interval begins at $M$. Since the upper-tail probability is $(M/u{)}^{a+n}$, the required upper endpoint satisfies $(M/u{)}^{a+n}=0.05$. The Pareto highest density interval is therefore

For $x\le c$, the cumulative distribution function is zero. For $x>c$, integrate the normalized probability density function:It is continuous at $c$, increases to one, and is the cumulative distribution function of the stated Pareto distribution.

Call the normalization constant $K$ to distinguish it from the moment index used later. A probability density function must have total integral one, soTherefore $K=\theta c^{\theta }$, the value denoted $k$ in the density. This is the Pareto distribution with scale $c$ and shape $\theta$.

For $k\ge 1$, the event $\{\lfloor X\rfloor \ge k\}$ equals $\{X\ge k\}$. Integrating the Pareto distribution's density givesThe mass formula follows by subtracting consecutive tail probabilities, or by integrating over $[k,k+1)$. With the paper's convention $0\in \mathbb{N}$, include the separate endpoint values $\mathbb{P}(\lfloor X\rfloor \ge 0)=1$ and $\mathbb{P}(\lfloor X\rfloor =0)=0$, since $X\ge 1$ almost surely. Substituting $k=0$ into the power expression would be invalid.

The expected value is finite because $t>2$:The condition on $t$ is precisely the first moment threshold for this parameterization of the Pareto distribution.

Normalization of the probability density function requires $1=c{\int }_1^{\infty }{x}^{-t}dx=c/(t-1)$. Hence $c=t-1$. It is positive for the given $t>2$, and the density is nonnegative, so this does define a probability distribution. It is a Pareto distribution with scale $1$ and shape $t-1$.

A power law is a relation $f(x)=C{x}^p$, with fixed constants $C$ and $p$. For positive $x$, any real number $p$ is allowed. Its useful feature is a simple scaling rule:The same proportional change of input produces the same proportional change of output, independently of the starting value. The exponent controls that response; the coefficient fixes the magnitude.

For example, the area of a circle is $A(r)=\pi r^2$. Doubling the radius multiplies its area by four, and tripling it multiplies the area by nine. The volume of a sphere is $V(r)=4\pi r^3/3$: the same changes multiply its volume by eight and twenty-seven. The distinction explains why similarly shaped objects do not preserve their ratio of surface area to volume when their size changes.

When the output is positive, a logarithm turns the power law into a straight-line relation. Using positive reference values $x_0$ and $f_0=f(x_0)$,These ratios are dimensionless quantities, so the formula also applies to physical measurements with units. Plotting both ratios on a logarithmic scale gives a line of slope $p$. Changing a unit changes the numerical coefficient, while the exponent stays the same. Positive $p$ describes increasing values; negative $p$ describes decreasing values; $p=0$ gives a constant. The zero coefficient is a degenerate case with no useful logarithmic slope.

Exact scaling is stronger than regular variation. The function $x^2\log x$ has a doubling ratio tending to four, but its logarithmic correction keeps growing. The diagram contrasts it with the exact power law $x^2$.

A power law for a probability density and a power law for a tail probability have different exponents. For a Pareto distribution with $\alpha >0$ and lower endpoint one,The tail probability has exponent $-\alpha$; differentiating it and changing sign gives the probability density exponent $-\alpha -1$. Confusing them changes the interpretation of a fitted exponent. The normalizing coefficient and lower endpoint also matter: a decreasing power law is not automatically a valid probability density over the entire positive real line.

The appearance of a line over a finite logarithmic range is evidence to investigate, rather than a proof of an exact power law or its mechanism. In their study of power-law distributions, Clauset, Shalizi and Newman combine maximum likelihood estimation, goodness-of-fit tests and comparisons with alternative models; their abstract explains why ordinary least-squares fitting can give misleading results for tail data. See Aaron Clauset, Cosma Rohilla Shalizi and M. E. J. Newman, Power-law distributions in empirical data, SIAM Review 51 (2009), 661–703. The scaling rule is mathematically exact when it defines a model; whether a dataset supports that model is a question of statistical inference.

A power-law distribution uses a power law for a probability mass function or a probability density on a specified range. For a continuous model with $p_X(x)=C{x}^{-\beta }$ on $x\ge x_{\min }>0$ and zero below it, the total integral is finite exactly when $\beta >1$. Normalization then requires $C=(\beta -1)x_{\min }^{\beta -1}$, giving a Pareto distribution with shape parameter $\alpha =\beta -1$. Discrete models instead normalize a sum. The support and normalizing coefficient are part of the model, and a power law for its tail probability uses a different exponent from its probability density.

By Mark Newman, credited as M. E. J. Newman. Published in Contemporary Physics 46, 323–351 (2005). The inspected arXiv version 3 was revised 29 May 2006; version 1 was submitted 1 December 2004. These preprint dates are distinct from the journal publication year.

Preprint: arxiv.org/pdf/cond-mat/0412004

Cited passages: §II, pp. 4–5, equations (3)–(5), density and survival exponents and fitting caveat; §III.D, pp. 10–11, equations (24)–(28), fraction of a total held by an upper population fraction. Its density exponent is one greater than the positive survival exponent used for the Pareto distribution.

A power law $q(t)=t^{\rho }$ responds to every fixed rescaling $t\mapsto ut$ by the same factor $u^{\rho }$. Regular variation keeps this scaling behaviour in the limit $t\to \infty$, while allowing departures from an exact power law. It is useful when the scaling exponent is stable but a smaller correction continues changing. The correction can even grow without bound.

An eventually positive measurable function $q$ has regular variation at infinity with index $\rho \in \mathbb{R}$ ifThe argument $ut$ is in the domain of a function for sufficiently large $t$. Each $u$ is held fixed as the limit is taken; allowing $u$ to change with $t$ is a different assertion. This is the scalar definition recalled by ^[ref]. The index is unique: taking $u=2$ determines $2^{\rho }$.

For example, on $t>1$ the function $q(t)=t^2\log t$ hasThus $q$ has regular variation of index two. Yet $q(t)/t^2=\log t\to \infty$: regular variation does not require asymptotic equivalence to a constant times a power law. Doubling $t$ gives $q(2t)/q(t)=4(1+\log 2/\log t)$. At $t=100,10^4,10^6$ this is approximately $4.60206,4.30103,4.20069$, respectively. The ratio approaches four slowly, even though the exponent is exactly two in the limiting scaling law.

The slowly varying function isolates this correction. Set $\ell (t)=q(t)/t^{\rho }$. ThenConversely, multiplying a slowly varying function by $t^{\rho }$ gives regular variation of index $\rho$. Hence the decomposition $q(t)=t^{\rho }\ell (t)$ is exact on its eventual domain of a function; it is the scaling relation, rather than the decomposition itself, that is asymptotic.

A Pareto distribution gives a decreasing example with a direct interpretation. For $x\ge 1$, take the tail probability $\mathbb{P}(X>x)=x^{-2}$. Its regular variation index is $-2$. Doubling a threshold divides its tail probability by four, andFor a tail probability with regular variation of index $-\alpha$, the same conditional probability tends to $2^{-\alpha }$. An exponential distribution behaves differently: its corresponding conditional probability is $e^{-\lambda t}\to 0$. This comparison explains why scaling limits are useful in extreme value theory; it does not establish that an observed dataset follows a particular tail model. The role of regular variation in heavy-tailed distributions is discussed in Lindskog, Resnick and Roy, introduction, p. 1.

An exponential function $q(t)=e^t$ is not regularly varying with any finite index: at $u=2$, $q(2t)/q(t)=e^t\to \infty$, rather than $2^{\rho }$. Boundedness alone is also insufficient, as the oscillating counterexample in slowly varying function shows. Checking a few rescaling ratios at finite arguments can illustrate a proposed law, but cannot prove its limiting definition.

Jovan Karamata developed this theory in his 1930 paper and returned to its fundamental results in 1933. The opening of his 1933 paper identifies the earlier work; pp. 57–58 connect the rescaling limit with a power law and a slowly varying factor. See J. Karamata, Sur un mode de croissance régulière. Théorèmes fondamentaux, Bulletin de la Société Mathématique de France 61 (1933), 55–62. This gives a precise language for “almost a power” without discarding corrections that remain large.

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 target density proportional to $x^{\alpha -1}{e}^{-\beta x}$ on $x>a>0$, use the Pareto distribution proposal $g(x)=b{a}^b{x}^{-b-1}$ with $b>0$. Its density ratio is proportional to $x^{\alpha +b}{e}^{-\beta x}$, maximized at $(\alpha +b)/\beta$ if that point exceeds $a$. The tight envelope yields the highest acceptance probability for this fixed proposal. An inverse draw is $a{U}^{-1/b}$ with $U$ uniformly distributed on $(0,1)$.

A Pareto distribution prior for the upper limit of independent uniform distributions remains Pareto after observing data. The likelihood adds the sample size to the shape and replaces the lower bound by the larger of the prior bound and sample maximum.

For positive observations and $m=\max (\beta ,{\max }_j{y}_j)$, the Bayesian model evidence follows by integrating the uniform distribution likelihood against a Pareto distribution prior. For independent groups with shared fixed hyperparameters, these evidences multiply. The resulting hyperparameter log likelihood is $I\log \alpha -{\sum }_i\log (\alpha +n_i)-\alpha {\sum }_i\log (m_i/\beta )$ up to a constant.

For a Pareto distribution of shape three and minimum $d>0$ with count rate $\lambda >0$, the sum of insurer and reinsurer aggregate variances is $V(M)=\lambda (3d^2-3dM+2M^2)$ for $0\le M\le d$, and $\lambda (3d^2-d^3/M)$ for $M\ge d$. The first branch has its unique minimum at $3d/4$; the second is increasing. The branches agree in value and slope at $d$, giving the displayed global minimum. This criterion minimizes the sum of party variances; the total portfolio variance includes their covariance and remains $3\lambda d^2$.