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Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with expected value $\mu =\mathbb{E}[X_1]$ and finite, strictly positive variance ${\sigma }^2=\mathbb{E}[(X_1-\mu {)}^2]$. Write $S_n={\sum }_{j=1}^n{X}_j$ and ${\overline{X}}_n=S_n/n$ for their sum and sample mean. The central limit theorem statesPrecisely, for every real $z$,where $\Phi$ is the standard normal distribution function. The conclusion is convergence in distribution to the standard normal distribution. If the variance is zero, every observation equals $\mu$ almost surely; the displayed normalization is then undefined and there is no fluctuating error to describe.

The striking feature is that the observations need not have a normal distribution. They may be binary, bounded, or strongly skewed. Their detailed probability distribution disappears from this limit except through the expected value and variance. The subtraction removes predictable growth: $\mathbb{E}[S_n]=n\mu$. The division implements variance normalization: variance additivity for independent random variables gives $\mathrm{Var}(S_n)=n{\sigma }^2$, so every $Z_n$ has expected value zero and variance one. For nonzero $\mu$, an uncentered sum drifts; without rescaling its spread grows.

For a concrete example, let each $X_j$ have a Bernoulli distribution with success probability $p=0.2$. Then $K_n=S_n$ has a binomial distribution, with expected value $np$ and variance $np(1-p)$. Its exact probability mass function and its normalized count areHere $\left(\genfrac{}{}{0ex}{}{n}{k}\right)$ is a binomial coefficient. Individual trials are conspicuously skewed, yet their centered, rescaled counts approach the same standard normal distribution as sums of fair trials.

At $n=100$, $K_n$ has expected value $20$ and standard deviation $4$. The exact binomial distribution gives $\mathbb{P}(16\le K_n\le 24)=0.740141\dots$. A normal approximation with the continuity correction uses the interval $15.5$ to $24.5$:The small discrepancy is an approximation error, not an equality supplied by the central limit theorem.

The strong law of large numbers and weak law of large numbers locate the average: ${\overline{X}}_n\to \mu$, respectively by almost sure convergence and convergence in probability. The central limit theorem magnifies that shrinking error and describes its shape. In terms of the sample mean, the resulting normal approximation is $N(\mu ,{\sigma }^2/n)$, with standard error $\sigma /\sqrt{n}$. This explains why four times as many independent observations halve the standard error. When $\sigma$ is known, ${\overline{X}}_n\pm 1.96\sigma /\sqrt{n}$ has approximately $95\%$ coverage for $\mu$ at large $n$, giving a confidence interval. This coverage refers to repeated samples, not a guarantee for an individual interval.

A useful application is the Monte Carlo method: estimate a success probability by the proportion of successful simulated trials. Nicholas Metropolis and S. Ulam's original paper, "The Monte Carlo Method" (1949), illustrates this with games of solitaire and estimation of a region's volume by randomly selected points, and discusses why estimates require probabilistic error control. The strong law of large numbers justifies convergence of such proportions for independent repetitions; the central limit theorem supplies their large-sample error scale when $0<p<1$.^[ref]

There are three useful limits to this picture:

A historical original-paper reference is J. W. Lindeberg's "Eine neue Herleitung des Exponentialgesetzes in der Wahrscheinlichkeitsrechnung" (1922), published in Mathematische Zeitschrift 15, pp. 211–225. The Lindeberg condition and Lindeberg-Feller central limit theorem below describe a version for sums whose summands need not share a probability distribution.^[ref]

A brief proof outline uses the characteristic function of $Y=(X_1-\mu )/\sigma$. Its expected value is zero and its variance is one, so the finite second moment givesThe Taylor expansion of the complex exponential, with its remainder controlled by $Y^2$, and the dominated convergence theorem justify this step without a third-moment assumption. Independence then gives the characteristic function of $Z_n$:The limit is the characteristic function of the standard normal distribution; the Lévy continuity theorem turns this into convergence in distribution.

Let $X$ take integer values with probabilities $p_n$, and let $Y_{\sigma }$ be an independent centered normal distribution with positive standard deviation $\sigma$. Conditioning on $X$ gives the probability density function of their sum as $g_{\sigma }(z)={\sum }_n{p}_n{\phi }_{\sigma }(z-n)$, where ${\phi }_{\sigma }$ is the normal density. Nonnegativity and Tonelli's theorem show that this density integrates to one. For small $\sigma$, isolated atoms become narrow smooth peaks with their original probability masses.

For each fixed real $t$, dominated convergence givessince $X/\sigma \to 0$ almost surely and ${\phi }_1$ is bounded. Thus every fixed integer-valued distribution acquires a Gaussian central profile on this expanding scale, without any moment assumption on $X$. This is not a claim about uniform relative error in the tails or about finite moments. For example, if $p_n=3/({\pi }^2{n}^2)$ for $n\ne 0$ and $p_0=0$, the expectation of $|X|$ is infinite. Since $\mathbb{E}|Y_{\sigma }|<\infty$, the triangle inequality forces $\mathbb{E}|X+Y_{\sigma }|=\infty$ too. Symmetry then gives equal infinite positive and negative first moments, so there is no ordinary mean. The second moment is also infinite. A Gaussian-looking center can therefore coexist with a heavy-tailed distribution.

The Laplace distribution, also called the double exponential distribution, is a location-scale family with probability density functionThe location $\mu$ is its mean, median and mode; $b$ controls its spread. In the centred rate convention, the same law has $\mu =0$ and $\theta =1/b>0$, giving $f(x)=\theta e^{-\theta |x|}/2$. Keeping track of whether a parameter is a scale or its reciprocal prevents reversing the effect of increasing it. ^[ref]

The two halves of the probability density function are reflected exponential decay curves, each with total probability $1/2$. Their meeting point has a sharp corner: the left and right derivatives are $1/(2b^2)$ and $-1/(2b^2)$. This finite peak is not a positive probability mass at $\mu$. Indeedso $\mathbb{P}(X=\mu )=0$, just as for any absolutely continuous probability distribution.

The figure compares a Laplace distribution with $\mu =0$, $b=1$ against a normal distribution with mean zero and standard deviation $\sigma =\sqrt{2}$: both have variance 2. The right panel uses a logarithmic scale for the two-sided tail probability $\mathbb{P}(|X|>r)$. Its straight Laplace distribution curve comes from $e^{-r}$; the normal distribution tail falls faster. Thus values far from the center can be more plausible under a Laplace distribution even after matching the variance. These are theoretical curves, not measured frequencies. The tails still have exponential decay; this is not a heavy-tailed distribution in the convention requiring every positive exponential moment to diverge.

Integrating either half of the probability density function gives the cumulative distribution functionIn particular $F(\mu )=1/2$. The basic moments areThus the scale is the expected value of the absolute distance from the center, and is not the standard deviation. ^[ref] A short way to obtain the moments is to note that $R=|X-\mu |$ has an exponential distribution with rate $1/b$: $\mathbb{P}(R>r)=e^{-r/b}$ for $r\ge 0$. Integrating gives $\mathbb{E}R=b$ and $\mathbb{E}{R}^2=2b^2$; the symmetric distribution gives the centered first moment zero.

A normal Q-Q plot compares order statistics with standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. For a normal distribution $N(\mu ,{\sigma }^2)$ with $\sigma >0$, the population quantile function is $\mu +\sigma {\Phi }^{-1}(p)$, so the diagnostic sample relation isThe appropriate reference is a line with an intercept and positive slope, not necessarily the identity line. A line can be fitted through two central sample quantiles; R's qqline uses the quantile levels $p=0.25$ and $p=0.75$ by default. Neither that line nor its fitting rule is part of the definition of a normal Q-Q plot. ^[ref]

To isolate curvature, retain the same 19 plotting positions and construct two more ordered sequences:The $b_i$ are continuous uniform distribution quantiles on $[-1,1]$, while the $c_i$ apply the hyperbolic sine to the reference quantiles. Both examples are symmetric about zero. At ranks $1,10,19$, $b_i$ is approximately $-0.947368,0,0.947368$ and $c_i$ is approximately $-1.812618,0,1.812618$. Again, these are constructed values rather than random draws or real data.

Read these shapes relative to their central slope. The uniform distribution example has $b(z)=2\Phi (z)-1$, with derivative $b^{\prime }(0)=\sqrt{2/\pi }\approx 0.797885$; at the upper endpoint the tangent predicts $1.546246$, while $b_{19}=0.947368$. Its lower end lies above the tangent and its upper end below it: the outer quantiles are compressed. The hyperbolic sine example has $c^{\prime }(0)=0.7$; its tangent predicts $1.356552$ at the upper endpoint, while $c_{19}=1.812618$. Its lower end lies below the tangent and its upper end above it: the outer quantiles are stretched. Unequal departures at the two ends can suggest skewness. The direction of these descriptions assumes that reference quantiles are horizontal and observed values vertical. The graphs illustrate finite-range shape differences; a sample pattern cannot establish an asymptotic tail probability law or membership in a heavy-tailed distribution class.

A nearly straight normal Q-Q plot is a diagnostic, not a proof of a normal distribution or a formal hypothesis test. Random statistical samples fluctuate, particularly at extreme order statistics, and apparent curvature depends on sample size, plotting positions and the reference line. A calibrated hypothesis test or uncertainty band requires its own assumptions and must account for parameters estimated from the same observations. Sorting also discards the original observation order: rearranging a time series leaves its Q-Q plot unchanged, so the plot cannot establish independence.

For a normal linear model, R's diagnostic normal Q-Q plot uses standardized regression residuals, also called internally studentized residuals. ^[ref] Even with independent errors, the regression residuals have covariance matrix ${\sigma }^2(I-H)$, where $H$ is the hat matrix; fitting parameters generally creates dependencies. Dividing by a shared estimated standard deviation creates further dependence, and the distribution of an internally studentized Gaussian residual is not exactly the standard normal distribution. The normal Q-Q plot therefore helps assess error shape alongside other diagnostics rather than certifying independent normal random variables.

For independent $Z\sim N(0,1)$ and $\Lambda \sim {\chi }_{\nu }^2$, the displayed ratio has the Student t-distribution with $\nu >0$ statistical degrees of freedom. Equivalently, conditional on $\Lambda =\lambda >0$, $T$ has normal distribution $N(0,\nu /\lambda )$. This is a normal scale mixture representation of Student t: a small latent denominator occasionally produces a much larger spread. The result is a heavy-tailed distribution even though every conditional probability distribution is a normal distribution.

For a short derivation, independence lets us multiply the conditional normal density and the chi-squared distribution's probability density function, then integrate over the unobserved $\lambda$:The resulting gamma function integral has shape $(\nu +1)/2$ and rate $(1+t^2/\nu )/2$, yielding the normalized probability density function above. Also, for $\nu >2$, conditional expectation gives $\mathbb{E}[T^2]=\nu \mathbb{E}[{\Lambda }^{-1}]=\nu /(\nu -2)$. Thus unit scale does not mean unit variance. Finally, $\mathbb{E}[\Lambda /\nu ]=1$ and $\mathrm{Var}(\Lambda /\nu )=2/\nu$ imply convergence in probability to 1 by Chebyshev's inequality; Slutsky's theorem then gives the standard normal distribution limit.

Olivier Catoni develops methods for robust statistics. His work on the mean and variance constructs M-estimators with deviation guarantees for heavy-tailed distributions under moment assumptions.^[ref]

The sample mean lets a single very large observation exert arbitrarily large influence. Catoni's construction replaces a linear estimating equation by a controlled nonlinear function of each residual. With an appropriate tuning parameter, a mean estimator can achieve Gaussian-order deviation bounds at a chosen confidence level under bounded variance, even when the observations are not normally distributed. The tuning and assumptions are part of the guarantee; the result does not assert that one fixed setting automatically works at every confidence level.^[ref]

The Pareto distribution describes a positive random variable whose tail probability falls as a power law. In the shape–minimum convention used here, $X\sim \mathrm{Pareto}(\alpha ,\beta )$ has shape parameter $\alpha >0$ and minimum $\beta >0$, with probability support $[\beta ,\infty )$. Its probability density function isThe minimum $\beta$ has the same units as $X$; the shape parameter $\alpha$ is dimensionless. At fixed $\alpha$, changing $\beta$ gives a scale family, since $X/\beta \sim \mathrm{Pareto}(\alpha ,1)$. This convention is also used in uniform-Pareto conjugacy.^[ref]

The survival function and cumulative distribution function areFor $0<p<1$, solving $F(x)=p$ gives the quantile functionIts right-hand limit at $p=0$ is $\beta$; it tends to infinity as $p\uparrow 1$.^[ref] There is no probability mass at $\beta$: changing a probability density function's value at that single point changes no probabilities.

Choose $\alpha =2$ and $\beta =10$ as an original theoretical example. Then $\mathbb{P}(X>20)=1/4$, $\mathbb{P}(X>100)=1/100$, the median is $10\sqrt{2}\simeq 14.142$, and the 0.99 quantile is $Q(0.99)=100$. These chosen values and all figures below describe mathematical probability distributions, not actual wealth observations.

On this log-log plot, write $r=x/\beta$ to obtain $\log \overline{F}(\beta r)=-\alpha \log r$. The slope is $-\alpha$, whereas the probability density function has power exponent $\alpha +1$. Multiplying a threshold by ten divides its tail probability by $10^{\alpha }$. A larger $\alpha$ therefore concentrates the Pareto distribution nearer $\beta$; a smaller $\alpha$ makes extreme values more plausible. This exact scaling is useful when modelling large positive losses or sizes over several orders of magnitude.

The Pareto distribution is a heavy-tailed distribution: for every $t>0$, the exponential moment $\mathbb{E}[e^{tX}]$ diverges because $e^{tx}{x}^{-\alpha -1}$ eventually grows without bound. In contrast, the Laplace distribution has exponentially decreasing tails and some finite positive exponential moments. The Student's t-distribution also has power tails, but on both sides of the real line, with a smooth central region rather than a positive minimum.

For large $\sigma$, neighboring peaks overlap and the density has a broad, smooth, symmetric central hump, approximately $(\sqrt{2\pi }\sigma {)}^{-1}{e}^{-z^2/(2{\sigma }^2)}$ on the scale $z=O(\sigma )$. A precise central-profile statement follows fromfor each fixed $t$, $X/\sigma \to 0$ almost surely and the integrand is bounded, so dominated convergence applies. The right panel of the preceding figure shows this Gaussian-looking core.

This approximation does not remove the heavy-tailed distribution. Here $\mathbb{E}|X|=(6/{\pi }^2){\sum }_{n\ge 1}1/n=\infty$, so the symmetric distribution has no ordinary mean and no finite variance. Adding the independent Gaussian, which has finite absolute moments, leaves these conclusions unchanged. In particular a central Gaussian sketch cannot justify assigning variance ${\sigma }^2$ to $Z$.

The central limit theorem gives $\sqrt{n}({\hat{\theta }}_n^{\mathrm{MC}}-\theta )\Rightarrow N(0,1)$, so, with $\Phi$ the cumulative distribution function of the standard normal distribution,The error has two tails. Using the supplied $2.58$ quantile, whose two-sided coverage is approximately $0.9902$, givesThe supplied $2.33$ quantile would be appropriate for a one-sided $99\%$ calculation; it would give only approximately $98.02\%$ two-sided coverage. Using the exact $0.995$ quantile instead of $2.58$ would give $66,349$ draws; we consistently use the paper's rounded value $N=66,564$ below.

For routine Monte Carlo integration, the central limit theorem calculation is preferable as a practical budget: it uses about one fifteenth as many draws as the Chebyshev's inequality bound. However, its coverage is approximate. Finite variance alone supplies no universal rate of convergence to the normal distribution; an integrand with large skewness or a heavy-tailed distribution can make the approximation poor even at a large sample count. If an unconditional finite-sample $99\%$ guarantee is required under only the stated assumptions, use the Chebyshev's inequality budget.

For $0<q\le 1$, put probability mass $1-q$ at zero and $q/2$ at each of $-1$ and $1$. This symmetric probability distribution has mean zero. Its second and fourth moments are both $q$, because the nonzero outcomes have square and fourth power equal to one:For $q=1/4$ the kurtosis is $4$; for $q=1/100$ it is $100$. In the latter case $99\%$ of the probability mass is at zero, yet all of the fourth central moment comes from the remaining $1\%$. The standard score $Z=X/\sqrt{q}$ takes values $0$ and $\pm q^{-1/2}$, with variance one. Each nonzero outcome contributesto kurtosis. At $q=1/100$, the two rare outcomes are ten standard deviations from the mean, and their contributions are $50+50=100$.

Every member has finite support of a probability distribution, so all its moments are finite. Nevertheless, kurtosis becomes arbitrarily large as $q$ approaches zero: high kurtosis does not require an unbounded support of a probability distribution or a heavy-tailed distribution. At $q=0$ the variance vanishes and kurtosis is undefined. At $q=1/3$ this discrete probability distribution even has the normal distribution value $\kappa =3$ and excess kurtosis zero, so those values do not establish a normal 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.

The Student's t-distribution describes a standard normal random variable divided by an independent estimate of its scale. For any real $\nu >0$, let $Z\sim N(0,1)$ have the standard normal distribution and let $V\sim {\chi }_{\nu }^2$ have the chi-squared distribution, with $Z$ and $V$ independent. ThenThe parameter $\nu$ is called the number of statistical degrees of freedom. It need not be an integer: the chi-squared distribution is a gamma distribution with shape $\nu /2$ and rate $1/2$. The independence assumption is essential; specifying only the two marginal distributions does not determine the probability distribution of their ratio.

Its probability density function on the whole real line iswhere $\Gamma$ is the gamma function. Every member is a symmetric distribution about zero. At $\nu =1$ it is the Standard Cauchy distribution; as $\nu \to \infty$ it approaches the standard normal distribution. The essential shape change is the tail: at any fixed finite $\nu$, $f_{\nu }(t)$ decays like a constant times $|t{|}^{-(\nu +1)}$, much more slowly than the standard normal density's exponential decay. Thus it is a heavy-tailed distribution. ^[ref]

The left panel shows the centers; the right panel uses a logarithmic scale to reveal the tails that a linear plot hides. All four curves have location zero and scale one. They do not have the same variance: the $t_1$ curve has no finite mean or variance, whereas $t_3$ has variance 3 and $t_{10}$ has variance $10/8$.

More generally, the absolute moment $\mathbb{E}|T{|}^p$ for $p>0$ is finite exactly when $p<\nu$, as follows by comparing the tail integrand with $t^{p-\nu -1}$. ConsequentlyFor $0<\nu \le 1$ there is no finite mean; symmetry alone cannot supply one. For $1<\nu \le 2$ the mean exists but the variance is infinite. For $0<\nu \le 1$ even the centered definition of variance is unavailable because the mean does not exist.

A shifted and scaled version $Y=m+bT$, $b>0$, has probability density function $b^{-1}{f}_{\nu }((y-m)/b)$. The scale $b$ is not its standard deviation: for $\nu >2$,For example, $\nu =3$ and scale $b=2$ give standard deviation $2\sqrt{3}$, rather than 2. The distinction matters when using a Student's t-distribution to model data with occasional extreme values, as well as when interpreting a Student t confidence interval.