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Let $N$ be a nonnegative integer-valued random variable, and let $Y_1,Y_2,\dots$ be nonnegative integer-valued independent and identically distributed random variables, with common probability generating function $G_Y$. Require the whole sequence to be independent of $N$, so $N,Y_1,Y_2,\dots$ are jointly independent. Set the random sumwith the empty sum equal to zero. Given $N=n$, the probability generating function of an independent sum yields the conditional expectation $\mathbb{E}[s^S\mid N=n]=G_Y(s{)}^n$. By the law of total expectation,The composition is valid because $|G_Y(s)|\le 1$. Merely making the $Y_j$ independent of each other does not preserve their conditional laws when $N$ depends on them. This is the probability generating function version of transform composition for an independent random sum.

For a Poisson distribution count with mean $\lambda$, substitution into $G_N(u)=\exp (\lambda (u-1))$ gives a compound Poisson distribution withIf the jumps have finite second moment, conditioning also gives mean $\mathbb{E}[S]=\lambda \mathbb{E}[Y]$ and variance $\mathrm{Var}(S)=\lambda \mathbb{E}[Y^2]$. The extra spread comes from randomness in the count as well as randomness in the jumps.^[ref]

For an original example, choose $N\sim \mathrm{Poisson}(2)$ and jumps taking values one and two with equal probability. Keep the independence assumptions above. ThenHere $\mathbb{E}[Y]=3/2$, $\mathrm{Var}(Y)=1/4$, and $\mathbb{E}[Y^2]=5/2$. The law of total expectation and law of total variance giveThus the aggregate is not itself a Poisson distribution, whose mean and variance would coincide.

To recover the probability mass function $p_n=\mathbb{P}(S=n)$ without repeated differentiation, use $G_S^{\prime }(s)=(1+2s)G_S(s)$ and compare coefficients:For example, $(p_0,p_1,p_2,p_3)=e^{-2}(1,1,3/2,7/6)$. As an independent check, expanding $e^{s+s^2-2}=e^{-2}{e}^s{e}^{s^2}$ gives

The plotted probability masses stop at 20, although the support of a probability distribution is unbounded. The omitted tail probability isThis model has bounded jumps and an unbounded count. Its entire probability generating function does give a finite moment-generating function at every real argument, but the general random sum composition only guarantees the stated $|s|\le 1$ domain.

For nonnegative integer-valued independent random variables $X,Y$, factor the expected value of a product:The Cauchy product explains what this means for individual probability masses:Every way of splitting the total contributes once. This is convolution of independent random variables; multiplying probability generating functions performs the bookkeeping automatically.

For an original example, take independent $X\sim \mathrm{Bernoulli}(0.3)$ and $Y\sim \mathrm{Binomial}(2,0.5)$, using the Bernoulli distribution and binomial distribution. ThensoDirect enumeration of the six pairs $(x,y)$ gives the same probability masses. For example, total one collects $(0,1)$ and $(1,0)$, with contributions $0.7\cdot 0.5+0.3\cdot 0.25=0.425$. The coefficients sum to one. The expected value is $G_{X+Y}^{\prime }(1)=1.3$; since $G_{X+Y}^{\prime \prime }(1)=1.1$, the variance is $1.1+1.3-1.3^2=0.71$, agreeing with $0.3\cdot 0.7+2\cdot 0.5\cdot 0.5$.

Uncorrelated random variables are insufficient for the product identity. For a concrete counterexample, let $U$ be uniform on $\{-1,0,1\}$ and set $A=U+1$, $B=U^2$. Then $\mathrm{Cov}(A,B)=\mathbb{E}[U^3]=0$, but $\mathbb{P}(A+B=0)=0$, whereas $G_A(0)G_B(0)=1/9$. Zero covariance controls one mixed moment, not the full joint probability mass function.

Probability inequalities bound event probabilities from limited information, such as moments, without requiring a complete probability distribution. Markov's inequality uses a nonnegative random variable's expected value; Chebyshev's inequality uses a variance to bound deviations from a mean. The attraction is a guarantee that applies to many possible probability distributions. Their bounds can be conservative for a particular model, and distributions attaining equality show when the available information really cannot give more.

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.

Write $m_k=\mathbb{E}{S}^k$, $m_0=1$, and ${\mu }_j=\mathbb{E}{X}^j$. Apply the Poisson size-bias identity with $h(s)=s^{k-1}$, and expand $(S+X{)}^{k-1}$ using the binomial theorem and independence. A finite $k$th severity moment suffices. In particular, the aggregate expected value, variance, and third central moment are $\lambda {\mu }_1$, $\lambda {\mu }_2$, and $\lambda {\mu }_3$, respectively; the latter two use raw severity moments.

The second moment of a real random variable is $\mathbb{E}[X^2]$. Its variance is $\mathbb{E}[X^2]-(\mathbb{E}X{)}^2$ when these quantities are finite.

Skewness is the standardized third central moment of a probability distribution. A symmetric distribution with finite third moment has zero skewness. Cosmology also uses the reduced skewness $S_3=⟨{\delta }^3⟩/{\sigma }^4$, which differs from the standardized skewness by a factor of $\sigma$.

A finite nonzero Saffman integral gives $E(k)=L{k}^2/(4{\pi }^2)+o(k^2)$ in the standard isotropic spectrum normalization. This follows by expanding the radial spectral transform at small wavenumber. A further analytic term requires additional correlation moments.

The space-time Probabilists' Hermite polynomial $H_n(x,t)=t^{n/2}{\mathrm{He}}_n(x/\sqrt{t})$ extends polynomially to $t=0$ and solves ${\partial }_t{H}_n+\frac{1}{2}{\partial }_x^2{H}_n=0$. Therefore $H_n(B_t,t)$ is a martingale for a standard Brownian motion, by the Itô formula and Gaussian moments.

A measurable function is square-integrable when the integral of its squared absolute value is finite. Its equivalence class belongs to the L2 space; on a probability space, this means a finite second moment.

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.

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.

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.

If $s>0$ and $\Theta$ has the continuous uniform distribution on $(-\pi /2,\pi /2)$, then $Y=s\tan \Theta$ has the Cauchy distribution with location zero and scale $s$. The tangent function is strictly increasing on this interval, so its cumulative distribution function isDifferentiating gives the probability density function $s/[\pi (s^2+y^2)]$. The probability of either endpoint is zero, so including endpoints in the uniform interval does not alter this probability distribution. The slow quadratic decay makes the expected value undefined: its positive and negative parts both have infinite integrals. Its second moment is also infinite, so it has no finite standard deviation.

Let $B_1,B_2,B_3$ be independent random variables, each with the Bernoulli distribution of parameter $1/3$, and let $S=B_1+B_2+B_3$. Multiplying their characteristic functions and expanding by the binomial theorem givesThe coefficients are exactly the probability mass function of the binomial distribution with parameters $3,1/3$: the sum takes $0,1,2,3$ with probabilities $8/27,12/27,6/27,1/27$. Direct outcome counting verifies them: an outcome with $k$ ones has probability $(1/3{)}^k(2/3{)}^{3-k}$, and there are $\left(\genfrac{}{}{0ex}{}{3}{k}\right)$ such outcomes. The coefficients sum to one, recovering ${\phi }_S(0)=1$.

At $t=\pi$, ${\phi }_S(\pi )=(1/3{)}^3=1/27$. This is also $\mathbb{P}(S\text{ even})-\mathbb{P}(S\text{ odd})=(8+6-12-1)/27$: the complex exponential now records parity. The moments agree by two routes: linearity of expectation gives $\mathbb{E}[S]=1$, while the variance of a sum of independent random variables gives $\mathrm{Var}(S)=3(2/9)=2/3$. Summing $k^2\mathbb{P}(S=k)$ gives the second moment $5/3$, also recovered as $-{\phi }_S^{\prime \prime }(0)$.

Approximate a compound Poisson distribution of positive claims by a shifted gamma distribution whose first three cumulants agree. Here $m_j=\mathbb{E}{X}^j$ are raw severity moments, while the compound cumulants are $\lambda m_j$. Solve $k+\alpha /\nu =\lambda m_1$, $\alpha /{\nu }^2=\lambda m_2$, and $2\alpha /{\nu }^3=\lambda m_3$ to obtain the displayed parameters. The approximation captures positive skewness, but not the atom $e^{-\lambda }$ at zero. Its shift can be negative; with exponential claims of mean $\mu$ it is $-\lambda \mu /3$. Matching three cumulants alone supplies no general tail-error bound.

A turbulence closure approximates the unclosed higher-order moments in evolution equations for lower-order turbulent statistics. For example, an equation for the velocity correlation tensor involves triple correlations, whose evolution involves fourth-order correlations and pressure. A closure must be checked against conservation laws, realizability and boundary conditions; matching an energy spectrum alone does not establish all these properties.

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.

For independent and identically distributed random variables $U_1,\dots ,U_n$ with the continuous uniform distribution on $(0,1)$, the $r$th order statistic has Beta distributionIts probability density function, mean, and variance areThe probability density function is zero outside $[0,1]$; values assigned to its endpoints do not change any probability. Peter Kempthorne's MIT lecture notes derive the uniform order statistic law and these moments. ^[ref].

For $n=5$, the minimum, sample median, and sample maximum correspond to $r=1,3,5$. Their probability density functions are $5(1-t{)}^4$, $30t^2(1-t{)}^2$, and $5t^4$, respectively. The sample median therefore has Beta distribution $\mathrm{Beta}(3,3)$, mean $1/2$, and variance $1/28\approx 0.0357143$. Its cumulative distribution function isFor instance, $\mathbb{P}(U_{(3)}\le 1/4)=53/512\approx 0.103516$. The mean of the minimum is $1/6$, and the mean of the sample maximum is $5/6$. The fixed observed sample median $0.4$ in the order statistic example is one possible outcome; $1/2$ is the average over repeated statistical samples from this uniform distribution.

Each curve has area one. The extreme order statistics favor the ends, while the sample median favors the center. The endpoint density height $5$ is legitimate: the probability density function gives area per unit length, rather than point probability. In particular, a continuous uniform distribution has zero probability of producing an endpoint.

For a short proof, the binomial distribution counting identity under order statistic gives $\mathbb{P}(U_{(r)}\le t)={\sum }_{k=r}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k(1-t{)}^{n-k}$. Taking its derivative cancels consecutive terms, leaving the displayed probability density function. Recognize its normalizing factorial coefficient as that of the Beta distribution, then substitute $\alpha =r$, $\beta =n+1-r$ into the Beta distribution mean and variance formulas. Although the original observations are independent, their order statistics generally are dependent; see the positive covariance of two uniform order statistics.