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For a random variable with finite mean $\mu$, its $n$th central moment is $\mathbb{E}[(X-\mu {)}^n]$ when defined. The first is zero; the second is the variance; the standardized third gives skewness. Translating $X$ by a constant leaves central moments unchanged, because the mean translates by the same amount. For equally likely values one and three, the second central moment is one, whereas the second raw moment is five. Odd central moments can cancel between positive and negative deviations, so an absolute moment provides a different integrability check.

The cumulant expansion is the Taylor expansion of the logarithm of a moment-generating function:If the moment-generating function is finite on a neighbourhood of zero, the logarithm is analytic on a sufficiently small neighbourhood and this is a convergent local expansion. The first terms areA formal series of moments is not automatically a convergent representation of a moment-generating function. Finitely many moments can instead justify finite-order expansions of the characteristic function and its logarithm near zero; exponential integrability is a separate hypothesis.

For a centered normal distribution of variance ${\sigma }^2$, $K_X(t)={\sigma }^2{t}^2/2$, so all cumulants beyond order two vanish. For a Poisson distribution of mean $\lambda$, $K_X(t)=\lambda (e^t-1)$, so every cumulant equals $\lambda$. independent sums add their cumulant-generating functions, making cumulants additive.

For a fluctuating interaction $V$, expanding a Boltzmann weight gives $\log \mathbb{E}{e}^{-\beta gV}=-\beta g\mathbb{E}V+{\beta }^2{g}^2\mathrm{Var}(V)/2+\cdots$ whenever the exponential moment exists near $g=0$. The logarithm thus organizes the expansion into cumulants rather than raw moments.

For an independent statistical sample from the beta distribution with parameters $(\theta ,1)$, $\theta >0$, the probability density function is $\theta x^{\theta -1}$ on $(0,1)$. Its raw moments are $\mathbb{E}{X}^r=\theta /(\theta +r)$ for $r>-\theta$, by direct integration. Consequently the method of moments gives $T_1=\bar{X}/(1-\bar{X})$, whereas the maximum-likelihood estimator is $T_2=-n/{\sum }_i\log X_i$.

The second formula follows by setting the score function $n/\theta +{\sum }_i\log X_i$ to zero; its derivative is negative. Also $-\log X_i$ has an exponential distribution of rate $\theta$, since $\mathbb{P}(-\log X_i>u)=e^{-\theta u}$. The central limit theorem and delta method giveThus the asymptotic relative efficiency of $T_1$ relative to $T_2$ is $1-(\theta +1{)}^{-2}$. The transformed sum has a gamma distribution of shape $n$ and rate $\theta$, which also gives $\mathbb{E}{T}_2=n\theta /(n-1)$ for $n>1$ and $\mathrm{Var}(T_2)=n^2{\theta }^2/[(n-1{)}^2(n-2)]$ for $n>2$.

For $0<p<1$, $\beta >0$ and $x>0$, this probability density function has cumulative distribution function $F(x)=1-\log (1-(1-p)e^{-\beta x})/\log p$. Differentiating gives the density, while the endpoint limits are zero and one. Its quantile function is $F^{-1}(u)={\beta }^{-1}[\log (1-p)-\log (1-p^{1-u})]$ for $0<u<1$, enabling inverse transform sampling.

Expanding the denominator as a geometric series givesTermwise integration, justified by the monotone convergence theorem, and the gamma function integral yield raw moments $\mathbb{E}[X^j]=-j!{\mathrm{Li}}_{j+1}(1-p)/({\beta }^j\log p)$. Thus the polylogarithm expresses its mean, variance and higher moments. Here $p$ is the complement of the series parameter $1-p$; parameter conventions must be checked when comparing formulas.

A moment integrates a power of a quantity against a specified measure. For a real random variable $X$, the $n$th raw moment is $\mathbb{E}[X^n]$ when defined; the $n$th central moment instead measures powers of deviation from its mean. For equally likely values one and three, the first raw moment is two and the second raw moment is five, while the second central moment is one, its variance. Moments summarize features of a law and can be easier to compute than its full probability distribution. Even knowing every moment does not always determine the law: that additional property is moment determinacy.

Let $m=\mathbb{E}X$ and $v=\mathrm{Var}(X)$, evaluated from the given polylogarithm formulas. Generate a large sample of independent and identically distributed random variables with r_sample_f, and compute its sample mean ${\overline{X}}_n$ and sample variance $S_n^2$ using mean and var. The central limit theorem gives the first standardized check:For the second check, the correct asymptotic variance of the sample variance involves the fourth central moment ${\mu }_4=\mathbb{E}[(X-m{)}^4]$:The denominator is not $\sqrt{2v^2}$ unless the probability distribution has the requisite fourth central moment, as does the normal distribution. Here the probability density function decays exponentially, so all required moments are finite.

To compute ${\mu }_4$, expand the denominator of the probability density function as a geometric series and integrate each nonnegative term. The gamma function integral gives the raw momentsThus $v=m_2-m_1^2$ and ${\mu }_4=m_4-4m_1{m}_3+6m_1^2{m}_2-3m_1^4$. The following R code implements both checks for chosen interior parameters and large n:

p <- 0.4
beta <- 1.3
n <- 100000
raw_m <- vapply(1:4, function(j) {
  -factorial(j) * polylog(j + 1, 1 - p) / (beta^j * log(p))
}, numeric(1))
m <- raw_m[1]
v <- raw_m[2] - m^2
mu4 <- raw_m[4] - 4 * m * raw_m[3] +
  6 * m^2 * raw_m[2] - 3 * m^4
x <- r_sample_f(n, p, beta)
c(mean_check = sqrt(n) * (mean(x) - m) / sqrt(v),
  variance_check = sqrt(n) * (var(x) - v) / sqrt(mu4 - v^2))

For each check separately, an absolute value below $1.96$ is consistent with its approximate $95\%$ confidence interval. Repeat with many independent samples: the two standardized statistics should each resemble a standard normal distribution, with about $5\%$ outside that range. Also vary $p$, $\beta$ and n; a systematic discrepancy that grows with $\sqrt{n}$ suggests a bias. The two checks from the same sample are generally dependent, so passing both is not itself a joint $95\%$ statement, nor can a finite test prove that a generator is correct.

The second central limit theorem follows from the central limit theorem for the sample variance: $\sqrt{n}(S_n^2-v)$ differs by a quantity tending to zero in probability from $n^{-1/2}{\sum }_i[(X_i-m{)}^2-v]$.

Using the raw moments from part (i), the variance of one observation isThe central limit theorem for the independent statistical sample givesFor $g(x)=x/(1-x)$, its derivative at the mean is $g^{\prime }(\mu )=(1-\mu {)}^{-2}=(\theta +1{)}^2$. The delta method therefore givesEquivalently, the large-sample normal approximation isHere $v$ denotes the leading asymptotic variance, rather than an assertion about an exact finite-sample variance.

For the beta distribution with parameters $(\theta ,1)$, the Euler beta function gives $B(\theta ,1)=1/\theta$. Its probability density function is therefore $\theta x^{\theta -1}$ on $(0,1)$. Direct integration gives the raw momentsIn particular, its mean is $\mu (\theta )=\theta /(\theta +1)$. Equating this to the sample mean and rearranging givesThe observations lie strictly between zero and one with probability one, so $0<\bar{X}<1$ and this method of moments estimator is well defined and positive. This is an instance of estimation of a beta shape parameter.