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This turbulence closure assumes the velocity increment skewness is the same throughout the equilibrium range, with its value fixed by the Kolmogorov two-thirds law and Kolmogorov four-fifths law. It yields a simple crossover between differentiable and inertial-scale increments, but suppresses the scale dependence of skewness and has no independent spectral-transfer dynamics.

The Cornish-Fisher expansion inverts an Edgeworth expansion to approximate quantiles. For a standardized iid sum, let ${\gamma }_1$ be the skewness of one summand and $z_{\alpha }$ the standard normal quantile. The first distribution function correction is $-{\gamma }_1(z^2-1)\varphi (z)/(6\sqrt{n})$. Substituting $y_{\alpha }=z_{\alpha }+c/\sqrt{n}$ and expanding $\Phi$ gives $c={\gamma }_1(z_{\alpha }^2-1)/6$. This is a fixed-quantile asymptotic, with $\alpha$ fixed in $(0,1)$.

A Gaussian mixture distribution is a weighted sum of normal distribution densities, with nonnegative weights summing to one. It need not itself be a normal distribution: components with different means and variances can produce skewness or several modes. Its raw moments are weighted sums of the component moments.

A histogram groups numerical observations into adjacent intervals called bins and draws one rectangle per bin. Its attraction is that a long list of values becomes a visible pattern of concentration, gaps and skewness. Specify the bin edges, the treatment of values on those edges, and what the vertical axis means: count, proportion, or density. Those choices are part of the graphic. ^[ref]

Let $n>0$ observations be covered by disjoint bins $B_j$ with positive widths ${\Delta }_j$ and counts $n_j$, so ${\sum }_j{n}_j=n$. For example, use left-closed, right-open bins, closing the final right edge too. The three common height choices have different meanings:

For the density convention, a histogram is a piecewise constant probability density function estimate:Its normalization follows by adding rectangular areas:If the observations have units, density has reciprocal units; a density height is not a probability and can exceed $1$. With equal bin widths, count heights and density heights differ only by a common scale factor. With unequal widths, count-height areas $n_j{\Delta }_j$ give wider bins extra visual weight unrelated to their counts. Dividing by width removes that distortion. R's histogram documentation explicitly describes area-one normalization for unequal bins. ^[ref]

Reuse the twelve invented values under box plot, with bins $[0,2),[2,4),[4,8),[8,22]$. The counts are $(1,3,6,2)$ and the widths are $(2,2,4,14)$. Hence the density heights and areas are exactlyThe second and third bars have equal density heights even though the third contains twice as many observations: it is also twice as wide. The last bar contains only two observations; its count-height area would be $2\cdot 14=28$, larger than the six-observation bar's $6\cdot 4=24$. Its correct density area is only $1/6$.

Bin choice trades detail against stability. Broad bins merge distinctions: the toy bin $[8,22]$ hides the empty stretch between $8$ and $20$. Narrow bins expose gaps but can make each individual observation dominate a bar. With one-unit bins starting at $0$, the bins $[2,3),[4,5),[6,7)$ each contain two values and appear equally prominent; that apparent pattern depends on grouping the repeated integers together. Moving the bin origin also changes membership without changing the observations. For density estimation, this is a form of the bias-variance tradeoff: averaging over broad bins loses local detail, while narrow bins give more variable density estimates across repeated statistical samples. ^[ref] A useful practice is to compare a few plausible widths and offsets, keeping common edges when comparing groups. Automatic bin selectors are starting points, rather than guarantees of a unique correct picture. ^[ref]

A histogram retains only bin membership. Its density estimate spreads each bin's fraction uniformly across that bin, so our last bar assigns density throughout $[8,22]$ even though the only toy values there are $8$ and $20$. It does not recover within-bin positions or observation order, and a visually convincing shape does not establish a population probability distribution. A box plot provides a complementary compact summary of quartiles; neither graphic replaces the original observations.

For isotropic incompressible flow, directional averaging gives $⟨({\partial }_x{u}_x{)}^3⟩=(8/105)⟨\mathrm{tr}{S}^3⟩=-(2/35)⟨{\omega }_i{\omega }_j{S}_{ij}⟩$. Positive mean vortex stretching therefore implies negative longitudinal derivative skewness. Statistical homogeneity alone does not determine the sign in a fixed direction.

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.

Let the summands have mean mu, variance ${\sigma }^2>0$, and cumulants ${\kappa }_j$. For the standardized sum $Z_n=(S_n-n\mu )/(\sigma \sqrt{n})$, expansion of its log-characteristic function gives a normal leading term and successively smaller cumulant terms. Fourier inversion turns these terms into Probabilists' Hermite polynomials. Under sufficient moments and a smooth nonlattice condition, the Edgeworth expansion beginsfor bounded z. Higher orders include the fourth cumulant and products of lower cumulants. This is a central expansion in powers of $n^{-1/2}$; it need not remain positive, normalized, or accurate in far tails after truncation.

For a saddlepoint density approximation, require the summand cumulant-generating function K to be finite on an interval about zero. At a sum value $s=ny$, find the interior saddle $\hat{t}$ satisfying $K^{\prime }(\hat{t})=y$. Exponential tilting makes the tilted mean y, so the tilted sum is approximated locally by its normal probability density at its mean. Undoing the tilt givesEquivalently this follows by deforming the transform-inversion contour through its stationary point. Suitable analytic and nonlattice regularity gives relative error $O(n^{-1})$ at fixed interior y. This approximation incorporates a nonquadratic tail exponent and is positive, while an Edgeworth truncation uses a normal probability density times a polynomial. It uses more information from the distribution and stronger exponential-moment assumptions. Its integral may differ from one by a higher-order amount, so positivity does not imply exact normalization.

For the centered Laplace distribution, $K(t)=t^2+t^4/2+\cdots$, hence ${\kappa }_2=2$ and ${\kappa }_3=0$. Its skewness vanishes, so the order-$n^{-1/2}$ Edgeworth correction is zero. With $z=s/\sqrt{2n}$, the requested expansion, omitting terms of order $n^{-1}$ in the standardized correction, isFor bounded z the exact probability density function is this expression times $1+O(n^{-1})$. The prefactor is $n^{-1/2}$ because this is the probability density function of the unstandardized sum; that change of units should not be confused with the orders inside the Edgeworth expansion.

For the saddlepoint, $K^{\prime }(t)=2t/(1-t^2)$. Write $y=s/n$ and $d=\sqrt{1+y^2}$. The root in $(-1,1)$ isThese formulas include $s=0$ without a division by y. Substitution yieldsThis is the leading saddlepoint approximation, without its order-$n^{-1}$ relative correction. Expanding d near one recovers the same central normal approximation, while retaining d captures the Laplace-sum large-deviation exponent away from the center.

Write $M(t)=\mathbb{E}{e}^{tX}$, with $M(0)=1$, and differentiate $\kappa (t)=\log M(t)$:Using $M^{(j)}(0)=\mathbb{E}{X}^j$ and $\mu =\mathbb{E}X$ givesThus the first three cumulants are the mean, variance and third central moment. A finite moment-generating function near zero justifies these differentiations. For positive claims with only three finite ordinary moments, the same calculations are valid as left derivatives at zero of the logarithm of the Laplace transform of a nonnegative random variable; positive exponential moments are not needed for the resulting moment identities.

For the independent Poisson distribution count, the random-sum transform identity givesConsequently the compound Poisson distribution hasThese are raw severity moments, not severity cumulants.

The printed gamma density uses rate $\nu$, not scale. Direct integration against $e^{ty}$ givesTherefore the required matching equations for the shifted gamma distribution areFor $\lambda >0$ and nonzero positive claims with finite $m_3$, these equations give positive $\alpha ,\nu$ and the unique solutionThis is the three-cumulant shifted gamma approximation. In particular, its skewness $2/\sqrt{\alpha }$ equals $m_3/(\sqrt{\lambda }{m}_2^{3/2})$, the skewness of $S$.

The competing normal approximation to a compound Poisson aggregate isTo justify its large-count limit, let ${\phi }_X$ be the claim characteristic function. For fixed severity law with $m_2<\infty$, its expansion is ${\phi }_X(u)=1+i{m}_{1}u-m_2{u}^2/2+o(u^2)$. The characteristic function of $(S-\lambda m_1)/\sqrt{\lambda m_2}$ isThus it converges in distribution to a standard normal distribution as $\lambda \to \infty$.

The normal approximation is simple, uses only $m_1,m_2$, and works well in the centre when the count is large and standardized skewness is small. It imposes symmetry, whereas positive compound-Poisson claims have positive third cumulant. The shifted-gamma approximation reproduces that third cumulant and can better describe moderate right skewness. It requires a finite and reliably known third moment, and matching three cumulants alone does not guarantee accurate extreme tails.

Neither continuous approximation reproduces the genuine atom $\mathbb{P}(S=0)=e^{-\lambda }$, which matters at small $\lambda$. The normal distribution also assigns probability $\Phi (-\sqrt{\lambda }{m}_1/\sqrt{m_2})$ to negative losses. The gamma approximation is bounded below by $k$, but its fitted shift need not be nonnegative: exponential claims of mean $\mu$ give $k=-\lambda \mu /3$, so it too can assign mass to negative losses. If $k>0$, it instead excludes some genuinely possible small losses. These support and atom discrepancies prevent either approximation from being a universal replacement for the exact aggregate law.

Let $c=g^{\prime \prime }(\tilde{y})>0$. Near the interior minimum, Taylor expansion givesOn the scale $y=\tilde{y}+z/\sqrt{nc}$, the leading exponent is $ng(\tilde{y})+z^2/2$ and $dy=dz/\sqrt{nc}$. If the integral is localized at this minimum, the rescaled integration endpoints tend to infinity in both directions, so the Gaussian integral gives Laplace's method:Formally expanding the cubic term gives an order-$n^{-1/2}$ correction odd in $z$, whose Gaussian integral is zero. Quartic terms and the square of the cubic term produce the order-$n^{-1}$ correction. Sufficient smoothness near the minimum and control of the exterior integral justify this formal calculation.

The printed local assumptions alone do not guarantee localization. A useful localization condition for Laplace's method is that $g(y)\ge g(\tilde{y})+{\delta }_{\epsilon }$ outside each sufficiently small fixed neighbourhood of $\tilde{y}$, for some ${\delta }_{\epsilon }>0$, and that ${\int }_a^b{e}^{-n_{\ast }g(y)}dy<\infty$ for some $n_{\ast }>0$. For $n>n_{\ast }$, the exterior contribution is at mostwhich is exponentially smaller than the displayed local term. To see why something of this kind is needed, take $g(y)=y^2(1-y{)}^4$ on $(-1,1)$. Its unique minimum is at zero and $g^{\prime \prime }(0)=2$, yet on $1-n^{-1/4}<y<1$ we have $ng(y)\le 1$. This boundary interval alone contributes at least $e^{-1}{n}^{-1/4}$, dominating the proposed $n^{-1/2}$ local approximation. The Laplace formula therefore has its usual implicit tail assumption, rather than following from uniqueness of the interior minimum alone.

For the gamma function, set $y=nu$ to obtainHere $g(u)=u-\log u$ has its unique minimum at $u=1$, with $g(1)=1$ and $g^{\prime \prime }(1)=1$. It tends to infinity at both endpoints, and its integral is finite for positive exponents, so localization holds. Laplace's method yields Stirling's formula:

For a prior density $p$ and independent observations, Bayes' theorem writes the desired posterior expectation as a ratio of integrals:Suppose the log-posterior kernel has a unique concentrating interior mode $\tilde{\theta }$, with curvature $J_n=-H_n^{\prime \prime }(\tilde{\theta })>0$ of order $n$, and the smoothness and tail conditions hold for both integrals. Expanding $H_n$ quadratically gives a common Gaussian factor $e^{H_n(\tilde{\theta })}\sqrt{2\pi /J_n}$. Replacing the smooth amplitude $g(\theta )$ by $g(\tilde{\theta })$ givesThis approximation does not require $g$ to be positive: it treats $g$ as an amplitude, without taking its logarithm. For more accuracy, expand the amplitude and the cubic term in the log-posterior kernel. With $s=\theta -\tilde{\theta }$, the Gaussian moments are $\mathbb{E}{s}^2=J_n^{-1}$ and $\mathbb{E}{s}^4=3J_n^{-2}$. Terms multiplying $g(\tilde{\theta })$ cancel between numerator and denominator; the two remaining first corrections areunder the stronger smoothness, derivative-scaling and integrability assumptions needed for this expansion. The first term accounts for the curvature of the amplitude and the second for posterior skewness. This is posterior expectation by Laplace approximation. If there are several modes, boundary modes, or nonconcentrating tails, the appropriate separate contributions must be included; a single interior Gaussian approximation cannot then be assumed.

Combining the Kolmogorov two-thirds law with the Kolmogorov four-fifths law gives the velocity increment skewness in the inertial range:The constant-skewness turbulence closure assumes this same value throughout the universal equilibrium range, including the dissipative crossover. This is an extra modeling assumption; neither similarity law establishes it near $r=0$.

Write $A=\beta (15\beta {)}^{1/2}$, $B=(15\beta {)}^{3/4}$, $S_2=A{v}^{2}h$ and $r=B\eta x$. Since $\nu =v\eta$ and $ϵ\eta =v^3$, substituting $S_3=-4S_2^{3/2}/(5{\beta }^{3/2})$ into the Kolmogorov equation for structure functions givesDividing by $2B/5$ yields the normalized constant-skewness structure-function equation:At small $x$, $h\sim x^2$, recovering $S_2\sim ϵr^2/(15\nu )$ for differentiable velocity increments. At large $x$, the dominant balance gives $h\sim x^{2/3}$, recovering the assumed Kolmogorov two-thirds law.

This turbulence closure is inexpensive: a single ordinary differential equation interpolates between the two regimes while respecting the local third-order balance. Its prescribed skewness cannot independently predict scale-dependent derivative statistics, transient spectral evolution or internal intermittency.

EDQNM closure instead evolves the turbulent energy spectrum through modeled spectral triad interactions. Its quasi-normal fourth-order approximation is supplemented by eddy damping and a Markovian treatment of correlation memory. It retains more information about scale-dependent transfer and evolving spectra, at the cost of spectral integrals and modeled damping times. It is still approximate and does not automatically describe coherent structures or internal intermittency. The simple constant-skewness model is therefore useful as an equilibrium interpolation; EDQNM closure addresses a broader spectral evolution problem.

Put $W=⟨{\omega }^2⟩$, $C=⟨|\nabla \times \mathbf{\omega }{|}^2⟩$ and $M=⟨G^3⟩$, where $G={\partial }_x{u}_x$. The small-$r$ Taylor series giveIn the Kármán-Howarth equation, the coefficient of $r^2$ on the left is $-\dot{W}/30$. On the right, the triple-correlation term contributes $7M/6$ and the viscous term contributes $\nu C/15$. HenceComparing with the mean enstrophy balance identifies the mean vortex stretching:Also $S_2=2(u^2-F)=W{r}^2/15+\cdots$, whereas differentiability gives $S_2=⟨G^2⟩r^2+\cdots$. Thus $⟨G^2⟩=W/15$. Homogeneity makes $⟨G⟩=0$, so the longitudinal velocity-gradient skewness is $S_0=M/(W/15{)}^{3/2}$. Substitution givesPositive mean vortex stretching therefore requires negative skewness. A centered Gaussian distribution has zero third moment, so the derivative statistics cannot be Gaussian. This conclusion uses the physically positive mean stretching, not homogeneity alone.

The probability density function has mean zero but a longer or stronger negative tail, representing intermittent compression. The illustration uses a standardized Gaussian mixture distribution solely to sketch this sign of skewness; it is not turbulence simulation data or a fitted universal distribution.

Write $m_j=\mathbb{E}{Y}^j$ and $M=M_Y$. The derivatives of the cumulant-generating function areSince $M(0)=1$ and $M^{(j)}(0)=m_j$, the first three cumulants areThe last equality follows by expanding the cube and taking expectations. We use the paper's unstandardized third central moment throughout. The usual skewness coefficient divides this quantity by $\mathrm{Var}(Y{)}^{3/2}$; its sign is the same whenever the variance is positive.

There is a minor domain qualification: finite ordinary moments do not guarantee a finite moment-generating function for positive arguments. The differentiations are valid in a neighborhood of zero when a positive exponential moment exists. For positive variables with the stated finite moments, all formulas also follow from left derivatives at zero, because $Y^j{e}^{tY}\le Y^j$ for $t\le 0$ and dominated convergence applies. Thus no extra exponential-moment assumption is needed for the moment identities themselves.

For the random sum of independent claims, condition on $N=k$. Independence of the claims and independence from the count giveTaking expectations and then logarithms yields the random-sum transform identity and the cumulant composition for a random sum:Here $G_N$ is the probability generating function. These identities hold wherever the transforms are finite, in particular for $t\le 0$ with positive claims.

For reuse in the three cases, write $m_j=\mathbb{E}{X}_1^j$ and $a_j={\kappa }_{N,j}$. Applying the chain rule three times givesor equivalentlyThis expresses the requested third central moment in raw claim moments, while keeping the contribution of claim-count variability explicit.

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.

A gamma distribution translated by a real constant $k$. With shape $\alpha >0$ and rate $\nu >0$, its support is $(k,\infty )$ and its first three cumulants are $k+\alpha /\nu$, $\alpha /{\nu }^2$, and $2\alpha /{\nu }^3$. Its skewness is $2/\sqrt{\alpha }$. A negative shift permits negative values; introducing a shift does not automatically respect the nonnegative support of an insurance loss.

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.

For Gaussian initial conditions and the Einstein-de Sitter second-order standard perturbation theory density kernel, the unsmoothed density contrast obeysThere are three choices of the second-order field and two connected Wick contractions; the angular average of $F_2$ is $5/7+(2/7)/3=17/21$. Thus the reduced skewness is $6(17/21)=34/7$. Smoothing changes this result, and the unsmoothed calculation requires a finite variance.