 Finite inverse moments in thickness models

Finite inverse moments in thickness models determine whether a positive but uncertain layer thickness permits a finite mean or variance after conversion to density. With fixed positive mass and area, write $\rho =K/\delta$, where $K=M/A$. A finite thickness mean does not ensure that $E({\delta }^{-1})$ or $E({\delta }^{-2})$ exists.

Suppose the thickness probability density function behaves near zero as $p(\delta )\sim C{\delta }^{a-1}$ with $C>0$. The small-thickness part of the inverse moment is proportional toFor $q>0$, this converges exactly when $a>q$. Thus a finite mean density requires $a>1$, while finite density variance requires $a>2$, assuming the rest of the distribution is well behaved.

An exact example uses a gamma distribution with shape $a$ and scale $\theta$. Substitution into its density and the gamma function integral givesThe distribution and parameter conventions are specified by NIST; the inverse-moment calculation here follows directly from the integral.^[ref] For $a=2$ and $\theta =d/2$, mean thickness is $d$ and mean density is $2K/d$, yet its second moment diverges. A simulation can return finite sample variances without establishing a finite population variance.

A lognormal distribution provides a different positive-tail model. If $\log \delta \sim N(\mu ,s^2)$, then $E({\delta }^{-q})=\exp (-q\mu +q^2{s}^2/2)$, finite for every finite positive $q$. NIST defines this logarithmic-normal parameterization and its coefficient of variation.^[ref] With mean thickness $d$ and coefficient of variation $c$, mean density is $K(1+c^2)/d$. At $c=0.20$, it exceeds $K/d$ by 4%.

Tail-model dependence of inferred density uncertainty is therefore experimentally consequential. Choosing a convenient positive distribution changes the implied inverse moments. Justify near-zero behaviour from dimensional evidence or physical bounds rather than treating positivity or a stable finite simulation as sufficient validation.