= Component-elastance ambiguity in total recoil measurements
\b[Component-elastance ambiguity in total recoil measurements] means that the stiffness of the <lung> and <chest wall> cannot be recovered separately from their sum alone. At a common passive volume increment, let
$$
E_T=E_L+E_W,\qquad E_L=\frac1{C_L},\quad E_W=\frac1{C_W}.
$$
Every positive pair $(E_L,E_W)$ with the same sum gives the same local total pressure-volume slope. Measuring that total more precisely does not create a second independent equation.
An original example has total elastance $10\,\mathrm{cm\,H_2O\,L^{-1}}$. Both $(E_L,E_W)=(5,5)$ and $(8,2)$ fit it, while their component compliances differ substantially. The <respiratory-system compliance> is 0.10 in either case. To determine the partition, an additional measurement must constrain one component.
With body-surface pressure fixed and flow absent, an increment in <pleural pressure> supplies wall-pressure information:
$$
E_W=\frac{\Delta P_{\rm pl}}{\Delta V},\qquad
E_L=\frac{\Delta P_{\rm alv}-\Delta P_{\rm pl}}{\Delta V}.
$$
An esophageal balloon is one experimental route to an internal pressure estimate, but the measurement's relation to pleural pressure needs validation. The primary respiratory text describes balloon-wall, esophageal and mediastinal contributions to the recorded signal.<Pulmonary Function Testing by Warren M. Gold and Laura L. Koth>[Lung Elasticity, intraesophageal pressure and its additional contributions]{ref}
<Pleural-pressure calibration and compensating component errors> can preserve an apparently correct total. Suppose known $\Delta V=0.5$ L gives total pressure increment 5 cm water and true pleural increment 2 cm water. Then $E_W=4$ and $E_L=6$. A pressure-surrogate gain of 1.5 would instead report 3 cm water, yielding estimated $E_W=6$ and $E_L=4$. Both estimates are positive and still sum to 10. Agreement with the total curve cannot validate that pressure gain.
The estimates also share uncertainty. If measured total elastance $T$ and inferred wall elastance $W$ have independent errors, then $\widehat E_L=T-W$ and $\widehat E_W=W$, so
$$
\operatorname{cov}(\widehat E_L,\widehat E_W)=-\operatorname{Var}(W).
$$
Including this negative covariance returns $\operatorname{Var}(\widehat E_L+\widehat E_W)=\operatorname{Var}(T)$, as the algebra requires. Treating the component estimates as independent would count the internal-pressure uncertainty twice. The GUM's correlated-input rule retains this common-input contribution.<Guide to the expression of uncertainty in measurement by JCGM>[§5.2.2, equation 13; §5.2.4, shared inputs]{ref}
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