 Optical trap calibration uncertainty

Write $s_V^2$ for the detector's position-related voltage variance, after characterizing independent detector noise and the acquisition response. The equipartition theorem estimate is $k=k_{B}T/({\beta }^2{s}_V^2)$. For small independent input standard uncertainties, first-order propagation givesIn this expression $u_{s_V^2}$ is the uncertainty of the variance itself. If the assumed independent relative uncertainties are $1\%$, $2\%$, and $3\%$, respectively, the resulting relative standard uncertainty is $\sqrt{26}\%\simeq 5.10\%$. This invented budget excludes model bias. Shared calibration inputs introduce covariance terms: a force estimate $F=k\beta \overline{V}$ using this same thermal estimate has $F=k_{B}T\overline{V}/(\beta s_V^2)$, so treating $k$ and $\beta$ as independent would count their sensitivity incorrectly.

Correlated Brownian motion also makes sample count a poor proxy for independent information. For a long continuous stationary Gaussian record of duration $\mathcal{T}$, estimating ${\sigma }^2$ from ${\mathcal{T}}^{-1}\int x(t{)}^{2}dt$ gives, to leading order,The Gaussian distribution fourth-moment identity gives $\mathrm{Cov}(x(t{)}^2,x(0{)}^2)=2C_x(t{)}^2$; integrating this covariance over the observation interval gives the result. Subtracting a fitted mean changes higher-order terms when $\mathcal{T}\gg \tau$. The ideal $10\mathrm{s}$ example gives about $0.97\%$ relative uncertainty in the variance, even though $200000$ samples were recorded. This ideal statistical estimate does not include drift, detector noise or incorrect drag.