= 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 $k=k_BT/(\beta^2s_V^2)$. For small independent input <standard uncertainties>, first-order propagation gives
$$
\left(\frac{u_k}{k}\right)^2\simeq
\left(\frac{u_T}{T}\right)^2+
4\left(\frac{u_\beta}{\beta}\right)^2+
\left(\frac{u_{s_V^2}}{s_V^2}\right)^2.
$$
In 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 $\sqrt{26}\%\simeq5.10\%$. 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 $F=k_BT\overline V/(\beta s_V^2)$, 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 $\mathcal T^{-1}\int x(t)^2dt$ gives, to leading order,
$$
\operatorname{Var}(\widehat{\sigma^2})\simeq
\frac{2\sigma^4\tau}{\mathcal T},\qquad
\frac{u_{\widehat{\sigma^2}}}{\sigma^2}\simeq
\sqrt{\frac{2\tau}{\mathcal T}}.
$$
The <Gaussian distribution> fourth-moment identity gives $\operatorname{Cov}(x(t)^2,x(0)^2)=2C_x(t)^2$; 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.
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