Offset uncertainty

Let $x_i\pm\sigma_i$ be the $i=1\ldots n$ results of independent measurements and ${\bf VF}_X$ the (diagonal) covariance matrix. Let us assume that they are all affected by the same calibration constant $c$, having a standard uncertainty $\sigma_c$. The corrected results are then $y_i = x_i + c$. We can assume, for simplicity, that the most probable value of $c$ is 0, i.e. the detector is well calibrated. One has to consider the calibration constant as the physical quantity $X_{n+1}$, whose best estimate is $x_{n+1} = 0$. A term $V_{X_{n+1,n+1}} = \sigma^2_c$ must be added to the covariance matrix.

The covariance matrix of the corrected results is given by the transformation

$${\bf V}_Y = {\bf M}{\bf V}_X{\bf M}^T\,,$$ (14.14)

where $M_{ij}= \left.\frac{\partial Y_i}{\partial X_j} \right\vert _{x_j}$. The elements of ${\bf V}_Y$ are given by

$$V_{Y_{kl}} = \sum_{ij} \left. \frac{\partial Y_k}{\partial X_i} \... ...x_i} \left. \frac{\partial Y_l}{\partial X_j} \right\vert _{x_j} V_{X_{ij}}\, .$$ (14.15)

In this case we get

$$\sigma^2(Y_i) = \sigma_i^2+\sigma_c^2,$$ (14.16)

$$(Y_i,Y_j) = \sigma_c^2 \hspace{1.3 cm} (i\ne j),$$ (14.17)

$$\rho_{ij} = \frac{\sigma_c^2} {\sqrt{\sigma_i^2+\sigma_c^2} \,\sqrt{\sigma_j^2+\sigma_c^2}}$$ (14.18)

$$= \frac{1} {\sqrt{1+\left(\frac{\sigma_i}{\sigma_c}\right)^2} \,\sqrt{1+\left(\frac{\sigma_j}{\sigma_c}\right)^2}}\, ,$$ (14.19)

reobtaining the results of Section . The total uncertainty on the single measurement is given by the combination in quadrature of the individual and the common standard uncertainties, and all the covariances are equal to $\sigma^2_c$. To verify, in a simple case, that the result is reasonable, let us consider only two independent quantities $X_1$ and $X_2$, and a calibration constant $X_3 = c$, having an expected value equal to zero. From these we can calculate the correlated quantities $Y_1$ and $Y_2$ and finally their sum ( $S\equiv Z_1$) and difference ( $D\equiv Z_2$). The results are

$${\bf V}_Y = \left( \begin{array}{cc} \sigma_1^2+\sigma_c^2 & \sigma_c^2\, \\ \sigma_c^2 & \sigma_2^2+\sigma_c^2 \end{array}\right) \, ,$$ (14.20)

$${\bf V}_Z = \left( \begin{array}{cc} \sigma_1^2 + \sigma_2^2+ 4\,\sigma_c^2 &... ...2 \\ \sigma_1^2-\sigma_2^2 & \ \sigma_1^2 + \sigma_2^2 \end{array}\right) \, .$$ (14.21)

It follows that

$$\sigma^2(S) = \sigma_1^2 +\sigma_2^2 +(2\,\sigma_c)^2,$$ (14.22)

$$\sigma^2(D) = \sigma_1^2 + \sigma_2^2\, ,$$ (14.23)

as intuitively expected.