Offset uncertainty

Let $x_1\pm\sigma_1$ and $x_2\pm\sigma_2$ be the two measured values, and $\sigma_c$ the common standard uncertainty:

$$\chi^2 = \frac{1}{D}\, \left[ (x_1-k)^2\, (\sigma_2^2+\sigma_c^2) +(x_2-k)^2\, (\sigma_1^2+\sigma_c^2)\right.$$

$$\hspace{0.7 cm} \left. -2\, (x_1-k)\, (x_2-k)\,\sigma_c^2 \right]\, ,$$ (6.48)

where $D=\sigma_1^2\,\sigma_2^2+ (\sigma_1^2+\sigma_2^2)\,\sigma_c^2$ is the determinant of the covariance matrix.

Minimizing $\chi^2$ and using the second derivative calculated at the minimum we obtain the best value of $k$ and its standard deviation:

$$\widehat{k} = \frac{x_1\,\sigma_2^2+x_2\,\sigma_1^2} {\sigma_1^2+\sigma_2^2} \hspace{0.3 cm}(= \overline{x}),$$ (6.49)

$$\sigma^2(\widehat{k}) = \frac{\sigma_1^2\,\sigma_2^2} {\sigma_1^2+\sigma_2^2} + \sigma_c^2\, .$$ (6.50)

The most probable value of the physical quantity is exactly that which one obtains from the average $\overline{x}$ weighted with the inverse of the individual variances. Its overall uncertainty is the quadratic sum of the standard deviation of the weighted average and the common one. The result coincides with the simple expectation.