Normalization uncertainty

Let us consider now the case where the calibration constant is the scale factor $f$, known with a standard uncertainty $\sigma_f$. Also in this case, for simplicity and without losing generality, let us suppose that the most probable value of $f$ is 1. Then $X_{n+1} = f$, i.e. $x_{n+1} = 1$, and $V_{X_{n+1,n+1}} = \sigma^2_f$. Then

$$\sigma^2(Y_i) = \sigma_i^2 + \sigma_f^2\, x_i^2 \, ,$$ (6.35)

$$(Y_i,Y_j) = \sigma_f^2\, x_i\, x_j \hspace{1.3 cm}(i\ne j) \, ,$$ (6.36)

$$\rho_{ij} = \frac{x_i\, x_j} {\sqrt{x_i^2+\frac{\sigma_i^2}{\sigma_f^2}} \,\sqrt{x_j^2+\frac{\sigma_j^2}{\sigma_f^2}}}\, ,$$ (6.37)

$$\vert\rho_{ij}\vert = \frac{1} {\sqrt{1+\left(\frac{\sigma_i}{\sigma_f\, x_i}\right)^2} \,\sqrt{1+\left(\frac{\sigma_j}{\sigma_f\, x_j}\right)^2} }\, .$$ (6.38)

To verify the results let us consider two independent measurements $X_1$ and $X_2$; let us calculate the correlated quantities $Y_1$ and $Y_2$, and finally their product ( $P\equiv Z_1$) and their ratio ( $R\equiv Z_2$):

$${\bf V}_Y = \left( \begin{array}{cc} \sigma_1^2+\sigma_f^2\, x_1^2 & \sigma_f... ... & \\ \sigma_f^2\, x_1\, x_2 & \sigma_2^2+\sigma_f^2\, x_2 \end{array}\right),$$ (6.39)

$${\bf V}_Z = \left( \begin{array}{cc} \sigma_1^2\, x_2^2 + \sigma_2^2\, x_1^2 ... ...rac{\sigma_1^2}{x_2^2} + \sigma_2^2\,\frac{x_1^2}{x_2^4} \end{array}\right)\, .$$ (6.40)

It follows that

$$\sigma^2(P) = \sigma_1^2\, x_2^2 + \sigma_2^2\, x_1^2 + (2\,\sigma_f\, x_1\, x_2)^2 \, ,$$ (6.41)

$$\sigma^2(R) = \frac{\sigma_1^2}{x_2^2} + \sigma_2^2\,\frac{x_1^2}{x_2^4} \, .$$ (6.42)

Just as an unknown common offset error cancels in differences and is enhanced in sums, an unknown normalization error has a similar effect on the ratio and the product. It is also interesting to calculate the standard uncertainty of a difference in the case of a normalization error:

$$\sigma^2(D) = \sigma_1^2+\sigma_2^2 +\sigma_f^2\,(x_1-x_2)^2\, .$$ (6.43)

The contribution from an unknown normalization error vanishes if the two values are equal.