Indirect calibration

Let us use the result of the previous section to solve another typical problem of measurements. Suppose that after (or before, it doesn't matter) we have done the measurements of $x_1$ and $x_2$ and we have the final result, summarized in (), we know the ``exact'' value of $\mu_1$ (for example we perform the measurement on a reference). Let us call it $\mu_1^\circ$. Will this information provide a better knowledge of $\mu_2$? In principle yes: the difference between $x_1$ and $\mu_1^\circ$ defines the systematic error (the true value of the ``zero'' $Z$). This error can then be subtracted from $x_2$ to get a corrected value. Also the overall uncertainty of $\mu_2$ should change, intuitively it ``should'' decrease, since we are adding new information. But its value doesn't seem to be obvious, since the logical link between $\mu_1^\circ$ and $\mu_2$ is $\mu_1^\circ\rightarrow Z \rightarrow \mu_2$.

The problem can be solved exactly using the concept of conditional probability density function $f(\mu_2\,\vert\,\mu_1^\circ)$ [see ()-()). We get

$$\Large {\mu_{2\,\vert\,\mu_1^\circ} \sim} {\Large\cal N}\left(x_2... ...ma_2^2+\left( \frac{1}{\sigma_1^2}+\frac{1}{\sigma_Z^2} \right)^{-1}}\right)\,.$$ (5.83)

The best value of $\mu_2$ is shifted by an amount $\Delta$, with respect to the measured value $x_2$, which is not exactly $x_1-\mu_1^\circ$, as was naï vely guessed, and the uncertainty depends on $\sigma_2$, $\sigma_Z$ and $\sigma_1$. It is easy to be convinced that the exact result is more reasonable than the (suggested) first guess. Let us rewrite $\Delta$ in two different ways:

$$\Delta = \frac{\sigma_Z^2}{\sigma_1^2+\sigma_Z^2}\,(\mu_1^\circ-x_1)$$ (5.84)

$$= \frac{1}{\frac{1}{\sigma_1^2}+\frac{1}{\sigma_Z^2}} \,\left[\frac{1}{\sigma_1^2}\cdot(x_1-\mu_1^\circ) + \frac{1}{\sigma_Z^2}\cdot 0 \right]\,.$$ (5.85)

• Eq. () shows that one has to apply the correction $x_1-\mu_1^\circ$ only if $\sigma_1=0$. If instead $\sigma_Z=0$ there is no correction to be applied, since the instrument is perfectly calibrated. If $\sigma_1\approx \sigma_Z$ the correction is half of the measured difference between $x_1$ and $\mu_1^\circ$.
• Eq. () shows explicitly what is going on and why the result is consistent with the way we have modelled the uncertainties. In fact we have performed two independent calibrations: one of the offset and one of $\mu_1$. The best estimate of the true value of the ``zero'' $Z$ is the weighted average of the two measured offsets.
• The new uncertainty of $\mu_2$ [see ()] is a combination of $\sigma_2$ and the uncertainty of the weighted average of the two offsets. Its value is smaller than it would be with only one calibration and, obviously, larger than that due to the sampling fluctuations alone:

  $$\sigma_2 \le \sqrt{\sigma_2^2+\frac{\sigma_1^2\,\sigma_Z^2} {\sigma_1^2+\sigma_Z^2}} \le \sqrt{\sigma_2^2+\sigma_Z^2}\,.$$ (5.86)