Combination of several measurements

Let us imagine making a second set of measurements of the physical quantity, which we assume unchanged from the previous set of measurements. How will our knowledge of $\mu$ change after this new information? Let us call $x_2 = \overline{q}_{n_2}$ and $\sigma_2 = \sigma^\prime/\sqrt{n_2}$ the new average and standard deviation of the average ( $\sigma^\prime$ may be different from $\sigma$ of the sample of $n_1$ measurements), respectively. Applying Bayes' theorem a second time we now have to use as initial distribution the final probability of the previous inference:

$$f(\mu\,\vert\,x_1,\sigma_1, x_2, \sigma_2, {\cal N}) = \frac{\fra... ...2}{2\,\sigma_2^2}} f(\mu\,\vert\,x_1,{\cal N}(\cdot,\sigma_1))\,\rm {d}\mu}\, .$$ (5.17)

The integral is not as simple as the previous one, but still feasible analytically. The final result is

$$f(\mu\,\vert\,x_1,\sigma_1, x_2, \sigma_2, {\cal N}) = \frac{1}{\sqrt{2\,\pi}\,\sigma_A} \,e^{-\frac{(\mu-x_A)^2}{2\,\sigma_A^2}}\, ,$$ (5.18)

where

$$x_A = \frac{x_1/\sigma_1^2 + x_2/\sigma_2^2} {1/\sigma_1^2 + 1/\sigma_2^2}\, ,$$ (5.19)

$$\frac{1}{\sigma_A^2} = \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\, .$$ (5.20)

One recognizes the famous formula of the weighted average with the inverse of the variances, usually obtained from maximum likelihood. There are some comments to be made.

• Bayes' theorem updates the knowledge about $\mu$ in an automatic and natural way.
• If $\sigma_1 \gg \sigma_2$ (and $x_1$ is not ``too far'' from $x_2$) the final result is only determined by the second sample of measurements. This suggests that an alternative vague a priori distribution can be, instead of uniform, a Gaussian with a large enough variance and a reasonable mean.
• The combination of the samples requires a subjective judgement that the two samples are really coming from the same true value $\mu$. We will not discuss this point in these notes^5.3, but a hint on how to proceed is to: take the inference on the difference of two measurements, $D$, as explained at the end of Section  and judge yourself whether $D=0$ is consistent with the probability density function of $D$.