next up previous contents
Next: Measurements close to the Up: Normally distributed observables Previous: Final distribution, prevision and   Contents


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:

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

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

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

where
$\displaystyle x_A$ $\displaystyle =$ $\displaystyle \frac{x_1/\sigma_1^2 + x_2/\sigma_2^2}
{1/\sigma_1^2 + 1/\sigma_2^2}\, ,$ (5.19)
$\displaystyle \frac{1}{\sigma_A^2}$ $\displaystyle =$ $\displaystyle \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2}\, .$ (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.
next up previous contents
Next: Measurements close to the Up: Normally distributed observables Previous: Final distribution, prevision and   Contents
Giulio D'Agostini 2003-05-15