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Final distribution, prevision and credibility intervals of the true value

The first application of the Bayesian inference will be that of a normally distributed quantity. Let us take a data sample $ \underline{q}$ of $ n_1$ measurements, of which we calculate the average $ \overline{q}_{n_1}$. In our formalism $ \overline{q}_{n_1}$ is a realization of the random variable $ \overline{Q}_{n_1}$. Let us assume we know the standard deviation $ \sigma$ of the variable $ Q$, either because $ n_1$ is very large and can be estimated accurately from the sample or because it was known a priori (We are not going to discuss in these notes the case of small samples and unknown variance5.2.)$ _2$ The property of the average (see Section [*]) tells us that the likelihood $ f(\overline{Q}_{n_1}\,\vert\,\mu,\sigma)$ is Gaussian:

$\displaystyle \overline{Q}_{n_1} \sim {\cal N}(\mu, \sigma/\sqrt{n_1}).$ (5.11)

To simplify the following notation, let us call $ x_1$ this average and $ \sigma_1$ the standard deviation of the average:
$\displaystyle x_1$ $\displaystyle =$ $\displaystyle \overline{q}_{n_1},$ (5.12)
$\displaystyle \sigma_1$ $\displaystyle =$ $\displaystyle \sigma/\sqrt{n_1}\,.$ (5.13)

We then apply ([*]) and get

$\displaystyle f(\mu\,\vert\,x_1, {\cal N}(\cdot,\sigma_1)) = \frac{\frac{1}{\sq...
...\sigma_1} \,e^{-\frac{(x_1-\mu)^2}{2\,\sigma_1^2}}f_\circ(\mu)\,\rm {d}\mu}\, .$ (5.14)

At this point we have to make a choice for $ f_\circ(\mu)$. A reasonable choice is to take, as a first guess, a uniform distribution defined over a ``large'' interval which includes $ x_1$. It is not really important how large the interval is, for a few $ \sigma_1$ away from $ x_1$ the integrand at the denominator tends to zero because of the Gaussian function. What is important is that a constant $ f_\circ(\mu)$ can be simplified in ([*]), obtaining

$\displaystyle f(\mu\,\vert\,x_1, {\cal N}(\cdot,\sigma_1)) = \frac{\frac{1}{\sq...
...sqrt{2\,\pi}\,\sigma_1} \,e^{-\frac{(x_1-\mu)^2}{2\,\sigma_1^2}}\rm {d}\mu}\, .$ (5.15)

The integral in the denominator is equal to unity, since integrating with respect to $ \mu$ is equivalent to integrating with respect to $ x_1$. The final result is then

$\displaystyle f(\mu) = f(\mu\,\vert\,x_1,{\cal N}(\cdot,\sigma_1)) = \frac{1}{\sqrt{2\,\pi}\,\sigma_1}\, e^{-\frac{(\mu-x_1)^2}{2\,\sigma_1^2}}\, :$ (5.16)


next up previous contents
Next: Combination of several measurements Up: Normally distributed observables Previous: Normally distributed observables   Contents
Giulio D'Agostini 2003-05-15