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 variance^5.2.)$_2$ The property of the average (see Section ) tells us that the likelihood $f(\overline{Q}_{n_1}\,\vert\,\mu,\sigma)$ is Gaussian:

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

To simplify the following notation, let us call $x_1$ this average and $\sigma_1$ the standard deviation of the average:

$$x_1 = \overline{q}_{n_1},$$ (5.12)

$$\sigma_1 = \sigma/\sqrt{n_1}\,.$$ (5.13)

We then apply () and get

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

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

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

• the true value is normally distributed around $x_1$;
• its best estimate (prevision) is E$[\mu]=x_1$;
• its variance is $\sigma_{\mu}=\sigma_1$;
• the ``confidence intervals'', or credibility intervals, in which there is a certain probability of finding the true value are easily calculable:
  
  
  

  Probability level	credibility interval
  (confidence level)	(confidence interval)
  $(\%)$
  68.3	$x_1$	$\pm$	$\sigma_1$
  90.0	$x_1$	$\pm$	$1.65\sigma_1$
  95.0	$x_1$	$\pm$	$1.96\sigma_1$
  99.0	$x_1$	$\pm$	$2.58\sigma_1$
  99.73	$x_1$	$\pm$	$3\sigma_1$