Combing Distributions

I am interested to know how risk analysts handle:

1. the revision of prior belief by experts perhaps on the basis of a series of 1-to-1 interviews.
2. the revision of belief over time as in the case of project risk where perhaps quarterly reports are produced, thereby, updating the perceived risk of the project.

Berry's book - Berry, Donald A. Statistics: A Bayesian Perspective 1996
pp347-354 - gives a risk weighted averaging process which appears to be the same as the Normal Triple Notation method given below but with a different rationale.

In both of these, I am not clear as to whether the sample size should = the number of iterations. Or when does a small sample change become a 'not small' sample as this affects the sample precision either sample size/variance or 1/variance. Berry seems to indicate small samples as 30 or less. Perhaps a statistician could answer this.

I have both the Normal Triple Notation method and Berry's method on Excel if anybody wants it - just ask.

Normal Triple Notation
The Normal distribution is represented by:

x ~ N (m,u) where m is the mean and u is the variance or standard
deviation^2

The precision is 1/u = p


{ x, ?; p } = exp ( -1/2 p (x - ? )^2

Therefore when multiplying two Normal distributions the prior and the
sample:
{x, a; p} x (x, b; q) = {(x, (pa + qb)/(p + q); p + q) x { a, b; pq/(p +q)}

General example:

A sample of n values from N( ?, ?^2 )

With a sample mean = m

Distributed as N(?, ?^2/n)

The prior has the distribution N(? , p^2 )

Let the prior precision be: 1/p2 = p

Let the sample precision be: 1/ ?^2=P


Specific example (small sample)

Sample of n=9 values from N(?, 4)

Sample mean = 20

Prior has distribution N( 25, 10)
i.e. ? = 25 and u = 10

Let the sample precision p= 1/4
Let the prior precision P= 1/10


By Bayes Theorem: { ? , (npm + P ?) / (np+P); np +P} x { m, ?; npP)/(np+P)}

{ ?, 9 x 1/4 x 20 +1/10 x 25) / (9 x 1/4 +1/10); 9 x 1/4 +1/10} x {20, 25; (9x 1/4 x 1/10)/ 9 x 1/4 +1/10)} = {?, (45+ 2.5)/( 2.25+0.1); 2.25+0.1} x {20.25; (0.225/2.35)} = (?, (47.5/2.35); 2.35) x {20, 25; 0.09} = (?, 20.21; 2.35) x {20, 25; 0.09}

Therefore: ? ~ N(20.21, 1/ 2.35) = N(20.21, 0.43) To calculate the 95% confidence interval for ?:

(npx / np+ P) +/- 1.96 ? ( 1/ (np+ P)

= {( 9 x 1/4 x 20.21) / (9 x 1/4 + 1/10)} +/- 1.96 ? 1/((9 x 1/4) + 1/10)

= (45.4725 / 2.35 ) +/- 1.96 x ? 1/ 2.35

= 19.35 +/- 1.96 x 0.6523

= 19.35 +/- 1.279

= {18.92, 21.50} i.e. 20.21 +/- 1.29