Refining Monte Carlo predictions using observed measurements I'm trying to build a monte-carlo simulation that can revise it's distribution of outcomes of a project based on observed measurements after the project has started.
I have a few questions about the best way to do this. I'm not a statistician, so please correct me if I am doing something wildly wrong.
For example, let's say I've observed that task x has been selected by person y (whose original 90% CI estimate for the task was [l,h]) , and that y has logged w hours of work to the task.
I can use that data to re-simulate the project under new constraints and compute a new, more accurate, distribution of outcomes.
For example, if w > l, then I know that the lower bound for the time to complete x is now w, not l, and can adjust the distribution used accordingly. However, w is not a 5% lower bound. It's a 0% lower bound (i.e. the limit), so using [w, h] as a 90% CI didn't quite seem correct. As a result I was thinking I could just pick some arbitrarily small number for p(w), say 0.0001, and continue using .05 for p(h) and then generate a new distribution for [w, h] (of course, I would just use the number of deviations for h and w rather than the probabilities).
Is that sound?
What's not immediately clear is what I would do in the case where w > h. I have calibrated estimates with a 90% CI, so I should expect to see this 5% of the time. If I ask: "what do I know in that case", I come up with the following:


*

*I know w and I know my arbitrarily low p(w)

*From the original confidence interval (which assumes a normal distribution), I can determinep(w + sigma).

*So, I could produce: [w, w + sigma] as an interval, using p(w) and p(w + sigma), and then derive a normal distribution from that (again, just using the z-values).


Is that sound as well?
 A: Here's my attempt at a (self) answer, based on user777's hints.
At first I was confused by the P(E|H) term in the Bayesian inference formula, thinking P(E|H) is 0 whenever H > E, and I only care about cases where H > E when I'm computing P(H|E).
However, then I reformulated the problem slightly and said "what is the probability that it will take at least H given that it takes at least E". The probably of at least E given at least H is one, so that yields:
$\int {p(H|E)}{dH} = \frac{\int{p(H)}{dH}}{\int^E_0{p(x)}{dx}}$
I believe that means that if I compute a random number using the original normal distribution that I should just be able to multiply it by $\int^E_0{p(x)}{dx}$ to get my new random number.
Rationale:
The normal random number should be $inv(\int{p(x)}{dx})(rand())$. To get a random number corresponding to $p(H|E)$ I need $inv\bigl({\int {p(H|E)}{dH}}\bigr)(rand())$, which should be equal to $\int^E_0{p(x)}{dx}*inv\bigl({\int{p(H)dH}}\bigr)$.
That should be exactly equal to normal_rand(sigma, mu)*cdf(w).
Update:
This needs to be truncated somehow, as the df is only valid when x > w.  I'm not quite sure how to do that.
Update 2:
I ended up using the info here: http://en.wikipedia.org/wiki/Truncated_distribution
to produce a truncated normal distribution. I think I can use that, without the Bayesian update. 
