# Accounting for unobserved events in baysian learning

I wanted to use Bayes theorem to help me automate the task of deciding if I should ignore events, but I am not sure how to update the posterior if I do

The simple story goes like this:

An event $y_i$ can take values 0 and 1, if it takes values 0 I can ignore it but with 1 it requires some actions.

For simplicity we will say that $P(Y|\theta) = \theta^y(1-\theta)^{1-y}$ and $\theta$ starts of as a $Beta(1,1)$

Having observed n events where y=0 I update my posterior of $\theta$ and check the probability for next event to be 0. Its high enough and I decide to ignore it.

The question is how to proceed from here? My first though was to count it as both 0 and 1 and update the posterior to $Beta(n+2, 2)$

It feels fairly intuitive, as we keep ignoring events our certainty decreases. However I am not sure how to make this rigour enough to generalize. For instance how would I proceed if I had an Normal distribution?

Any help would be greatly appreciated.

• I believe this belongs on Cross Validated SE. Jun 30 '17 at 0:06

After observing $n$ events where $y=0$, I would have thought that your posterior distribution for $\theta$ should be $\operatorname{Beta}(1,n+1)$ which has expected value $\frac{1}{n+2}$ and so might give you a point estimate of $1-\theta$, the probability of the next observation being a $0$, of $\frac{n+1}{n+2}$.
But until you actually have information about the value of the next observation, you cannot justifying updating your posterior for $\theta$.
As an alternative approach with the same result, you think there is a probability of $\frac{n+1}{n+2}$ that you might next update the posterior to $\operatorname{Beta}(1,n+2)$ and a probability of $\frac{1}{n+2}$ that you might next update the posterior to $\operatorname{Beta}(2,n+1)$. A weighted mixture like this will have a combined distribution of $\operatorname{Beta}(1,n+1)$, the same as you already had as a result of the observations you had actually seen.