# Bayesian statistics with a probability observation is wrong

I have data coming from a real system taking measurements, and it occasionally gives a number completely wrong (ranging system with multiple returns). I'm using bayesian inference to accumulate measurements and I'd like to incorporate this chance into my model.

Now Bayes theorem gives us $$P(H|E) = \frac{P(E|H)P(H)}{P(E)}.$$ If $$T$$ is the event that the observation $$E$$ is "true" then $$P(E|H) = P(E|H,T)P(T) + P(E|H,!T)P(!T).$$ So the posterior becomes $$P(H|E) = \frac{P(H)}{P(E)}\big(P(E|H,T)P(T) + P(E|H,!T)P(!T)\big).$$

In the case that $$P(!T)=1$$ we would like $$P(H|E) = \frac{P(H)}{P(E)}\big(0 + P(E|H,!T)\big) =P(H),$$ that is, the posterior is equal to the prior, as we have gained no information. This gets us to $$\frac{P(E|H)}{P(E)} = 1$$ which says the probability of the observation is independent of the hypothesis, which is true in the model. But I don't know how to generalise it to the case where $$0 < P(T) < 1$$ and I think I'm missing something obvious.

Do I just set $$P(E|H,!T) = P(E)?$$

The exact formula for the theorem that you used here is a bit misleading unless you have a neat bisection of probabilities into two classes. It's not wrong, but you need to remember to account for all the combinations.

A more general form is:

$$p(H|E) = \frac {p(E|H)p(H)} {\sum_i { p(E|S_i) } }$$ Where $$S_i$$ are all the possible realizations of $$E_i$$ and so $$H \in S$$.

(an integral form would be even more general, but I do not want to wrestle with MathJax for it).

For $$T_{yours} = S = \{0, 1\}$$ in the case you wrote up, you wind up with exactly the formula you wrote.

If you had a weird experiment, with $$S = \{0, 0, 1\}$$ (e.g. you reject any outcome less than 100%), you'll get: $$p(E|H) =\\ = \frac {p(H)} {p(E)} ( (p(E|H, T=1)*(1/3) + p(E|H,T=0)*(1/3) + p(E|H,T=0)*(1/3) ) p(E|H) = \frac {p(H)} {p(E)} ( p(P(E|H, T=1)*(1/3) + p(E|H,T=0)*(2/3) )$$

For a simple Bernoulli-distributed binary $$T$$, the parenthesized sum is: $$(p(E|H, T=1)*p(T) + p(E|H,T=0)*(1-p(T))$$

If $$S$$ is any set of $$\alpha$$ accept-E and $$\beta$$ reject-E real-valued distributions of a binary $$T$$, you get: $$(p(E|H, T=1)*Beta(\alpha, \beta) + p(E|H,T=0)*(1-Beta(\alpha, \beta))$$

If you want to go even crazier, and accept, say, $$T \in \{0, 1, 2\}$$, the sum expands to three terms and you'd have to draw the probabilities to multiply each by from some Categorical distribution.