Bayes Theorem: $$Pr(A|B) = \frac{Pr(A)*Pr(B|A)}{Pr(B)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(0)$$
Furthermore: $$Pr(A) = Pr(A|B)*Pr(B) + Pr(A|\neg B)*Pr(|\neg B) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$$
Scenario:
Suppose we are not given that an event $B$ has occured.
Assuming we are given a base Probability of $A$: $(Pr(A_1)$ (this is our prior probability), subsequently, we are now given a $Pr(A|B_1)$, a $Pr(B_1)$, a $Pr(A|\neg B_1)$, and a $Pr(\neg B_1)$ as additional information. Assuming, that I now want to calculate a $Pr(A_2)$ that is the posterior probability in light of the new information.
From $(1)$: $$Pr(A_2) = Pr(A|B_1)\cdot Pr(B_1) + Pr(A|\neg B_1)\cdot Pr(\neg B_1)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$$
From $(0)$: $Pr(A|B_1) = \frac{Pr(A_1)*Pr(B_1|A)}{Pr(B_1)} \,\,:\,\, Pr(A|\neg B_1) = \frac{Pr(A_1)*Pr(\neg B_1|A)}{Pr(\neg B_1)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(3)$$
Rewriting $(2)$ in light of $(3)$ $$Pr(A_2) = \frac{Pr(A_1)*Pr(B_1|A)}{Pr(B_1)} \times Pr(B_1) + \frac{Pr(A_1)*Pr(\neg B_1|A)}{Pr(\neg B_1)} \times Pr(\neg B_1)$$
Which gives us: $$Pr(A_2) = Pr(A_1)*Pr(B_1|A) + Pr(A_1)*Pr(\neg B_1|A)$$ Simplifying to: $$Pr(A_2) = Pr(A_1)\cdot\left(Pr(B_1|A) +Pr(\neg B_1|A)\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(4)$$
Generalising from $(4)$: $$Pr(A_{l+1}) = Pr(A_l)\cdot\left(Pr(B_l|A) +Pr(\neg B_l|A)\right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(5)$$
My Question:
- Is $(5)$ valid bayesian inference?
- If not, how do I perform valid inference in the specified scenario?