# Bayesian (In)Decision

Let $$A_j$$ be the action of person $$j$$, $$A_k$$ be the action of person $$k$$, and $$p(A)$$ be the probability of an action. Using Bayes Rule, $$p(A_j=x|A_k=y)=\frac{p(A_k=y|A_j=x)p(A_j=x)}{p(A_k=y)}$$

If $$p(A_k=y|A_j=x)=1$$, $$p(A_j=x)=\frac{2}{3}$$, and $$p(A_k=y)=\frac{1}{2}$$, then the posterior $$p(A_j=x|A_k=y)=\frac{4}{3}>1$$!

Similarly, if $$p(A_j=x|A_k=y)=\frac{1}{2}$$, $$p(A_k=y|A_j=x)=0$$, and $$p(A_k=y)=\frac{1}{2}$$, then this yields $$(1/4)=0$$!

Importantly, I'm finding total probability is not satisfied, such that $$p(A_j=x)\ne p(A_j=x|A_k=y)p(A_k=y)+p(A_j=x|A_k\ne y)p(A_k\ne y)$$, because $$\frac{1}{3}\ne(\frac{1}{2})(\frac{1}{2})+0(\frac{1}{2})$$.

However, the norming axiom is met, such that $$\frac{p(A_j=x,A_k=y)}{p(A_k=y)}\le\frac{p(A_j=x)}{p(A_k=y)}$$, because $$\frac{(\frac{2}{3})(\frac{1}{2})}{\frac{1}{2}}\le\frac{\frac{2}{3}}{\frac{1}{2}}$$.

I am surely missing something fundamental. Why is the posterior grater than one? Why does the formula yield a falsehood? Why is the total probability not satisfied? Why is the norming axiom met even though the total probability is not?

The marginal and conditional distributions you have given are incompatible.. For example, if $$p(A_k=y|A_j=x)=1$$, $$p(A_j=x)=\frac{2}{3}$$, then it is impossible to have $$p(A_k=y)=\frac{1}{2}$$:
$$p(A_k=y) = \sum_x' p(A_k=y|A_j=x')p(A_j=x') \geq p(A_k=y|A_j=x)p(A_j=x)=\frac23.$$
$$p(A_j = x) = 2/3, p(A_k = y) = 1/2$$
we know that $$p(A_j = x, A_k = y) \le 1/2$$, since it is the measure of intersection of two events, $$A_j = x$$ and $$A_k = y$$.
Thus $$p(A_j = x | A_k =y ) = \frac{p(A_j = x, A_k = y) }{p(A_k = y)} \le 1$$, which is never contradictory to the commen sense