Bayesian (In)Decision

Question

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?

Answer 1

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 $p(A_k=y)=\frac{1}{2}$:

$$p(A_k=y) = \sum_x' p(A_k=y|A_j=x')p(A_j=x') \geq p(A_k=y|A_j=x)p(A_j=x)=\frac23.$$

Answer 2

From

$$ p(A_j = x) = 2/3, p(A_k = y) = 1/2 $$

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