Why is the definition of conditional probability recursive

Question

The definition of conditional probability after Kolmogorov seems to be recursive without a base case. It says

$\text{P}(A|B) = \frac{\text{P}(A \cap B)}{\text{P}(B}$

Now, $\text{P}(A \cap B)$ is $\text{P}(A)\text{P}(B | A)$

So we get

$\newcommand{\prob}{\text{P}} \begin{align*} \prob(A | B) = & \frac{\text{P}(A \cap B)}{\text{P}(B}\\ = & \frac{\prob(A)\prob(B|A)}{\prob(B)}\\ = & \frac{\prob(A)\left[ \frac{\prob(B)\prob(A|B)}{\prob(A)} \right]}{\prob(B)}\\ = & \frac{\prob(A)\left[ \frac{\prob(B) \left[ \frac{\prob(A)\prob(B|A)}{\prob(B)} \right]}{\prob(A)} \right]}{\prob(B)}\\ = & \frac{\prob(A)\left[ \frac{\prob(B) \left[ \frac{\prob(A) \left[ \frac{\prob(B)\prob(A|B)}{\prob(A)} \right] }{\prob(B)} \right]}{\prob(A)} \right]}{\prob(B)}\\ & \vdots \end{align*}$

And the recursion goes on.

If one does not actually calculate $\prob(A| B)$ like this for instance:

$$\prob(A | B) = \frac{\#(B, A)}{\sum_{e_i}\#(B,e_i)} $$ you never get a base case for the recursion. Why is that?

Answer 1

There is no recursion. $P(A \cap B)$ and $P(B)$ are both well-defined and then you define $P(A |B) = \frac{P(A \cap B)}{P(B)}$. If you look closely at your calculations, then you will see that the equation

$$P(A | B) = \frac{P(A) \frac{P(B) P(A|B)}{P(A)}}{P(B)}$$ simplifies to $P(A|B) = P(A|B)$.

Answer 2

I will answer with an ironic (though hopefully enlightening) example:

Let $y=2x$, hence $x=y/2$.

Therefore $y=2(y/2)$.

Therefore $y=2(2x/2)$.

Therefore $y=2(2y/2/2)$.

Therefore $y=2(2(2x)/2/2)$.

Therefore $y=2(2(2y/2)/2/2)$.

Therefore $y=2(2(2(2x)/2)/2/2)$.

Therefore $y=2(2(2(2y/2)/2)/2/2)$.

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