Probability density, Bayes and units

Question

The variable $X$ has a continuous probability density function (for example, it may be normally distributed) with mean $a+b$ and a constant variance, say $1$.

I try to find the following:

$P(ab \geq 0 | X=x)$

Using the Bayes' theorem, can I turn it into this:

$\frac {p(X=x | ab \geq 0) P(ab \geq 0)} {p(X=x)}$?

I find that for it to make sense, I could not have had $P(X=x | ab \geq 0)$ and $P(X=x)$ because otherwise both the numerator and the denominator would be 0, so it has to be $p(X=x | ab \geq 0)$.

I also compared the units of the LHS and the RHS. Probability is a pure number, and let $X$ have the unit, "$unit$", then probability density has the unit "$\frac {1} {unit} = unit^{-1}$".

For the LHS,

$P(ab \geq 0 |X=x) = \frac {P(ab \geq 0, X=x)} {p(X=x)}$. The numerator would have the unit $1•unit^{-1} = unit^{-1}$ and the denominator would have the unit "$unit^{-1}$". These cancel out to give us a pure number as expected.

For the RHS,

The numerator again has the unit $unit^{-1}$ and the denominator has the unit $unit^{-1}$, cancelling out again to give us a pure number.

This hints that the expression is correct.

Is it?

Answer 1

Yes, the expression (and your intuition) is correct. To put it more clearly, let $A$ be an arbitrary event: $$P(A|X=x)=\frac{p_X(x|A)P(A)}{p_X(x)}$$

It's generally useful to think about the approximation case where $P(X=x)\approx p_X(x)dx$, where you can find the above expression algebraically.