# Probability density, Bayes and units

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?

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.