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Conditioning on an event of probability zero

If P (B) = 0 {\displaystyle P(B)=0}, then according to the definition, P (A ∣ B) {\displaystyle P(A\mid B)} is undefined.

The case of greatest interest is that of a random variable Y, conditioned on a continuous random variable X resulting in a particular outcome x. The event B = { X = x } {\displaystyle B=\{X=x\}} has probability zero and, as such, cannot be conditioned on.

Instead of conditioning on X being exactly x, we could condition on it being closer than distance ϵ {\displaystyle \epsilon } away from x. The event B = { x − ϵ < X < x + ϵ } {\displaystyle B=\{x-\epsilon <X<x+\epsilon \}} will generally have nonzero probability and hence, can be conditioned on.We can then take the limit

For example, if two continuous random variables X and Y have a joint density f X, Y (x, y) {\displaystyle f_{X,Y}(x,y)}, then by L'Hôpital's rule and Leibniz integral rule, upon differentiation with respect to ϵ {\displaystyle \epsilon }:

The resulting limit is the conditional probability distribution of Y given X and exists when the denominator, the probability density f X (x 0) {\displaystyle f_{X}(x_{0})}, is strictly positive.

It is tempting to define the undefined probability P (A ∣ X = x) {\displaystyle P(A\mid X=x)} using limit (1), but this cannot be done in a consistent manner. In particular, it is possible to find random variables X and W and values x, w such that the events { X = x } {\displaystyle \{X=x\}} and { W = w } {\displaystyle \{W=w\}} are identical but the resulting limits are not:

The Borel–Kolmogorov paradox demonstrates this with a geometrical argument.