# Two events, both of probability zero, have caused an outcome of probability zero. Which of them did happen?

Let $$X$$ and $$Y$$ both be standard normal distributions - so with mean 0 and variance 1, and independent of each other.

Now let $$Z = XY$$. We know that $$P(Z=0) = P(XY=0) = 0$$, because the set $$\{0\}$$ is of measure zero, and this is a continuous probability.

However, let's say that we know that $$Z$$ is equal to 0. This is perfectly possible, even though it happens with probability zero. What is the probability that $$X$$ is zero? Intuition suggests $$1/2$$, but when I'm trying to apply the Bayes rule: $$P(X=0|Z=0) = \frac {P(Z=0|X=0)P(X=0)} {P(Z=0)} = \frac 0 0$$ The conditional probability formula cannot be applied as well:
$$P(X=0|Z=0) = \frac {P(X=0 ,Z=0)} {P(Z=0)} = \frac {P(X=0 ,Z=0)} {0}$$

Is there anything I can reason about $$X$$ or $$Y$$? For me it's seems that $$P(X=0|Z=0)=1/2$$ but I do not know how to prove it.
The above problem has arisen as a part of a broader question. Let $$X$$, $$Y$$ and $$Z$$ be standard, independent, random normal variables. I want to prove that $$XY$$ is not independent from $$YZ$$ by showing that if $$XY$$ is equal to 0, there is a good chance that $$YZ$$ is equal to 0 as well.

• You will have a much easier time with a different approach. For instance, notice that $E[(XY)^2]=E[X^2]E[Y^2]=1$ and similarly with $E[(ZY)^2],$ yet $E[(XY)^2(ZY)^2]=E[X^2]E[Y^4]E[Z^2]=3\ne E[(XY)^2]E[(ZY)^2]$ shows $(XY)^2$ is not independent of $(ZY)^2$ and therefore $XY$ is not independent of $ZY.$ Working with expectations is one way to avoid the problems with conditioning on zero-probability events -- a circumstance that is nonintuitive and leads to many of the best-known probability paradoxes.
– whuber
Nov 14, 2023 at 23:49

Let me answer your "broader question" first because it is less technical. There are many ways to show $$XY$$ and $$YZ$$ are not independent given $$X, Y, Z \text{ i.i.d. } \sim N(0, 1)$$. For example, if $$XY$$ and $$YZ$$ are independent, then $$E[(XY)^2(YZ)^2] = E[(XY)^2]E[(YZ)^2]$$. However, by independence and normal distribution moments: \begin{align*} & E[(XY)^2(YZ)^2] = E[X^2Y^4Z^2] = E[X^2]E[Y^4]E[Z^2] = 1 \times 3 \times 1 = 3, \\ & E[(XY)^2] = E[X^2Y^2] = E[X^2]E[Y^2] = 1 \times 1 = 1, \\ & E[(YZ)^2] = E[Y^2Z^2] = E[Y^2]E[Z^2] = 1 \times 1 = 1, \\ \end{align*} a contradiction.

Now to understand the precise meaning of the notation "$$P[X = 0|Z = 0]$$" when $$\{Z = 0\}$$ is an event of probability zero, you will need some measure-theoretic probability knowledge, refer to good discussions in this thread. A short answer is that could be any pre-specified constant value because the conditional probability $$P[X = 0|\sigma(Z)]$$ can be defined arbitrarily on zero probability $$\sigma(Z)$$-sets. But two things are certain:

1. $$P(X = 0|Z = 0)$$ cannot be handled by the elementary conditional probability formula $$P(A|B) = P(A \cap B)/P(B)$$. This is as you observed.
2. $$P(X = 0 | Z = 0) = \frac{1}{2}$$ is technically OK (as I mentioned in above). But it is more like a choice/specification, instead of the result of deduction. I could also say $$P(X = 0|Z = 0) = 2023$$ for the same reason.
• +1 -- but your final comment is puzzling, because since you refer to "$P$" explicitly as a probability then axiomatically it cannot equal 2023.
– whuber
Nov 15, 2023 at 12:29
• @whuber That is why I said this concept might be too "technical" and I don't want to divert the topic too much -- the conditional probability $P(A|\mathscr{G})$ essentially can have different versions whose values can be arbitrary (but constant) on $0$-probability $\mathscr{G}$-sets. See Probability and Measure (3rd edition): "... In this case $P(A|\mathscr{G})$ will be taken to have any constant value on $B_i$; the value is arbitrary but must be the same over all of the set $B_i$". By "arbitrary", the author meant the value can be any real number, not necessarily in $[0, 1]$. Nov 15, 2023 at 12:59
• In the same reference, Billingsley presented the following example: Suppose that $B_1, \ldots, B_r$ is a partition of $\Omega$ into $\mathscr{F}$-sets, and let $\mathscr{G} = \sigma(B_1, \ldots, B_r)$. If $P(B_1) = 0$ and $P(B_i) > 0$ for the other $i$, then one version of $P[A \| \mathscr{G}]$ is $$P[A\|\mathscr{G}]_\omega = \begin{cases} 17 & \text{ if } \omega \in B_1, \\ \frac{P(A \cap B_i)}{P(B_i)} & \text{ if } \omega \in B_i, i = 1, 2, \ldots, r.\end{cases}$$. Nov 15, 2023 at 13:03
• My point in the answer is to remind OP that his claim "$P(X = 0 | Z = 0) = \frac{1}{2}$" is actually unverifiable. It may be too much to call it "wrong" because of the reason explained above, but it is essentially not a "real probability" that can be interpreted in the usual way. Nov 15, 2023 at 13:06
• I would dispute your interpretation of the author's meaning, because a basic axiom of probability is that it is a number in $[0,1].$
– whuber
Nov 15, 2023 at 13:32