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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.

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  • $\begingroup$ 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. $\endgroup$
    – whuber
    Nov 14, 2023 at 23:49

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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.
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    $\begingroup$ +1 -- but your final comment is puzzling, because since you refer to "$P$" explicitly as a probability then axiomatically it cannot equal 2023. $\endgroup$
    – whuber
    Nov 15, 2023 at 12:29
  • $\begingroup$ @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 $[0, 1]$. $\endgroup$
    – Zhanxiong
    Nov 15, 2023 at 12:59
  • $\begingroup$ 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 $$P[A\|\mathscr{G}]_\omega = \begin{cases} 17 & \text{ if } \omega \in B_1, \\ \frac{P(A \cap B_i)}{P(B_i)} & \text{ if } \omega \in B_i, i = 1, 2, \ldots, r.\end{cases}$$. $\endgroup$
    – Zhanxiong
    Nov 15, 2023 at 13:03
  • $\begingroup$ 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. $\endgroup$
    – Zhanxiong
    Nov 15, 2023 at 13:06
  • $\begingroup$ I would dispute your interpretation of the author's meaning, because a basic axiom of probability is that it is a number in $[0,1].$ $\endgroup$
    – whuber
    Nov 15, 2023 at 13:32

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