Law of total probability and absolute value paradox

Question

I am trying to reconcile a difference between two methods of finding $\mathbb{P}(|X|\geq y)$ for some RV $X$ and some $y > 0$.

Method 1

Consider the following line of reasoning: \begin{align*} &|X| \geq y\\ \iff& (X > 0\text{ and }X \geq y)\text{ or } (X < 0\text{ and } -X\geq y)\\ \iff& (X \geq y)\text{ or } (-X\geq y) &\text{since $y>0$}\\ \iff& (X \geq y) \text{ or } (X \leq -y) \end{align*}

Note that on the last line, the two events are disjoint. This leads us to believe $$ \mathbb{P}(|X| \geq y) = \mathbb{P}(X\geq y) + \mathbb{P}(X\leq -y) $$

Method 2

We have that $\{\{X > 0\}, \{X \leq 0\}\}$ is an event space. Then by the law of total probability, we have \begin{align*} \mathbb{P}(|X|\geq y) &= \mathbb{P}(|X|\geq y\:\vert\:X>0)\,\mathbb{P}(X>0) + \mathbb{P}(|X|\geq y\:\vert\:X\leq 0)\,\mathbb{P}(X\leq 0)\\ &= \mathbb{P}(X\geq y)\,\mathbb{P}(X > 0)+\mathbb{P}(-X\geq y)\,\mathbb{P}(X\leq 0)\\\\ \mathbb{P}(|X|\geq y)&= \mathbb{P}(X\geq y)\,\mathbb{P}(X>0)+\mathbb{P}(X\leq-y)\,\mathbb{P}(X\leq 0) \end{align*}

It seems that these two methods should always give the same answer, however they only coincide when further assumptions about the probabilities are made.

What is the resolution here?

Answer 1

You do not always have $\mathbb{P}(|X|\ge y | X > 0) = \mathbb{P}(X\ge y)$.

Instead, you have $\mathbb{P}(|X|\ge y | X > 0)\mathbb{P}(X > 0) = \mathbb{P}(X \ge y)$.

When you replace that in your formula (and similarly for the negative), you get exactly the same result as Method 1.

Answer 2

Since in general $$ \Bbb P(|X|\ge y\mid X>0)\ne \Bbb P(|X|\ge y\land X>0),$$ your method 2 is incorrect

Answer 3

$P(|X|\geq y | X>0) \ne P(X\geq y)$

Consider the problem where X=2 or X=-2 with probability $\frac 12$ and $y=1$:

$P(|X|\geq y | X>0) $ means : what is the probability of $|X|>1$ given that $X>0$. Answer : 1.

$P(X\geq y) $ means : what is the probability of $|X|>1$. Answer : $\frac 12$.

In $P(A|B)$, conditioning by B tells you the event B has happened. It is "deterministic". It truncates your universe. This event can not be consider at the same level as A which is still random and you want to access the probability that A happens in your smaller universe defined by B. You can not intersect them and simplify the probability. What I try to explain in Layman words is the Bayes formula.

Indeed, $P(|X|\geq y \cap X>0) = P(X\geq y)$ is true and I think it corresponds to the reasoning you did.