Since the normal distribution is symmetric, why *wouldn't* $Z$ equal $\sqrt{y}$ and $-\sqrt{y}$ with equal probabilities?

Question

My textbook, Introduction to Probability by Blitzstein and Hwang, provides the following example:

Example 4

Let $Z \sim N(0, 1)$ and $Y = Z^2$. Find $\mathbb{E}[Y | Z]$ and $\mathbb{E}[Z | Y]$.

Since $Y$ is a function of $Z$, $E(Y | Z) = E(Z^2 | Z) = Z^2$ by taking out what's known.

To get $E(Z | Y)$, notice that conditional on $Y = y$, $Z$ equals $\sqrt{y}$ or $- \sqrt{y}$ with equal probabilities by the symmetry of the standard normal, so $E(Z | Y = y) = 0$ and $E(Z | Y) = 0$.

My intuition is having a bit of trouble with the second explanation. Since the normal distribution is symmetric, why wouldn't $Z$ equal $\sqrt{y}$ and $-\sqrt{y}$ with equal probabilities? Naively speaking, it seems like the symmetry argument would be an argument in support of this fact.

I would appreciate it if someone could please take the time to clarify this.

EDIT:

I just came across another example that seems to allude to the same fact as above:

Example 6

Let $Z \sim N(0, 1)$ and $Y = Z^2$. Find $\text{Var}[Y | Z]$ and $\text{Var}[Z | Y]$.

$\vdots$

To get $\text{Var}(Z | Y)$, apply the definition:

$\text{Var}[Z | Y] = \mathbb{E}[Z^2 | Y] - \mathbb{E}[Z | Y]^2 = \mathbb{E}[ Y | Y] - \mathbb{E}[Z | Y]^2 = Y - 0 = Y$

Noted that $\mathbb{E}[Z | Y] = 0$ by symmetry.

Answer 1

You write:

Hmm, perhaps I misinterpreted what the author is saying here? I thought they were saying that because of the symmetry of the normal distribution, there is no value y such that $P(Z=\sqrt{y}) = P(Z=-\sqrt{y})$, and so it must be that $P(Z=\sqrt{y}) = P(Z=-\sqrt{y}) = 0$?

Nope, the authors are saying so such thing.

It sounds like you're reading this as a proof by contradiction, and so you're wondering where the contradiction is. But it's not a proof by contradiction.

The authors are simply saying that as a matter of fact, $Z$ equals $\sqrt{y}$ or $-\sqrt{y}$ with equal probability $\frac12$, so the expected value can be calculated as $\frac12 (\sqrt{y}) + \frac12 (-\sqrt{y}) = 0$.