If $Y|X \sim \mathcal{N}(0,1)$ then is $Y^2|X \sim \chi^2(1)$?

Suppose we have random variables $$X$$ and $$Y$$ such that $$Y|X \sim \mathcal{N}(0,1)$$. Can we then say that $$Y^2|X \sim \chi^2(1)$$?

If we can, then what about when $$Y|X \sim \mathcal{N}(0,\sigma^2/4)$$, can we use the results of this post to say that $$Y^2|X \sim \frac{\sigma^2}{4}\chi^2(1)$$?

Essentially I want to know if the results of squaring a normal distribution can be applied to the case where a random variable is conditionally normally distributed?

• I find $$Y|X \sim N(0,1)$$ a bit weird expression. It is confidential on $X$ but on the right side there is nothing that indictates how this conditional distribution depends on $X$. So effectively you just got $$Y \sim N(0,1)$$ – Sextus Empiricus Nov 27 '20 at 15:39

YES. You have given that $$Y \mid X=x \sim \mathcal{N}(0,1), ~~\text{for all x within the range of X.}$$ This implies that the random variables $$X$$ and $$Y$$ are independent, and the conclusion follows.
A more intuitive answer: You have given a distribution of $$Y$$ conditional on some $$X$$, but the condition is not used at all when stating the distribution. That means that the condition is irrelevant, and irrelevancies should be ignored$$^\dagger$$. So just ignore it, and the result is obvious.
$$^\dagger$$That might be easier in math than in real life ...
• What if $X$ is binary or otherwise not defined on all of $\mathbb{R}?$ – Dave Nov 17 '20 at 16:19
• @Dave I'll repeat my question to you, then: why would the distribution of $X$ have any relevance to this question? – whuber Nov 17 '20 at 16:21
• Re the edit: why do you even need to introduce a value "$x$"?? – whuber Nov 17 '20 at 16:23