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?

  • 1
    $\begingroup$ 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 $$Y \sim N(0,1)$$ $\endgroup$ – 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 ...

  • $\begingroup$ Why would independence be relevant, since the question asks only about conditional distributions? $\endgroup$ – whuber Nov 17 '20 at 16:19
  • $\begingroup$ What if $X$ is binary or otherwise not defined on all of $\mathbb{R}?$ $\endgroup$ – Dave Nov 17 '20 at 16:19
  • $\begingroup$ @Dave: I wil formulate more precisely! Thanks $\endgroup$ – kjetil b halvorsen Nov 17 '20 at 16:21
  • $\begingroup$ @Dave I'll repeat my question to you, then: why would the distribution of $X$ have any relevance to this question? $\endgroup$ – whuber Nov 17 '20 at 16:21
  • $\begingroup$ Re the edit: why do you even need to introduce a value "$x$"?? $\endgroup$ – whuber Nov 17 '20 at 16:23

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