3
$\begingroup$

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?

$\endgroup$
  • 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
2
$\begingroup$

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 ...

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.