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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?

kjetil b halvorsen
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Bertus101
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    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

1 Answers1

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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 ...

kjetil b halvorsen
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  • Why would independence be relevant, since the question asks only about conditional distributions? – whuber Nov 17 '20 at 16:19
  • What if $X$ is binary or otherwise not defined on all of $\mathbb{R}?$ – Dave Nov 17 '20 at 16:19
  • @Dave: I wil formulate more precisely! Thanks – kjetil b halvorsen Nov 17 '20 at 16:21
  • @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
  • @whuber: The distribution of $X$ is irrelevant, independence is relevant, since it gives the conclusion ... – kjetil b halvorsen Nov 17 '20 at 16:23
  • Independence is completely irrelevant. After all, consider the Bayesian point of view: *all distributions are conditional.* From this perspective the question is trivial: the square of a standard Normal variate has a chi-square distribution, period: there's nothing more to say or show. – whuber Nov 17 '20 at 16:25
  • @whuber: (I wrote this answer because I thought the existing answer was unneccesarily unclear. Introducing $x$ is only for clarity ... since otherwise the notation $Y\mid X$, clear enough for us, isn't necessarily so for everybody. – kjetil b halvorsen Nov 17 '20 at 16:27
  • I understood the motivation for your answer ;-). In this case I would suggest letting the OP's notation guide you. As a general principle, unless the OP's notation is confusing, the answer will be clearer for everyone when it uses the same notation. – whuber Nov 17 '20 at 16:29
  • @whuber: Yes, for us there is no need to show anything. But it was apparently necessary for the OP ... so this is more a question of how to explain the obvious to somebody for which it is not obvious---Do you have a better way? – kjetil b halvorsen Nov 17 '20 at 16:30
  • IMHO, a good approach has two parts. The first is to identify the key concept(s) that need elucidation. The second is to find a way to make the result obvious. Neither part is necessarily easy to achieve! In this case, my first concern is that independence is not the crux of the issue. The key here is that the question concerns *distributions.* Since that's the case, we needn't worry about whether the distribution is conditional or not. At that point, I think obviousness is assured and little more needs to be said. – whuber Nov 17 '20 at 16:36