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Suppose we have a random variable $X$ and its distribution given by $f_X(x)$ and $Y = X^2$. Say $X = 1/2$ then is $f_{X,Y}(1/2,y) = f_{X,Y}(1/2, 1/4)$? If not, then how does it work?

StubbornAtom
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Ray
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    It doesn't work, because $(X,Y)$ is not absolutely continuous with respect to the Lebesgue measure on $\Bbb R^2$. So no $f_{X,Y}$. –  Jan 21 '20 at 01:40
  • I haven't taken measure theory, this is from an intro to statistics class. One of the questions I was given was: given X, $f_X(x)$ and $Y = X^2$, is $X$ and $Y$ independent? (Of course $f_X(x)$ was explicitly given). How would I do this? – Ray Jan 21 '20 at 01:45
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    @RayOfHope They do not have a joint density, so you cannot use that. Do you think they are independent or not, because that will determine how you tackle the problem. – kccu Jan 21 '20 at 01:47
  • @kccu I don't know, I don't really get how $Y = g(X)$ works exactly. For reference, the function $f_X(x)$ was $1+x$ when $-1 \le x \le 0$ and $1-x$ when $0 \le x \le 1$, $0$ elsewhere – Ray Jan 21 '20 at 01:51
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    Think about it intuitively. Random variables are independent if knowing the value of one doesn't influence the value of the other. If you have $Y=X^2$, then does knowing the value of one influence the value of the other? – kccu Jan 21 '20 at 01:57
  • @kccu Yeah, I think so. But how would I show this if there is no joint? – Ray Jan 21 '20 at 02:08
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    You'd have to show that some property that must hold for random variables fails for $X$ and $Y$. What properties do you know must hold for independent random variables that don't have to do with their joint density? – kccu Jan 21 '20 at 02:42
  • The position is worse than $X$ and $Y$ not being independent. Knowing $X=x$ tells you $Y=x^2$ with no uncertainty. So in the same way that a constant random variable does not have a density, $Y$ does not have a conditional density given $X=x$ and that means there is no joint density for $X$ and $Y$ – Henry Jan 21 '20 at 10:37

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