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Assume $X$ to be standard normal random variable, and define $Y$ as$$Y=\begin{cases}X,&\text{if }⌊X⌋\text{ is even}\\-X,&\text{if }⌊X⌋\text{ is odd}\end{cases}.$$

I am trying to show that $X$ and $Y$ are mutually completely dependent. For this I want to find the support of the copula of $(X,Y)$: $$ C(u,v)=\mathbb{P}( X \leq F^{-1}(u), Y \leq G^{-1}(v)),$$ and conclude by finding the probability mass concentrated on some locus. Here $F(x) = \Phi(x)$ is the standard normal distribution of $X$, and $G(y)$ is the distribution of $Y$.

However, I am not sure how to technically tackle the $Y$, that is, how to properly split it between cases odd/even integer values.

My approach:
I am adding an excerpt from another question of mine (which is linked) with my approach. If my work is correct, I worked out that $$G(y)=\frac{1}{2}[1 + F(y) - F(-y)].$$ Further,\begin{align*} C(u,v) &= \mathbb{P}(X \leq F^{-1}(u), Y \leq G^{-1}(v))\\ &= \frac{1}{2}\left[\mathbb{P}(X \leq F^{-1}(u), X > -G^{-1}(v)) + \mathbb{P}(X \leq F^{-1}(u), X \leq G^{-1}(v)) \right]\\ &=\frac{1}{2}\left[\mathbb{P}(X \leq \min\{F^{-1}(u), G^{-1}(v)\}) + \mathbb{P}(X \in [-G^{-1}(u), F^{-1}(v)]) \right]\\ &= \frac{1}{2}\left[ v - F(G^{-1}(v)) + F(\min\{F^{-1}(u), G^{-1}(v)\}) \right]. \tag{1} \end{align*}

It seems I might be able to untangle this as a function of $(u,v)$ if I find $F(G^{-1}(v))$. However, here I'm not too sure that the usual approach of finding the inverse works.

$$ y = G(G^{-1}(y)) =\frac{1}{2}[1 + F(G^{-1}(y)) - F(-G^{-1}(y))] $$ $$ 2y -1 = F(G^{-1}(y)) - F(-G^{-1}(y)) $$ $$ F^{-1}(2y -1) = F^{-1}\left( F(G^{-1}(y)) -F(-G^{-1}(y)) \right) \tag{2}$$

And it seems I am stuck here. I am thinking that it is enough to simplify $(1)$ to some convenient form, here my idea is that finding the expression for $G^{-1}(y)$ would help, but I got stuck at $(2)$.

Maybe another approach is better?

Would it be easier to show that $\mathbb{P}(X = g(Y))= 1$, where $g$ is some function of $Y$? But I still need to find the support of $C$ for following exercises.

Would appreciate any hints or suggestions on how to proceed!

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  • $\begingroup$ @AlexFrancisco This is true. If my work is correct, the distribution function of $Y$ should be $G(y)=\frac{1}{2}[1 + F(y) - F(-y)]$, where $F$ is standard normal distribution of $X$. $\endgroup$
    – runr
    Oct 18, 2018 at 13:16
  • $\begingroup$ @AlexFrancisco Thanks, I've edited the question, that was a typo from my side. $\endgroup$
    – runr
    Oct 18, 2018 at 13:22
  • $\begingroup$ @AlexFrancisco I'm following this definition: (Lancaster 1963), X and Y are mutually completely dependent if there exists a one-to-one function $\varphi$ such, that $\mathbb{P}(Y = \varphi(X))=1$. On the other hand, in the textbook it is shown that X and Y is mutually completely dependent, if the support of their copula is singular. That is, the support then is the graph of the one-to-one function. $\endgroup$
    – runr
    Oct 18, 2018 at 13:43
  • $\begingroup$ @AlexFrancisco I think that might be useful. Still, the next exercise requires to show that the $C(x,y)$ is not a shuffle of $min(x,y)$ copula (see here, i.e., Definition 1.3, or Nelsen), for which I think the expression of $C$ would be essential. Unless it is possible to conclude this without the expression of $C$? $\endgroup$
    – runr
    Oct 18, 2018 at 13:51
  • $\begingroup$ Why don't you simply prove that $$\Pr\{X<x,Y<y\}=\Pr\{X<x\}\cdot\Pr\{Y<y\}$$? $\endgroup$ Oct 22, 2018 at 9:21

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