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A question on copula: If $s: \Bbb R\to \Bbb R$ is an increasing function, and $t: \Bbb R\to \Bbb R$ is a decreasing function, find the copula $C_{s(X),t(Y)}$ of $(s(X), t(Y))$ in terms of $C_{X,Y}$. (Assume $X$ and $Y$ to be continuous random variables)

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    what are your thoughts, have you tried anything yet? – 2017-01-10
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    I think if $s$ and $t$ are increasing functions, then $C_{s(X),t (Y )}(x, y) = C_{X,Y} (x, y)$. I am confused about the case where $t$ is a decreasing function. – 2017-01-10
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    this is only true if $X$ and $Y$ are continuous random variables. – 2017-01-10
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    Yes, X and Y are assumed to be continuous random variables. – 2017-01-10
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    Then put this in the question please and all other information that might be missing. It's not easy to guess the right question and then give an answer to it. Also put in ideas that you have or calculations that you already made. – 2017-01-10

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Let's introduce some notation, we have the cumulative distribution of $(X,Y)$, $X$ and $Y$ :

$$F_{(X,Y)}(x,y) = \mathbb{P}(X\leq x, Y \leq y) \; , \;\; F_X(x)=\mathbb{P}(X\leq x) \; \mbox{ and } \; F_Y(y)=\mathbb{P}(Y\leq y) \; .$$

Likewise for $(s(X),t(Y))$, $s(X)$ and $t(Y)$ :

$$F_{(s(X),t(Y))}(u,v) = \mathbb{P}(s(X)\leq u, t(Y) \leq v) \; , \;\; F_{s(X)}(u)=\mathbb{P}(s(X)\leq u) \; \mbox{ and } \\ \; F_{t(Y)}(v)=\mathbb{P}(t(Y)\leq v) \; .$$

Now, we first establish the relationship between $F_{(X,Y)}$ and $F_{(s(X),t(Y))}$ :

$$F_{(X,Y)}(x,y) = \mathbb{P}(X\leq x, Y \leq y) = \mathbb{P}(s(X)\leq s(x), t(Y) \geq t(y))$$

The last step is obtained by applying the functions $s$ and $t$ since $s$ preserves order and $t$ reverses it. This can be further transformed into

$$F_{(X,Y)}(x,y) = 1 - \mathbb{P}(s(X)\leq s(x), t(Y) \leq t(y)) = 1 - F_{(s(X),t(Y))}(s(x),t(y))$$

Since our random variables are continuous, we assume that the difference between $t(Y) \leq t(y)$ and $t(Y)

Now, to transform this into a statement about copulas, note that

$$C_{(X,Y)}(a,b) = F_{(X,Y)}(F_X^{-1}(a),F_Y^{-1}(b))$$

Thus, plugging $x=F_X^{-1}(a)$ and $y=F_Y^{-1}(b)$ into our previous formula, we get

$$F_{(X,Y)}(F_X^{-1}(a),F_Y^{-1}(b)) = 1 - F_{(s(X),t(Y))}(s(F_X^{-1}(a)),t(F_Y^{-1}(b)))$$

The left hand side is the copula $C_{(X,Y)}$, the right hand side still needs some work. Note that

$$F_{s(X)}(s(F_X^{-1}(a))) = \mathbb{P}(s(X)\leq s(F_X^{-1}(a))) = \mathbb{P}(X\leq F_X^{-1}(a)) = F_X(F_X^{-1}(a)) = a$$

and likewise

$$F_{t(Y)}(s(F_Y^{-1}(b))) = \mathbb{P}(t(Y)\leq t(F_Y^{-1}(b))) = \mathbb{P}(Y\geq F_Y^{-1}(b)) = 1-F_Y(F_Y^{-1}(b)) = 1-b$$

Combining all results we obtain for the relationship between the copulas

$$C_{(X,Y)}(a,b) = 1-C_{(s(X),t(Y))}(a,1-b) \; .$$