If s is an increasing function, and t is a decreasing function, find Cs (X), t (Y) in terms of CX, Y.

C (X, Y) (a, b) = 1-C (s (X), t (Y)) (a, 1-b).

Step-by-step explanation:

Let's introduce the cumulative distribution of (X, Y), X and Y:

F (X, Y) (x, y) = P (X≤x, Y≤y)

FX (x) = P (X≤x) FY (y) = P (Y≤y).

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

F (s (X), t (Y)) (u, v) = P (s (X) ≤u

t (Y) ≤v) Fs (X) (u) = P (s (X) ≤u) Ft (Y) (v) = P (t (Y) ≤v).

Now, First establish the relationship between F (X, Y) and F (s (X), t (Y)):

F (X, Y) (x, y) = P (X≤x, Y≤y) = P (s (X) ≤s (x), t (Y) ≥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-P (s (X) ≤s (x), t (Y) ≤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) ≤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-1X (a), F-1Y (b))

Thus, plugging x=F-1X (a) and y=F-1Y (b) into our previous formula,

we get

F (X, Y) (F-1X (a), F-1Y (b)) = 1-F (s (X), t (Y)) (s (F-1X (a)), t (F-1Y (b)))

The left hand side is the copula C (X, Y), the right hand side still needs some work.

Note that

Fs (X) (s (F-1X (a))) = P (s (X) ≤s (F-1X (a))) = P (X≤F-1X (a)) = FX (F-1X (a)) = a

and likewise

Ft (Y) (s (F-1Y (b))) = P (t (Y) ≤t (F-1Y (b))) = P (Y≥F-1Y (b)) = 1-FY (F-1Y (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).