Answer:
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)<t(y)) is just a set of zero measure.
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).
= 6.