Expressing a marginal probability using copulas

Please correct me if I am wrong and kindly provide me with the correct notations. I have two questions:

We know that for the variables $$(X,Y,Z)\in \mathbb{R}^3$$, the marginal joint density $$f(x,y)$$ can be expressed as

$$$$f(x,y)=\int_{z}f(x,y,z)dz$$$$

Furthermore, we know From Sklar's Theorem that

$$$$f(x,y,z)=f(x)f(y)f(z)c(F(x),F(y),F(z))$$$$

Q1: So would it be correct to express $$f(x,y)$$ as follows

$$$$f(x,y)=\int_{z}f(x)f(y)f(z)c(F(x),F(y),F(z))dz$$$$ and since $$f(z)=dF(z)/dz$$ (assuming $$F(z)$$ is differentiable)

$$$$f(x,y)=\int_{z}f(x)f(y)c(F(x),F(y),F(z))dF(z)$$$$

Q2: If yes, how can one go on about calculating the above integral.

The issue of notation seems crucial. I propose, therefore, to disambiguate the ubiquitous and overloaded "$$f$$" by means of subscripts. Thus, $$f_{XYZ}$$ will be the full density function and (therefore) the marginal density for $$(X,Y)$$ is

$$f_{XY}(x,y) = \int_{-\infty}^{\infty} f_{XYZ}(x,y,z)\,\mathrm{d}z.$$

If, for a sufficiently smooth version of $$f_{XYZ}$$ and real numbers $$(x,y,z)$$ you define a function $$c$$ on $$[0,1]^3$$ as

c\left(F_X(x),F_Y(y),F_Z(z)\right) = \left\{\begin{aligned}\frac{f_{XYZ}(x,y,z)}{f_X(x)f_Y(y)f_Z(z)} & & \text{if } f_X(x)f_Y(y)f_Z(z)\ne 0 \\ 0 && \text{otherwise,}\end{aligned}\right.

then indeed you may substitute this into the first expression for $$f_{XY}$$ to obtain

$$f_{XY}(x,y) = \int_{-\infty}^{\infty} f_X(x)f_Y(y)f_Z(z) c(F_X(x),F_Y(y),F_Z(z))\,\mathrm{d}z$$

and, because $$\mathrm{d}F_Z(z) = f_Z(z)\,\mathrm{d}z$$ by definition, substituting that into the foregoing does give

$$f_{XY}(x,y) = \int_{-\infty}^{\infty} f_X(x)f_Y(y)c(F_X(x),F_Y(y),F_Z(z))\,\mathrm{d}F_Z(z).$$

Concerning the calculation of such integrals, it comes down to what information you have and what form it's in; this is an unanswerable question in such generality.

Note that this $$c$$ is not the copula for $$f_{XYZ}.$$ The copula $$C$$ is given by

\begin{aligned} C(F_X(x),F_Y(y),F_Z(z)) &= \Pr(X\le x,\,Y\le y,\,Z \le z) \\ &= F_{XYZ}(x,y,z) \\ &= \int^x\int^y\int^z f_{XYZ}(x,y,z)\,\mathrm{d}z\mathrm{d}y\mathrm{d}x. \end{aligned}

Using a standard notation in literature on copulas,

$$DC(u,v,w) = \frac{\partial^3C(u,v,w)}{\partial u\partial v \partial w}$$

for $$(u,v,w)\in[0,1]^3.$$ Applying the Chain Rule (three times) we may relate that to the foregoing via

\begin{aligned} f_{XYZ}(x,y,z) &= \frac{\partial^3C(F_X(x),F_Y(y),F_Z(z))}{\partial x\partial y \partial z} \\ &= DC(F_X(x),F_Y(y),F_Z(z))f_X(x)f_Y(y)f_X(z), \end{aligned}

revealing $$c$$ as

$$c(u,v,w) = (DC)(u,v,w).$$

A simple example to contrast $$c$$ and $$C$$ is the case of independence of the variables $$(X,Y,Z),$$ for which $$C(u,v,w)=uvw$$ (the "independence copula") and $$c(u,v,w)=DC(u,v,w)=1.$$

Finally, to address the question in the title, a simple expression for the marginal probability in terms of the copula is

$$F_{XY}(x,y) = \Pr(X\le x,\,Y\le y) = \lim_{z\to\infty}\Pr(X\le x,Y\le y,Z\le z) = C(F(x),F(y),1).$$

Differentiate this with respect to $$(x,y)$$ to obtain the marginal density $$f_{XY}.$$

• This is an incredibly complete and useful response. I thank you greatly for sparing your time.
– Carl
Aug 24, 2020 at 21:39