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

\begin{equation} f(x,y)=\int_{z}f(x,y,z)dz \end{equation}

Furthermore, we know From Sklar's Theorem that

\begin{equation} f(x,y,z)=f(x)f(y)f(z)c(F(x),F(y),F(z)) \end{equation}

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

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

\begin{equation} f(x,y)=\int_{z}f(x)f(y)c(F(x),F(y),F(z))dF(z) \end{equation}

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

Thanks in advance.


1 Answer 1


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 $f_{XY}.$

  • 1
    $\begingroup$ This is an incredibly complete and useful response. I thank you greatly for sparing your time. $\endgroup$
    – Carl
    Aug 24, 2020 at 21:39

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