7
$\begingroup$

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.

$\endgroup$
0

1 Answer 1

9
$\begingroup$

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 $f_{XY}.$

$\endgroup$
1
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.