What is the copula of a variable with itself? In Sklar's theorem for joint probability functions,
$$f(x,y) = c(F_X(x), F_Y(y)) \cdot f(x) f(y)$$
the copula is $c(\cdot)$ of variables $X$ and $Y$, while $f(\cdot)$ are their marginal distributions.
How would this change if we, for whatever reason, try to get variable $X$'s joint distribution (and therefore its copula) with itself? Does the below formula have any meaning, significance? can it be reduced further into something else?
$$f(x,x) = c(F_X(x), F_X(x)) \cdot f(x) f(x)$$
 A: I see two issues with your question:

*

*You need to define more rigorously what you mean by "$X$'s joint distribution with itself"

*I think it is confusing or maybe even inappropriate to use densities in this context.

Definition
"$X$'s joint distribution with itself" can (at least) mean two different things:

*

*The random variable $(X,X):\Omega\rightarrow \mathbb{R}^2$ defined by $(X,X)(\omega)=(X(\omega), X(\omega))$

*A random variable $(X_1,X_2)$ defined on some probability space such that the distributions of $X_1, X_2$ and $X$ are the same.

In the first case the copula of $(X,X)$ is the comonotonic copula (see definition).
In the second case the definition only specifies the margins of the joint distribution and there is no restriction on the copula at all.
Since your question is void in the second case I will discuss the first case.
Joint distribution of $(X,X)$
Start by looking at the joint distribution of $(X,X)$:
$$ P((X,X) \leq (x_1, x_2)) = P(X\leq x_1 \text{ and } X\leq x_2) = 
\begin{cases}
P(X\leq x_1) \text{ if } x_1\leq x_2\\
P(X\leq x_2) \text{ if } x_2\leq x_1
\end{cases}$$
or in other words
$$  P((X,X) \leq (x_1, x_2)) = \min_{i=1,2} P(X\leq x_i).$$
Copula
Recall that a copula is simply a measure or CDF on $[0,1]^2$ with uniform margins and that the copula for $(X,X)$ is:
$$ C(u,v) = P((X,X)\leq (F^{-1}(u), F^{-1}(v))) = \min \{u,v\}.$$
Interpreting $C$ as a measure on $[0,1]^2$ and using the additivity of measures it is easy to see (draw a picture and apply the min property of $C$ above) that $C$ is concentrated on the diagonal of $[0,1]^2.$ This is why I think it is ill advised to use densities. Since $C$ is degenerate it does not have a proper density on $[0,1]^2.$
Interpretation of $c(F(x), F(x)) = \frac{1}{f(x)}$
This is just the fact that $C$ has uniform margins. If you restrict $C$ to the diagonal and using its definition you see that its density is the density of the uniform distribution, i.e. the constant 1. From this you get the result in the headline by differentiation of $ u = C(u,u)$ and the chain rule:
$$ 1 = \frac{d}{du}u = \frac{d}{du} C(u,u) = \frac{d}{dx} C(F(x), F(x)) = c(F(x), F(x)) \frac{d}{dx} F(x) = c(F(x), F(x)) f(x).$$
A: Let us refine the first formula to
$$
f(x,y) = c(F_X(x), F_Y(y)) \cdot f_X(x) f_Y(y)
$$
to show exactly what the PDFs/PMFs are. If we consider $X = Y$, we have $F_X=F_Y$ and $f_X=f_Y$. Then we obtain (instead of the second formula)
$$
f(x,y) = c(F_X(x), F_X(y)) \cdot f_X(x) f_X(y).
$$
Here are a couple of things we can say about $f(x,y)$:

*

*For $x \neq y$, $f(x,y) = 0$.

*For $x = y$, $f(x,y) = f(x,x) = f(x)$. Thus $f(x) = c(F_X(x), F_X(x)) \cdot f_X(x) f_X(x)$.
This yields
$$
c(F_X(x), F_X(x)) = \frac{1}{f_X(x)}
$$ when $f_X(x) \neq 0$. When $f_X(x) = 0$, $c(F_X(x), F_X(x))$ does not seem to be restricted anyhow.

