Does Sklar's Theorem work when $P(X_k \leq t)$ are not continuous?

Question

Wikipedia here describes for $X = (X_1,...,X_n): \Omega \rightarrow \mathbb{R}^n$ with each $F_k(t) = P(X_k \leq t)$ continuous, there is a pushforward measure induced on $[0,1]^n$ via $(F_1(X_1),...,F_n(X_n))$. Using this measure, one finds $Y(u_1,...,u_n)=(F_1^{-1}(u_1),...,F_n^{-1}(u_n)), \ \forall (u_1,...,u_n) \in [0,1]^n$, has the same distribution as $X$.

When Sklar's theorem is described, it appears that it is no longer assuming each marginal $F_k$ is continuous. This makes me wonder why they assumed it in the first place.

Does Sklar's theorem or a related theorem imply for $(X_1,...,X_n)$ (where the $F_k$ are not continuous) there exists a nice measure $\mu$ on $[0,1]^n$ and $Y: [0,1]^n \rightarrow \mathbb{R}^n$, $Y(s_1,...,s_n) = (Y_1(s_1),....,Y_n(s_n))$, with the same distribution as $X$?

Answer 1

Yes, Sklar's theorem holds for all distributions.

For every multivariate cumulative distribution function (cdf), there exits a copula with cdf $C: [0,1]^n \rightarrow [0,1]$ such that:

$$F(x_1,...,x_n)=C \left (F_{X_1}(x_1),..., F_{X_n}(x_n) \right).$$

A copula is any multivariate distribution whose univariate margins are uniform distributions on $[0,1]$.

When, the margins have discrete distributions, it is not easy to find the underlying copula (while there is one); see Section 2 of this paper. However, you can easily use any copula to construct a multivariate distribution with arbitrary univariate margins.

Answer 2

Copulas exist whether the marginals of $\mathbf{X}=[X_1,\ldots,X_n]^\intercal$ are continuous or not. The issue is that for noncontinuous the copula is not unique in general, and different choices may introduce artificial local dependence between the components of a random vector.

Constructions of copulas can be obtained by an extension of the "quantile transform" as follows:

• Let $(\Omega,\mathscr{F},\mathbb{P})$ be a probability space,
• Suppose $X$ is a real-valued random on $\Omega$, and let $F(x):=\mathbb{P}[X\leq x]$.
• The distributional transform of $X$ is the function $G:\mathbb{R}\times[0,1]\rightarrow\mathbb{R}$ defined as $$G(x,v)=F(x-)+v(F(x)-F(x-))$$
• Let $V\sim\operatorname{Unif}(0,1)$ independent from $X$ and define the random variable $$U:=G(X,V).$$ It can be shown that

1. $U\sim\operatorname{Unif}(0,1)$, and
2. that $X=Q(U)$ $\mathbb{P}$-a.s., where $Q$ is the quantile transform of $X$, that is $Q(t)=\inf\{x\in\mathbb{R}: F(x)\geq t\}$.
3. Recall that $F(x)\leq t$ iff $Q(t)\leq x$.
4. From (3) it follows that if $F$ is continuous at $Q(t)$, then $F(Q(t))=t$. Hence, when $F$ is contuinuous, then $U=F(X)\sim\operatorname{Unif}(0,1)$ which is a well known fact for continuous random variables.

Theorem: Suppose $\mathbf{X}=[X_1,\ldots ,X_n]^\intercal$ is an $\mathbb{R}^n$-valued random variable with distribution $F(x_1,\ldots,x_n)=\mathbb{P}[X_1\leq x_1,]\ldots,X_n\leq x_n]$. Denote by $F_j(x)=\mathbb{P}[X_j\leq x]$ the marginal distribution of the $j$-th component of $\mathbf{X}$. There exists a measure $\nu$ on $([0,1]^n,\mathscr{B}([0,1]^n))$ with $U(0,1)$-marginals such that
\begin{align} F(x_1,\ldots,x_n)=C(F_1(x_1),\ldots, F_n(x_n))\label{copula1} \end{align} where $C(u_1,\ldots,u_n)=\nu([0,u_1]\times\ldots\times[0,u_n])$. The function $C$ is said to be a copula for $\mathbf{X}$ (or for the distribution $F$).

Proof: Here is a short proof based on the extended quantile transform introduced above. Let $V\sim U(0,1)$ independent from $\mathbf{X}=[X_1,\ldots, X_n]$, and for each $1\leq j\leq n$ define $$U_j=F_j(X_j-)+\big(F_j(X_j)-F_j(X_j-)\big)V$$ where $F_j$ is the distribution function of $X_j$. Let $Q_j$ be the quantile function associated to $X_j$. Observations (1) and (2) above imply that $U_j\sim U(0,1)$ and that $Q_j(U_j)=X_j$ a.s. Let $C$ be the joint distribution of $\mathbf{U}=[U_1,\ldots, U_n]^\intercal$, that is $$ C(u_1,\ldots,u_n)=\mathbb{P}[U_1\leq u_1,\ldots, U_n\leq u_n). $$ It follows that \begin{align} \mathbb{P}[X_j\leq x_j,1\leq j\leq n]&=\mathbb{P}[Q_j(U_j)\leq x_j,1\leq j\leq n]\\ &=\mathbb{P}[U_j\leq F(x_j), \,1\leq j\leq n]=C(F(x_1),\ldots, F(x_n)) \end{align}

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Comments:

• If each marginal $X_i$ is continuous, then $C$ is unique as $C$ is the distribution of $[F_1(X_1),\ldots, F_n(X_n)]^\intercal$.

• If $\mathbf{X}$ is not continuous, different choices of the randomizations V at the jumps or in the components, i.e. choosing $U_i = G_i(X_i , V_i)$ give different copulas.

• In this case, from a given a choice of a copula is not possible to see whether some local positive or negative dependence is a real one or just comes from the particular choice of the copula.

• The distributional transform function $G(x,v)$ goes back to Ferguson, T. S., Mathematical Statistics: A Decision Theoretic Approach. Probability and Mathematical Statistics, Vol. 1. New York: Academic Press, 1967.

• The idea to use the distributional transform $G$ to prove Sklar's theorem appears also in Rüschendorf, L.,On the distributional transform, Sklar’s Theorem, and the empirical copula process., Journal of Statistical Planning and Inference, Volume 139, Issue 11, 1 November 2009, Pages 3921-3927