[proofplan] We first construct a copula for a given joint distribution by passing from each coordinate $X_j$ to a uniform random variable through the randomized distributional transform. The generalized inverse then recovers $X_j$ from this uniform variable almost surely, which gives the [representation formula](/theorems/39). Uniqueness follows because a continuous distribution function has range dense enough, together with right-continuity of copulas, to determine the copula on all of $[0,1]^p$. For the converse, we verify directly that composing a copula with univariate distribution functions preserves monotonicity, right-continuity, boundary limits, and nonnegative rectangle probabilities, and then compute the margins using the uniform marginal property of the copula. [/proofplan] [step:Construct uniform coordinates by the randomized distributional transform] Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space carrying a random vector \begin{align*} X: \Omega &\to \mathbb{R}^p,\\ \omega &\mapsto (X_1(\omega),\dots,X_p(\omega)), \end{align*} whose joint distribution function is $F$. For each $j \in \{1,\dots,p\}$, let \begin{align*} F_j(t-) := \lim_{s \uparrow t} F_j(s) \end{align*} denote the left limit of the marginal distribution function $F_j$ at $t \in \mathbb{R}$. Enlarge the probability space, if necessary, so that it also carries random variables \begin{align*} V_j: \Omega \to [0,1], \qquad j \in \{1,\dots,p\}, \end{align*} which are independent, each uniformly distributed on $[0,1]$, and independent of $X$. Define \begin{align*} U_j: \Omega &\to [0,1],\\ \omega &\mapsto F_j(X_j(\omega)-)+V_j(\omega)\bigl(F_j(X_j(\omega))-F_j(X_j(\omega)-)\bigr). \end{align*} Then each $U_j$ is uniformly distributed on $[0,1]$. [guided] The purpose of the variables $U_j$ is to replace each marginal coordinate by a uniform coordinate without assuming that the marginal distribution function is continuous. If $F_j$ were continuous, the simpler transform $F_j(X_j)$ would already be uniform. When $F_j$ has jumps, the expression \begin{align*} F_j(X_j-)+V_j\bigl(F_j(X_j)-F_j(X_j-)\bigr) \end{align*} spreads the mass at each atom uniformly across the jump interval. We verify the uniformity. Fix $j \in \{1,\dots,p\}$ and $u \in [0,1]$. Define the generalized inverse \begin{align*} Q_j: (0,1) &\to \mathbb{R},\\ r &\mapsto \inf\{t \in \mathbb{R}: F_j(t) \ge r\}. \end{align*} For $u=0$ and $u=1$, the identity $\mathbb{P}(U_j \le u)=u$ follows from the definition and from the fact that $V_j$ has no atoms at $0$ or $1$. Now assume $0<u<1$ and set $a:=Q_j(u)$. By definition of $Q_j$, we have \begin{align*} F_j(a-) \le u \le F_j(a). \end{align*} The event $\{U_j \le u\}$ is split according to whether $X_j<a$, $X_j=a$, or $X_j>a$. If $X_j<a$, then $F_j(X_j)\le F_j(a-)\le u$, so $U_j\le u$. If $X_j>a$, then $F_j(X_j-)\ge F_j(a)\ge u$, and the event $U_j\le u$ has probability zero except at the endpoint case, which is harmless because $V_j$ has no atoms. On the atom $\{X_j=a\}$, conditional on $X_j=a$, the random variable $U_j$ is uniformly distributed on the interval $[F_j(a-),F_j(a)]$. Therefore \begin{align*} \mathbb{P}(U_j \le u) &= \mathbb{P}(X_j<a) + \mathbb{P}(X_j=a)\frac{u-F_j(a-)}{F_j(a)-F_j(a-)}\\ &= F_j(a-) + \bigl(F_j(a)-F_j(a-)\bigr) \frac{u-F_j(a-)}{F_j(a)-F_j(a-)}\\ &= u, \end{align*} with the second term interpreted as $0$ when $F_j(a)=F_j(a-)$. Hence $U_j$ is uniformly distributed on $[0,1]$. [/guided] [/step] [step:Recover the original coordinates from generalized inverses] For each $j \in \{1,\dots,p\}$, define the generalized inverse \begin{align*} Q_j: (0,1) &\to \mathbb{R},\\ u &\mapsto \inf\{t \in \mathbb{R}: F_j(t) \ge u\}. \end{align*} Then \begin{align*} Q_j(U_j)=X_j \end{align*} almost surely. Indeed, if $X_j(\omega)=x$, then \begin{align*} F_j(x-) \le U_j(\omega) \le F_j(x), \end{align*} and, except on the null event where $V_j(\omega)$ is an endpoint value inside a jump interval, this places $U_j(\omega)$ strictly inside the jump interval corresponding to $x$ whenever $x$ is an atom. Hence the defining property of $Q_j$ gives $Q_j(U_j(\omega))=x$ almost surely. Moreover, for every $x_j \in \mathbb{R}$ and every $u \in (0,1)$, \begin{align*} Q_j(u)\le x_j \quad \Longleftrightarrow \quad u \le F_j(x_j). \end{align*} [guided] The generalized inverse is the device that turns the uniform variable back into a variable with distribution function $F_j$. Its basic order property is \begin{align*} Q_j(u)\le x_j \quad \Longleftrightarrow \quad u \le F_j(x_j). \end{align*} To prove it, first suppose $Q_j(u)\le x_j$. By the definition of $Q_j(u)$ as an infimum, choose a decreasing sequence $(t_n)_{n=1}^{\infty}$ with $t_n \downarrow Q_j(u)$ and $F_j(t_n)\ge u$ for every $n$. Since $F_j$ is right-continuous, \begin{align*} F_j(Q_j(u))=\lim_{n\to\infty}F_j(t_n)\ge u. \end{align*} Because $F_j$ is nondecreasing and $Q_j(u)\le x_j$, it follows that $u\le F_j(x_j)$. Conversely, if $u\le F_j(x_j)$, then $x_j$ belongs to the set \begin{align*} \{t \in \mathbb{R}: F_j(t)\ge u\}, \end{align*} so the infimum of that set satisfies $Q_j(u)\le x_j$. Now apply this to $u=U_j(\omega)$. By construction, \begin{align*} F_j(X_j(\omega)-)\le U_j(\omega)\le F_j(X_j(\omega)). \end{align*} On non-atomic values of $X_j$, the interval collapses and the inverse returns $X_j(\omega)$. On atomic values, the auxiliary uniform variable $V_j$ chooses a point inside the jump interval, and endpoint choices occur with probability zero. Thus \begin{align*} Q_j(U_j)=X_j \end{align*} almost surely. [/guided] [/step] [step:Define the copula and prove the representation formula] Define \begin{align*} C: [0,1]^p &\to [0,1],\\ (u_1,\dots,u_p) &\mapsto \mathbb{P}(U_1\le u_1,\dots,U_p\le u_p). \end{align*} Since $(U_1,\dots,U_p)$ is a random vector with values in $[0,1]^p$ and each $U_j$ is uniformly distributed on $[0,1]$, the function $C$ is a $p$-copula. For $x=(x_1,\dots,x_p)\in\mathbb{R}^p$, using $X_j=Q_j(U_j)$ almost surely and the equivalence $Q_j(u)\le x_j \Longleftrightarrow u\le F_j(x_j)$, we obtain \begin{align*} F(x_1,\dots,x_p) &= \mathbb{P}(X_1\le x_1,\dots,X_p\le x_p)\\ &= \mathbb{P}(Q_1(U_1)\le x_1,\dots,Q_p(U_p)\le x_p)\\ &= \mathbb{P}(U_1\le F_1(x_1),\dots,U_p\le F_p(x_p))\\ &= C(F_1(x_1),\dots,F_p(x_p)). \end{align*} This proves existence. [/step] [step:Prove uniqueness when the marginal distribution functions are continuous] Assume now that $F_1,\dots,F_p$ are continuous. Let $C$ and $\widetilde{C}$ be $p$-copulas satisfying \begin{align*} F(x_1,\dots,x_p) = C(F_1(x_1),\dots,F_p(x_p)) = \widetilde{C}(F_1(x_1),\dots,F_p(x_p)) \end{align*} for every $x=(x_1,\dots,x_p)\in\mathbb{R}^p$. For each $j$, continuity of $F_j$, together with \begin{align*} \lim_{t\to -\infty}F_j(t)=0, \qquad \lim_{t\to \infty}F_j(t)=1, \end{align*} implies that the range $F_j(\mathbb{R})$ is dense in $[0,1]$. Therefore the product set \begin{align*} F_1(\mathbb{R})\times \cdots \times F_p(\mathbb{R}) \end{align*} is dense in $[0,1]^p$. The preceding identity gives $C=\widetilde{C}$ on this dense product set. Since copulas are distribution functions on $[0,1]^p$, they are right-continuous in each coordinate; hence equality on a [dense subset](/page/Dense%20Subset) implies equality at every point of $[0,1]^p$. Thus $C=\widetilde{C}$. [guided] To prove uniqueness, we must show that the representation formula determines $C(u_1,\dots,u_p)$ for every $u\in[0,1]^p$. The formula directly determines $C$ only at points of the form \begin{align*} (F_1(x_1),\dots,F_p(x_p)). \end{align*} So the key question is: are these points dense in the whole cube? For each $j$, the distribution function $F_j$ is continuous, nondecreasing, and satisfies \begin{align*} \lim_{t\to -\infty}F_j(t)=0, \qquad \lim_{t\to \infty}F_j(t)=1. \end{align*} Fix $u_j\in(0,1)$ and $\varepsilon>0$. Choose $a,b\in\mathbb{R}$ such that \begin{align*} F_j(a)<u_j< F_j(b). \end{align*} By the intermediate value property for the continuous function $F_j$, there exists $x_j\in[a,b]$ with $F_j(x_j)=u_j$. Endpoint values $0$ and $1$ are obtained as limits as $x_j\to-\infty$ and $x_j\to\infty$. Thus $F_j(\mathbb{R})$ is dense in $[0,1]$. Now let $C$ and $\widetilde C$ be two copulas giving the same representation of $F$. For every $x\in\mathbb{R}^p$, \begin{align*} C(F_1(x_1),\dots,F_p(x_p)) = \widetilde C(F_1(x_1),\dots,F_p(x_p)). \end{align*} Hence $C$ and $\widetilde C$ agree on the dense product \begin{align*} F_1(\mathbb{R})\times\cdots\times F_p(\mathbb{R}). \end{align*} Copulas are joint distribution functions, so they are right-continuous in each coordinate. Given any $u\in[0,1]^p$, choose a sequence $(u_k)_{k=1}^{\infty}$ from the dense product with $u_k\downarrow u$ coordinatewise. Right-continuity gives \begin{align*} C(u)=\lim_{k\to\infty}C(u_k) =\lim_{k\to\infty}\widetilde C(u_k) =\widetilde C(u). \end{align*} Therefore $C=\widetilde C$ on $[0,1]^p$. [/guided] [/step] [step:Verify that composing a copula with marginal distribution functions gives a joint distribution function] Conversely, let $C:[0,1]^p\to[0,1]$ be a $p$-copula and let $F_1,\dots,F_p$ be univariate distribution functions. Define \begin{align*} G: \mathbb{R}^p &\to [0,1],\\ (x_1,\dots,x_p) &\mapsto C(F_1(x_1),\dots,F_p(x_p)). \end{align*} Since each $F_j$ is nondecreasing and $C$ is nondecreasing in each coordinate, $G$ is nondecreasing in each coordinate. Since each $F_j$ is right-continuous and $C$ is right-continuous, $G$ is right-continuous. Let $a_j,b_j\in\mathbb{R}$ satisfy $a_j<b_j$ for every $j$. The $G$-mass of the rectangle \begin{align*} (a_1,b_1]\times\cdots\times(a_p,b_p] \end{align*} is \begin{align*} \sum_{\varepsilon\in\{0,1\}^p} (-1)^{p-\varepsilon_1-\cdots-\varepsilon_p} G(y_{\varepsilon,1},\dots,y_{\varepsilon,p}), \end{align*} where $y_{\varepsilon,j}=b_j$ if $\varepsilon_j=1$ and $y_{\varepsilon,j}=a_j$ if $\varepsilon_j=0$. Substituting the definition of $G$ gives the corresponding $C$-mass of the rectangle \begin{align*} (F_1(a_1),F_1(b_1)]\times\cdots\times(F_p(a_p),F_p(b_p)] \end{align*} inside $[0,1]^p$, which is nonnegative because $C$ is a copula. Thus $G$ has nonnegative rectangle probabilities. Finally, \begin{align*} \lim_{x_1,\dots,x_p\to\infty}G(x_1,\dots,x_p)=C(1,\dots,1)=1, \end{align*} and if $x_j\to-\infty$ in any coordinate, then $F_j(x_j)\to0$, so the groundedness of the copula gives \begin{align*} G(x_1,\dots,x_p)\to0. \end{align*} Therefore $G$ is a joint distribution function on $\mathbb{R}^p$. [/step] [step:Compute the margins of the constructed joint distribution function] Fix $j\in\{1,\dots,p\}$ and $x_j\in\mathbb{R}$. The $j$-th marginal distribution function of $G$ is obtained by sending every coordinate except the $j$-th one to $+\infty$. Since each $F_i$ satisfies $\lim_{t\to\infty}F_i(t)=1$, and since a copula has uniform one-dimensional margins, we get \begin{align*} \lim_{\substack{x_i\to\infty\\ i\ne j}} G(x_1,\dots,x_p) &= C(1,\dots,1,F_j(x_j),1,\dots,1)\\ &=F_j(x_j). \end{align*} Thus the $j$-th marginal distribution function of $G$ is $F_j$. Since $j$ was arbitrary, all prescribed margins are recovered, completing the proof. [/step]