# Bivariate Gaussian copula with exponential margins

I got little bit lost in the formulas.

Assume to have two random variables distributed exponentially $X_i \sim Exp(\lambda_i)$ and $X_j \sim Exp(\lambda_j)$.

Thus, the distribution functions are $F_{X_i}(x_i)= 1-\exp(-\lambda_i x_i)$ and $F_{X_j}(x_j)=1-\exp(-\lambda_j x_j)$.

What is the formula for a Gaussian copula, $C(u,v)$, linking these exponential margins?

$$C(u,v) = \mathbb{P}\left(X\leq N^{(-1)}(u),\quad \rho X + \sqrt{1-\rho^2}X^\perp \leq N^{(-1)}(v)\right)$$
• +1, with $X$ and $X^{\perp}$ following a $\mathcal{N}(0,1)$. – Quantuple May 20 '16 at 12:48
• +1, that is correct. Let $\Phi_2$ be the two-dimensional Gaussian distribution function, then the copula is defined by $C(u, v) = \Phi_2(N^{-1}(u), N^{-1}(v))$. Moreover, $X_i$ and $X_j$ can be simulated by $X_i = F_i^{-1}(N(X))$ and $X_j = F_j^{-1}(N(\rho X + \sqrt{1-\rho^2}X^\perp))$, where $F_i$ and $F_j$ are the respective cumulative distribution function of $X_i$ and $X_j$. – Gordon May 20 '16 at 13:04
• What is $X^{\perp}$, @MJ73550? – CaffeRistretto May 21 '16 at 10:35
• @CaffeRistretto. People like to use $X^{\perp}$ to denote a standard normal random variable that is independent of $X$. – Gordon May 21 '16 at 21:40