2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

$$C(u,v) = \mathbb{P}\left(X\leq N^{(-1)}(u),\quad \rho X + \sqrt{1-\rho^2}X^\perp \leq N^{(-1)}(v)\right)$$

$\endgroup$
  • 1
    $\begingroup$ +1, with $X$ and $X^{\perp}$ following a $\mathcal{N}(0,1)$. $\endgroup$ – Quantuple May 20 '16 at 12:48
  • 1
    $\begingroup$ +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 $X_j$. $\endgroup$ – Gordon May 20 '16 at 13:04
  • $\begingroup$ What is $X^{\perp}$, @MJ73550? $\endgroup$ – CaffeRistretto May 21 '16 at 10:35
  • $\begingroup$ @CaffeRistretto. People like to use $X^{\perp}$ to denote a standard normal random variable that is independent of $X$. $\endgroup$ – Gordon May 21 '16 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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