New answers tagged

1

Note that the survival copula $C_{\theta_A, \theta_B}(u, v)$ and the non-survival copula $C(u, v)$ are related by \begin{align*} C_{\theta_A, \theta_B}(\hat{u}, \hat{v}) = \hat{u}+\hat{v}-1 + C(1-\hat{u}, 1-\hat{v}), \end{align*} where $\hat{u}=\bar{F}_A(x)=1 - F_A(x)$ and $\hat{v}=\bar{F}_B(y)=1-F_B(y)$. Then, \begin{align*} C(u, v) = C_{\theta_A, ...


1

$$\text{Pr}[\tau_1>t,\tau_2\leq t,\tau_3\leq t]=\text{Pr}[\tau_2\leq t,\tau_3\leq t] - \text{Pr}[\tau_1\leq t,\tau_2\leq t,\tau_3\leq t]$$ $$\text{Pr}[\tau_2\leq t,\tau_3\leq t]=C(1,q_2(t),q_3(t))$$


3

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


0

You need to know what your original conditional distribution was when you fitted the AR-GARCH(1,1). Assuming that you chose a student-t distribution, the reverse transformation after step 4 in R would look as follows: step 1: Fit Garch fit <- rugarchfit step 4: Simulate points sim <- 'simulated 100 points' step 5: Convert 100 uniformly ...


0

You need to estimate or assume a marginal distribution of the (u,v). Lets say you assume normality (don't do this), you would be able to perform a rosenblatt-transformation, to perform the task you describe. https://en.wikipedia.org/wiki/Inverse_transform_sampling This could be a useful resource.



Top 50 recent answers are included