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I am trying to do a Monte Carlo simulation and want to define a function for the conditional inverse function for the the Ali-Mikhail-Haq and the Farlie-Gumbel-Morgenstern Copula. Here is an example of the formula i am looking for enter image description here

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By definition, when $(U,V)$ has a copula distribution $C,$

$$C(u,v) = \Pr(U\le u,\ V \le v).$$

To find the distribution conditional on $u$ when $C$ is differentiable at $u$ with derivative $C_u(u,v)$ for all $0\le v\le 1,$ evaluate this in the infinitesimal strip between $u$ and $u+\mathrm du$ to find

$$\Pr(u\le U \lt u +\mathrm du,\ V \le v) = C_u(u,v)\mathrm du.$$

Since the copula is supported on $[0,1],$ the conditional distribution of $V$ is

$$C_{V\mid U=u}(v) = \Pr(V\le v\mid U = u) = \frac{C_u(u,v)}{C_u(u,1)}.\tag{*}$$


Let's test this on one of your formulas; say, the family of Clayton copulas. Apparently your parameterization is

$$C(u,v;\rho) = \left(u^{-\rho} + v^{-\rho} - 1\right)^{-1/\rho}.$$

Compute the derivatives and do the algebra to simplify $(*)$ to

$$C_{V\mid U=u}(v;\rho) = \left(1 + \left(\frac{u}{v}\right)^\rho - u^\rho\right) ^ {-1 - 1/\rho}.$$

The value of its inverse at some probability $0\le q\le 1$ solves the equation $$C_{V\mid U=u}(v;\rho) = q$$ for $v.$

Doing the algebra yields

$$C^{-1}_{V\mid U=u}(q;\rho) = \left(1 - u^{-\rho} + u^{-\rho}q^{-\rho/(1+\rho)}\right)^ {-1/\rho},$$

agreeing with your first formula. So we're on the right track.


For the Ali-Mikhail-Haq family parameterized by $-1 \le \theta \le 1$ we may write

$$C(u,v;\theta) = \frac{uv}{1 - \theta(1-u)(1-v)}.$$

Now $(*)$ tells us we must solve the equation

$$q = C_{V\mid U=u}(v;\theta) = \frac{\theta v (v-1) + v}{\left(\theta(u-1)(v-1) - 1\right)^2}.$$

Upon clearing the denominator, this is a quadratic equation in $z$ -- it's unnecessary in this forum to write the formula for its roots (and often the formula is not needed). You will, of course, select the non-negative root.


The same method applies to the F-G-M family. A parameterization of its copula is given at https://stats.stackexchange.com/questions/457294;

$$C(u,v;\theta) = u v ( 1 + \theta(1-u)(1-v) ).$$

Computing $C_u$ as before and applying $(*)$ tells us we need to solve

$$q = C_{V\mid U=u}(v;\theta) = \theta(2u-1)(v-1)v + v.$$

This again is a quadratic equation, whose solution you may express or compute as you see fit. Here are some illustrations of $C_{V\mid U=u}(\ ;\theta)$ for three values of $\theta$ where $u$ ranges from $0$ (blue curves) through $1$ (purple curves).

enter image description here

NB: this solution differs from the one presented at https://stats.stackexchange.com/a/457325/919, which quotes its reference accurately. Because I used exactly the same method in all three cases, I think my result is correct.

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