# Sampling from bivariate joint cumulative distribution function

Given two variable $$x,y$$, they are subjected to a joint probability density function:

$$f(x,y) = \dfrac{1}{3}(3x^2 + 4xy + 3y^2)\\ 0\leq x \leq 1;0\leq y \leq 1$$

Obviously, its corresponding cumulative distribution function (CDF) is:

$$F(x,y) = \dfrac{1}{3}(x^3y+x^2y^2+xy^3)$$

My objective is to generate random samples from this CDF using the inverse tranformation method. As far as I know, the first step is to get the marginal CDF of $$x$$:

$$F_x = F(x,y=1) = \dfrac{1}{3}(x^3 + x^2 + x)$$

Then we can readily get a random value of $$x$$, denoted as $$u_x$$.

Next, for sampling $$y$$, we have to derive its conditional CDF $$F_{y|x}$$ given $$x$$.

My question is how to derive the $$F_{y|x}$$? Can the well-known conditional rule still be adapted for this problem, such that:

$$F_{y|x} = \dfrac{F(x,y)}{F_x}$$

It is worth noting that the actual CDF I encountered is much more complicated than the above example, and is very difficult to perform integration on the PDF or CDF.

Thank you very much!

• "Copula" is a great search term. We have some threads about simulating from copulas.
– whuber
Aug 6, 2021 at 15:33
• @whuber: since the joint density is given, I am unsure copulas need be involved. Aug 6, 2021 at 15:40
• @Xi'an The approach taken in the question is that of sampling from a copula.
– whuber
Aug 6, 2021 at 16:52
• @whuber: Ah OK, sure! Aug 6, 2021 at 18:27

If$$f(x,y) = \dfrac{1}{3}(3x^2 + 4xy + 3y^2)$$ then $$f_{Y|X=x}(y)\propto f(x,y)\propto 3x^2 + 4xy + 3y^2$$ leads to$$f_{Y|X=x}(y)=\dfrac{3x^2 + 4xy + 3y^2}{\underbrace{\int_0^1 (3x^2 + 4xy + 3y^2)\,\text dy}_{3x^2+2x+1}}$$ from which the conditional cdf can be derived.