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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!

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

1 Answer 1

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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.

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