# How to generate data from this particular multivariate distribution?

Let $$(X_0,X_1)$$ be a random vector distributed according to the CDF $$F_{(X_0,X_1)}(x,y)= \min (F_{X_0}(x),F_{X_1}(y))$$ where $$F_{X_0}(x),F_{X_1}(x)$$ are the CDFs of $$X_0,X_1$$ respectively. We do not necessarily have densities.

I know how to sample from a one-dimensional CDF using inverse transform sampling. I want to sample from $$F_{(X_0,X_1)}$$

If $$F_{(X_0,X_1)}$$ had a density, then I could see how this might go:

• Generate $$x_0$$ from $$F_{X_0}$$

• Compute conditional density of $$X_1$$ given $$X_0=x_0$$,

• and generate $$x_1$$ from this density

But, here $$F_{(X_0,X_1)}$$ does not have a density.

What to do ?

If $$\mathbb P(X_0\le x,X_1\le y)=\min \{F_{X_0}(x),F_{X_1}(y)\}$$ then $$\mathbb P(F_0(X_0)\le F_0(x),F_1(X_1)\le F_1(y))=\min \{F_{X_0}(x),F_{X_1}(y)\}$$ Denoting $$U_0=F_0(X_0),\quad U_1=F_1(X_1)$$ this implies that $$\mathbb P(U_0\le u_0,U_1\le u_1) = \min(\mathbb P(U_0\le u_0,u_0 and therefore that, when $$u_1\ge u_0$$, $$\mathbb P(U_0\le u_0,U_1\le u_1) = u_0$$ meaning that $$\mathbb P(U_0\le u_0,U_1\le u_1) - \mathbb P(U_0\le u_0,U_1\le u_0) = \mathbb P(U_0\le u_0,u_0 Similarly, when $$u_0>u_1$$ $$\mathbb P(u_1 So $$U_1$$ cannot take values larger than $$U_0$$ and conversely, which implies that$$U_0=U_1$$with probability one.

Conclusion: To draw from this distribution,

1. generate a Uniform variate $$U_0$$
2. transform $$U_0$$ into $$X_0=F_0^{-1}(U_0)$$
3. transform $$U_0$$ into $$X_1=F_1^{-1}(U_0)$$
• You omitted the last step: return the larger of $X_0$ and $X_1.$
– whuber
Nov 17 at 18:05
• @whuber: I do not think so as one need return a pair of random variables. Assuming all inverse are well-defined, $F_0(F_0^{-1}(U_0))=U_0=F_1(F_1^{-1}(U_0))$. Nov 17 at 18:39
• Thank you for pointing that out: I had misinterpreted the question by not noticing it concerned a bivariate CDF rather than a univariate one. +1.
– whuber
Nov 17 at 18:42
• @whuber: However, the explanation you provided for the cdf remains valid within this question. Nov 17 at 18:44
• Yes, the joint distribution is supported by the curve $x_1=F_1^{-1}(F_0(x_0))$. Nov 18 at 5:36