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 $$\mathbb P(U_0\le u_0,U_1\le u_1) = \min(\mathbb P(U_0\le u_0,u_0<U_1\le u_1) =u_0,u_1)$$ 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 $$\mathbb P(U_0\le u_0,U_1\le u_1) - \mathbb P(U_0\le u_0,U_1\le u_0) = \mathbb P(U_0\le u_0,u_0<U_1\le u_1) =0$$ Similarly, when $u_0>u_1$ $$\mathbb P(u_1<U_0\le u_0,U_1\le u_1) =0$$ 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 $X_1=F_1^{-1}(U_0)$
  • $\begingroup$ You omitted the last step: return the larger of $X_0$ and $X_1.$ $\endgroup$
    – whuber
    Nov 17 '21 at 18:05
  • $\begingroup$ @whuber: I do not think so as one need return a pair of random variables. Assuming all inverse are well-defined, $F_0(F_0^{-1}(U_0))=U_0=F_1(F_1^{-1}(U_0))$. $\endgroup$
    – Xi'an
    Nov 17 '21 at 18:39
  • 2
    $\begingroup$ 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. $\endgroup$
    – whuber
    Nov 17 '21 at 18:42
  • $\begingroup$ @whuber: However, the explanation you provided for the cdf remains valid within this question. $\endgroup$
    – Xi'an
    Nov 17 '21 at 18:44
  • 1
    $\begingroup$ Yes, the joint distribution is supported by the curve $x_1=F_1^{-1}(F_0(x_0))$. $\endgroup$
    – Xi'an
    Nov 18 '21 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.