1

Reading a text on Transformation and Variate Generation that says that if

  • we want to simulate V,
  • but we know only its conditional distribution given U,
  • and we can simulate U

then we can simulate V as follows:

  1. Draw u_sim from the marginal distribution of U
  2. Draw v_sim from the conditional distribution of V given U=u_sim

then v_sim is a variate from the marginal distribution of V. This is the part I don't understand: why are we allowed to say we have simulated from the marginal distribution of V and not the conditional $p(v|u)$?

Maybe this is just an approximation, and this distribution converges on $p(v)$ for large samples?

  • $V$ is _both_ generated from $p_V$ and, conditional on $U$, from $p(\cdot|U)$, there is no contradiction. – Xi'an Feb 23 '20 at 09:23

1 Answers1

4

Maybe this is just an approximation, and this distribution converges on p(v) for large samples?

No, it's exact for any size of sample.

  1. The joint distribution is the product of the known marginal and the conditional:

    $f_{U,V}(u,v) = f_{V|U}(v)\cdot f_{U}(u)$

    So we can generate a sample of observations from the joint distribution of $(U,V)$, that is, a set of $(u_i,v_i), i=1, 2,\ldots,n$ by the scheme of generating from the marginal for $U$ and then the conditional for $V$ given $U$.

  2. If you have a sample from the joint, you have a sample from any of the marginals simply by taking the corresponding univariate sample.

    That is, if we have a set of values $(u_i,v_i), i=1, 2,\ldots,n$, just take the $v_i$ values. It's a very handy aspect of simulation.

Glen_b
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