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This question already has an answer here:

I have a MCMC sampler that targets $$\mathbb{P}(U_1,U_2,...U_n \mid G(U) \leq 0)$$ where $U=(U_1,U_2,...U_n)^T$. I realize now I am more interested in estimating the conditional density $$p_k = p(u_k \mid G(U)\leq 0)$$ for all $k=1,\ldots,n$. Is there a relatively easy way to achieve this either by using the sample that I already have from $\mathbb{P}(U_1,U_2,...U_n\mid G(U) \leq 0)$ or by designing another sampler ? I gather from this paper that (1) if the conditional distributions $\mathbb{P}(U_k\mid G(U) ,U_{j} \leq 0)$, $j\neq k$ are available then it is possible to derive density estimators for $p_k$. Another possibility (2) is if a functional form of the joint density $p( u_1,u_2,...u_n)$ is available and at least one $\mathbb{P}(U_k\mid G(U) ,U_{j} \leq 0)$ is straightforward to sample, then something can be done about it. I find myself in unknown territory since neither (1) nor (2) is true in my case.

Would anyone have a suggestion to tackle this problem ? Even a MCMC sampler yielding samples from $p_k$ for all $k$ would be nice, as i could try KDE afterwards to get the corresponding density. The only annoying thing is the relatively high dimension.

Thanks

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marked as duplicate by Nick Cox, kjetil b halvorsen, John, jbowman, COOLSerdash Jan 28 '15 at 12:50

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On a general level, if you can simulate samples from a joint distribution $p(u_1,\ldots,u_n)$, $(u_1,\ldots,u_n)^{(1)}$, $\ldots$, $(u_1,\ldots,u_n)^{(T)}$, say, then the first component of each simulation, $u_1^{(1)},\ldots,u_n^{(T)}$, is generated from the marginal of $p$, $$p_1(u_1) = \int p(u_1,\ldots,u_n) \text{d}u_2\cdots\text{d}u_n\,.$$Unless you have an easy solution, you do not need to aim at an algorithm generating from $p_1$.

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  • $\begingroup$ Yes thank you although it is obvious (!), for some reason when conditioning by $G(U)\leq 0$, I was under the impression that this was more complicated than it really is... $\endgroup$ – RobertoRonaldo Jan 20 '15 at 9:15

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