In my specific case, I have a pdf that has no closed form, and I want to generate random values ​​of this distribution. It depends on a summation that goes to infinity (coming from a poisson process) and two Lebesgue integrals. Can anyone tell me how to choose a suitable method to generate random values ​​from this distribution?

Here is my pdf for a finite collection of points $r=(s_1,\dots,s_n)$. $p(Y_r)=\sum^{\infty}_{|N|=0}\frac{{\rm e}^{-\lambda\mu(S)}\left[\lambda \right]^{|N|}}{|N|!}\int_{S^{|N|}}\int_{\mathbb{R}^{|N|}}f_N(Y_N\mid S_N,N)p(Y_r\mid Y_N,S_N)\text{d}Y_N\text{d}S_N$,

where $f_N(Y_N\mid S_N,N)$ is a normal distribution $|N|$-dimensional, and $p(Y_r\mid Y_N,S_N)$ is a conditional normal distribution $n$-dimensional. Here $S_N$ is the vector of locations of the points from my Poisson process ($N$).

I want generate values for $Y_r$.


closed as unclear what you're asking by Xi'an, user158565, Michael Chernick, mkt, kjetil b halvorsen May 16 at 9:18

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  • $\begingroup$ This remains unclear: any connection between $r$ and the index of $Y_r$? how are the normal densities conditional on $N$ and $S_N$ defined? Why is it impossible to integrate the inner integral since Normal x Normal is manageable? $\endgroup$ – Xi'an May 15 at 20:23
  • $\begingroup$ For your first question is yes. For the second, it's a normal distribution $|N|$-dimensional, where $|N|$ is the number of points of the Poisson process $N$. $Y_N|S_N,N\sim(a, \Sigma^2)$ where $a$ is a $|N|$-vector of constants and $\Sigma^2$ is the covariance matrix depending on the locations. It is 'impossible' to solve the last integral because it's with respect to locations. $\endgroup$ – Ga13 May 15 at 20:31

From the representation $$\sum^{\infty}_{|N|=0}\frac{{\rm e}^{-\lambda\mu(S)}\left[\lambda \right]^{|N|}}{|N|!}\int_{S^{|N|}}\int_{\mathbb{R}^{|N|}}f_N(Y_N\mid S_N,N)p(Y_r\mid Y_N,S_N)\text{d}Y_N\text{d}S_N$$one can spot $$\underbrace{\sum^{\infty}_{|N|=0}}_\text{marginal in $N$}\overbrace{\frac{{\rm e}^{-\lambda\mu(S)}\left[\lambda \right]^{|N|}}{|N|!}}_{\text{Poisson distributions on $N$}}\underbrace{\int_{S^{|N|}}\int_{\mathbb{R}^{|N|}}}_\text{marginal in $Y_N,S_N$}\overbrace{f_N(Y_N\mid S_N,N)}^\text{generation of $Y_N$}\underbrace{p(Y_r\mid Y_N,S_N)}_\text{generation of $Y_r$}\text{d}Y_N\text{d}S_N$$ Hence this means that, if one

  1. Generate a Poisson $|N|\sim\mathcal P(\lambda)$
  2. Generate $S_N$ conditionally on $N$ [I am unclear how]
  3. Generate $Y_N$ as a Normal conditionally on $S_N$
  4. Generate $Y_r$ as a Normal conditionally on $Y_N,S_N$

one produces a simulation from the original distribution. This is a typical example of a marginalisation technique where one simulates 4 variables to only keep one.

  • $\begingroup$ The output (r.v's from marginal Yr) is obtained by step 4? $\endgroup$ – Ga13 May 30 at 18:46
  • 1
    $\begingroup$ Yes, producing $Y_r$ from the marginal is equivalent to producing the quadruple from the joint and extracting $Y_r$. $\endgroup$ – Xi'an May 31 at 4:42

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