# Normalizing constant calculation of Strauss Process

Suppose that I have the following Strauss Process up to a proportionality constant

$$p(\mu_{1}, \mu_{2},..., \mu_{K},K)\propto \xi^{K}\prod_{i=1}^{K} I(\mu_{i}\in R) *a^{\sum_{i,j}|\mu_{i}-\mu_{j}|

where $$\mu_{i}\in \mathbb{R}, a\in [0,1]$$ and $$\xi, d>0$$. I know how to sample, and I acquired samples from the $$(*)$$ with the use of Birth and Death Algorithm. What I would like to do now is to calculate the normalizing constant of $$(*)$$ and I'm wondering because it seems quite straight forward, if the following is correct.

For, fixed $$a,d$$ and $$\xi$$, then the normalizing constant is calculated as

$$Z = \frac{\sum_{j=1}^{M}p(\mu_{1,j}, \mu_{2,j},..., \mu_{K^{j},j},K^{k})}{M}$$

where $$M$$ is the number of samples that I drawn from $$(*)$$.

To me the previous calculation seems correct, but every textbook says that in general the normalizing constant is intractable, that's why I doubt a little bit.

• The normalizing constant you are calculating isn't the normalizing constant of the Strauss process, which is indeed intractable, but simply the normalizing constant for resampling from your sample. Commented Sep 8, 2023 at 18:36
• Is there a missing indicator in the exponent of $a$ ? The notation $\sum_{ij}|\mu_{i}-\mu_{j}|<d$ is ambiguous. Commented Sep 9, 2023 at 12:13
• The summation$$\sum_{j=1}^{M}p(\mu_{1,j}, \mu_{2,j},..., \mu_{K^{j},j},K^{k})$$is not possible since $p$ is only known up to a normalising constant. Commented Sep 9, 2023 at 12:14
• @Xi'an Why the summation is not possible? This would be the normalising constant for resampling as jbowman said. If I'm not mistaken Commented Sep 9, 2023 at 12:28
• As written it involves $p(\cdots)$ but $p$ is missing its normalising constant. Commented Sep 9, 2023 at 12:33

• As I understand the inverse logistic regression described in Geyers paper. I'm gonna use the formulas (5) and (8) in the paper. Where $q(\theta_{i})$ is gonna be the unormilized version of the Strauss process, this can be tempered or not. Then I'm gonna calculate the pseudo prior and then I'll move forward to formula (8). Now, based on samples from $(*)$ I can do all the calculations. However, since the pseudo prior has indeed an unknown normalizing constant as well that's why we need recursion? Commented Sep 9, 2023 at 14:13
• No, you need two different samples, one from $p$ (unormalised) and one from an alternative $q$ (normalised) that is free to choose. Commented Sep 9, 2023 at 17:11
• Where the $q$ can be whatever we want, for example independent Gaussians over $\mu_{i}$. Commented Sep 9, 2023 at 17:21