I am working with the following integral :

$$ I(u_1,\ldots,u_M)=\sum_{\sigma\in S_M}\bigg(\prod_{i=1}^j\int_{u_{\sigma(i)}}^1\mathrm{d}y_i\bigg)\bigg(\prod_{l=j+1}^M\int_0^{u_{\sigma(l)}}\mathrm{d}y_l\bigg)n(y_1\ldots y_M)\\ = \sum_{\sigma\in S_M}\bigg(\prod_{i=1}^j\int_{u_{\sigma(i)}}^1\mathrm{d}y_i\bigg)\prod_{l=j+1}^M\bigg(\int_0^1\mathrm{d}y_l-\int_0^{u_{\sigma(l)}}\mathrm{d}y_l\bigg)n(y_1\ldots y_M) $$

Where $M$ is an integer (after all the calculations I would like to take $M\to\infty$ $S_M$ is the set of permutations of $\{1,\ldots , M\}$, $0\leq j\leq M$ and $u_i\in [0;1] \ \forall i\in\{1,\ldots , M\} $ and $n$ is a function that is symmetric over exchange of its arguments.

Provided I divide this by a normalization constant, I believe can see this as some sort of probability over the $u_i$ variables. For instance if $n$ is separable, i.e. $n(y_1,\ldots,y_M)=\prod_i n_i(y_i)$ then by writing (up to normalization) :

$$p_i\sim\int_{u_{\sigma(i)}}^1 \mathrm{d}y_i n_i(y_i)$$

the integral becomes something like :

$$I \sim \sum_{\sigma\in S_M}\prod_{i=1}^j p_i \prod_{l=j+1}^M(1-p_j) $$

And looks like the sum of bernoulli trials with different probabilities for each trial (here they would be independent), and I am summing over all the probabilities

However all I have over my probability distribution is not independence, but merely exchangeability over the $u_i$ variables. I did some research and came upon De Finetti's exchange theorem for exchangeable probability distributions. I didn't know the theorem previously and have no clue of how to apply it in this case.

Would there be a way to write this integral in the large $M$ limit, knowing what $n$ looks like, and by saying that the $u_i$'s are distributed along some distribution $g$ ?


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