# Large Deviation of Triple Poisson Product

Let $$X_i$$ with $$i=1,\ldots,n$$ be independent Poisson variables, $$X_i$$ with parameter $$\lambda_i.$$

Let $$\circ$$ be a group operation on a group of size $$n.$$

I would like to obtain a large deviation inequality on $$\Gamma=\sum_{1\leq i,j\leq n} X_i X_j X_{i \circ j},$$ but it seems that analytic computation of the moments $$\mathbb{E}(\Gamma^k)$$ is problematic.

Even the constant parameter $$\lambda$$ case is of interest.

Edit: For simplicity, assume $$\lambda$$ is the common parameter. By independence when $$i\neq j,$$ $$\mathbb{E}(X_iX_j)=\mathbb{E}(X_i)\mathbb{E}(X_j)$$ and this together with the fact that the third factorial moment is $$\mathbb{E}(X_i(X_i-1)(X_i-2))=\lambda^3,$$

can be used recursively to obtain the following (if my computations are correct) when $$i\neq j$$, which is the difficult case:

$$\mathbb{E}(X_i X_j X_{i \circ j})= \mathbb{E}(X_i)\mathbb{E}(X_j)\mathbb{E}(X_{i \circ j})= \mathbb{E}(X_i)^3=\lambda^3,\qquad i\neq e,j\neq e,$$ since the three indices are distinct, and $$\mathbb{E}(X_e X_j X_{e\circ j})=\mathbb{E}(X_e)\mathbb{E}(X_{j}^2)= (\lambda^2+\lambda)\lambda,\qquad i= e,j\neq e.$$ When $$i=j,$$ we get $$\mathbb{E}(X_i^2 X_{i \circ i})=\mathbb{E}(X_i^2) \mathbb{E}(X_{i \circ i})=(\lambda^2+\lambda)\lambda,\quad i\neq e,$$ and $$\mathbb{E}(X_i^2 X_{i \circ i})=\mathbb{E}(X_e^3) =\lambda^3+3\lambda^2+\lambda,\quad i=e.$$ This type of argument can be used to yield a complicated expression for $$\mathbb{E}(\Gamma^k),$$ for $$k=1,2.$$ In particular, it seems that the mean is $$\mathbb{E}(\Gamma)=n^2 \lambda^3+3n \lambda^2+\lambda.$$

However the question of a good bound still stands. Any references to related work will also be appreciated.