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.*