Let $x\_i, i=1, \ldots n$ be Poisson random variables with parameters $\lambda\_i$ correspondingly with condition that $\sum\_{i=1}^nx\_i=T$. Due to linearity of the expectation one can write:
$$
E\left(\left|\sum\_{i=1}^n a\_ix\_i\right|^{2k} \big| \sum\_{i=1}^nx\_i=T\right)\\
=\sum\_{k\_1+\ldots k\_n=2k}\frac{(2k)!}{k\_1!\ldots k\_n!}a\_1^{k\_1} \ldots a\_n^{k\_n}E\left(x\_1^{k\_1}\ldots x\_n^{k\_n}\big | \sum\_{i=1}^nx\_i=T\right)
$$
I would like to bound this expression from above. Ideally, I would like to get something like $C\times E\left(x\_1^{k\_1}\ldots x\_n^{k\_n}\big | \sum\_{i=1}^nx\_i=T\right)\times \|a\|\_1$ in the right hand side. Or, at least to understand in which cases this bound would hold.


But I am not sure on how to take into account all the possible cases for $k\_i \in \{0, \ldots, 2k\}$?



Bound for multinomial expansion involving Poisson random variables
For the following we only need that $X\_1,\ldots,X\_n$ are arbitrary random variables with values in $\mathbb{N}\_0$ such that $\mathbb{P}(X\_1+\ldots+X\_n = T) > 0$. (In particular the original situation is included.) Let $Q(A) := \mathbb{P}(A | X\_1+\ldots+X\_n = T)$ for measurable $A$. Let $E\_Q$ be the expectation w.r.t. $Q$. Then the question reduces to: Does $C > 0$ exist with
$$E\_Q\left( |\sum\_{i=1}^n a\_iX\_i|^{2k} \right) \leq C \cdot E\_Q\left( |\sum\_{i=1}^n X\_i|^{2k} \right) \cdot \|a\|\_1$$
First, such $C$ cannot exist if $2k > 1$. Let $a = (t,\ldots,t)$ with $t > 0$. Then $E\_Q\left( |\sum\_{i=1}^n a\_iX\_i|^{2k} \right) = \|a\|\_1^{2k}/n^{2k} \cdot E\_Q\left( |\sum\_{i=1}^n X\_i|^{2k} \right)$. Thus we have to replace at least $\|a\|\_1$, f.i. with $\|a\|\_1^{2k}$. But then the inequality it trivial. Given $a$, let $t := \max\{|a\_1|,\ldots,|a\_n|\} = \|a\|\_\infty$. Then
$$E\_Q\left( |\sum\_{i=1}^n a\_iX\_i|^{2k} \right) \leq E\_Q\left( |\sum\_{i=1}^n X\_i|^{2k} \cdot \|a\|\_\infty^{2k} \right) \leq E\_Q\left( |\sum\_{i=1}^n X\_i|^{2k} \right) \cdot \|a\|\_1^{2k},$$
since $\|a\|\_\infty \leq \|a\|\_1$. As simple examples show there is no better $C$.


