Bound for multinomial expansion involving Poisson random variables

Question

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 $k_i \in \{0, \ldots, 2k\}$?

Answer 1

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 $C$.