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

  • $\begingroup$ Do you assume independence of the random variables $x_1,\ldots,x_n$? $\endgroup$ Jun 27, 2020 at 13:37
  • $\begingroup$ The poisson variables are slightly dependent by the condition that their sum is equal to $T$. $\endgroup$
    – user124297
    Jun 27, 2020 at 13:46
  • $\begingroup$ @user124297: The term "slightly dependent" is unknown to me. I suppose that $x_1,\ldots,x_n$ are independent and that you then work with conditional probabilities. Right? $\endgroup$ Jun 27, 2020 at 13:51
  • $\begingroup$ Yes you can think about it in this way. In the beginning $x_i$ are independent, but when you put conditional probability they are not independent anymore. In any case, you are working with conditional probability. $\endgroup$
    – user124297
    Jun 27, 2020 at 13:55
  • $\begingroup$ Have you tried the case $n = 2$? In this case with respect to $P(.|x_1+x_2 = T)$ $x_1$ has the binomial distribution $Bin(T,\lambda_1/(\lambda_1+\lambda_2))$. (N.B.: Without independence of $x_1,x_2$ the distribution of $x_1$ can be rather arbitrary.) $\endgroup$ Jun 27, 2020 at 14:34

1 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 $C$.

  • $\begingroup$ Thank you, is it possible to actually find this constant $C$? Also, does it mean that the bound would hold for negative moments instead? And is it possible to find an upper bound for the original problem? $\endgroup$
    – user124297
    Jun 27, 2020 at 18:27
  • $\begingroup$ First, $C = 1$, since $\|a\|_\infty \leq \|a\|_1$ and there is no better $C$, since $\|(1,0,\ldots,0)\|_\infty = \|(1,0,\ldots,0)\|_1$. Further, as shown in the first part, there actually is no $C$ with the required properties. I've edited my answer. $\endgroup$ Jun 27, 2020 at 23:27
  • $\begingroup$ Thank you. How about situation when $C$ is dependent on $N$? Would it be possible to get such a $C$? I guess one would have to go through combination of all the possible $k_i$? $\endgroup$
    – user124297
    Jun 27, 2020 at 23:46
  • $\begingroup$ Hello, actually $C$ does not depend on $n$! $\endgroup$ Jun 28, 2020 at 8:38

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