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 $x_1,\ldots,x_n$? $\endgroup$ – Dieter Kadelka Jun 27 '20 at 13:37
  • $\begingroup$ The poisson variables are slightly dependent by the condition that their sum is equal to $T$. $\endgroup$ – user124297 Jun 27 '20 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? $\endgroup$ – Dieter Kadelka Jun 27 '20 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 '20 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.) $\endgroup$ – Dieter Kadelka Jun 27 '20 at 14:34

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 '20 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. $\endgroup$ – Dieter Kadelka Jun 27 '20 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 '20 at 23:46
  • $\begingroup$ Hello, actually $C$ does not depend on $n$! $\endgroup$ – Dieter Kadelka Jun 28 '20 at 8:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.