0
$\begingroup$

It is well known that the variance of the sum of independent random variables (not necessarily i.i.d.) is the sum of the variance of each random variable (i.e. $Var[X_1 + X_2 ... X_n] = \sum_{i=1}^{n} Var[X_i]$). What about the higher absolute moment? For instance, does the following equation hold? \begin{equation} E\left[\left| \sum_{i=1}^n X_i - E\left[\sum_{i=1}^n X_i\right] \right|^3\right] = \sum_{i=1}^n E\left[\left| X_i - E\left[X_i\right] \right|^3\right]. \end{equation}

$\endgroup$
1
  • 1
    $\begingroup$ You could try some easy example cases. For example, does it hold if $X_i$ are uniformly distributed on $[-1, 1]$? $\endgroup$ May 27 at 9:17
5
$\begingroup$

The generalization to higher powers of the additivity statement of the variance goes via the cumulants: If two variables $X$ and $Y$ are independent, then the cumulants $\kappa_n$ are additive, $\kappa_n(X+Y)=\kappa_n(X)+\kappa_n(Y)$.

It follows from this additivity that third central moments are additive, $$E\left[\left(\sum_i X_i-\sum_i E[X_i]\right)^3\right]=\sum_i E\left[\left(X_i-E[X_i]\right)^3\right],$$ iff the variables $X_i$ all have the same variance. No such simple additivity criterion exists for absolute values of third powers.

$\endgroup$
1
  • $\begingroup$ Thank you for the clarification! $\endgroup$
    – Bertrille
    May 27 at 11:11

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.