7
$\begingroup$

currently, I am trying to understand how one calculates the standard errors on higher moments using Rao's book [1].

On page 437, he defines

$$ O_r = \frac{1}{n} \sum x_i^r, \, v_r = E[x^r], \, \mathrm{and} \; \mu_r = E[(x-v_1)^r]. $$

I am stuck at computing $E[O_2 O_1^2]$. In the book it is claimed on p. 438 that

$$ E[O_2 O_1^2] = \frac{\mu_4 + (n - 1) \mu_2^2}{n^2} $$

assuming that the origin is where the population mean is; claiming that this is without loss of generality. However, when I try to compute this

\begin{eqnarray} E[O_2^{} O_1^2] &=& \frac{1}{n^3} E\left[\sum_i x_i^2 \sum_j x_j \sum_k x_k\right] \\ &=& \frac{1}{n^3} E\left[ \sum_i x_i^2 \left( \sum_l x_l^2 + \sum_{k \neq j} x_k x_j \right) \right] \\ &=& \frac{1}{n^3} E\left[\sum_m x_m^4 + \sum_{i \neq l} x_i^2 x_l^2 + \sum_{i} x_i^2 \sum_{k \neq j} x_k x_j \right] \\ &=& \frac{v_4 + (n - 1) v_2^2}{n^2} + \frac{1}{n^3} E\left[ \sum_{i} x_i^2 \sum_{k \neq j} x_k x_j \right] \end{eqnarray}

I cannot see how this would go over to the form Rao uses. If my calculations are correct the missing term is

\begin{eqnarray} E\left[ \sum_{i} x_i^2 \sum_{k \neq j} x_k x_j \right] &=& E \left[ 2\sum_{l \neq m} x_l^3 x_m + \sum_{i \neq j \neq k \neq i} x_i^2 x_j x_k\right] \\ &=& 2 n (n - 1) v_3 v_1 + (n^2 (n - 1) - 2 n (n - 1)) v_2 v_1^2 \end{eqnarray}

which is quite complicated. Are my calculations wrong here or do I miss something? Does this also mean the formula for the variance of the variance only holds for the assumption about the mean??

Thanks in advance!

[1] Linear statistical interference and its application

$\endgroup$
1
  • 1
    $\begingroup$ You seem not to be using the assumption that the mean of the $x_i$ is zero. $\endgroup$
    – whuber
    Commented Oct 29, 2021 at 14:06

1 Answer 1

10
$\begingroup$

Here is a quick check using the moments of moments functions from the mathStatica package for Mathematica. That uses power sum notation where $s_r = \sum_{i=1}^n X_i^r$. Your problem is to find $E[O_2 O_1^2]$, which denotes the 1st RawMoment of $\frac{s_2}{n} \frac{s_1^2} {n^2}$:

enter image description here

where the solution is expressed in terms of Central Moments $\mu_r$ of the population. In the special case where the mean of the population is 0, this simplifies to:

enter image description here

... as stated in Rao.

$\endgroup$
4
  • $\begingroup$ The question is not whether Rao is right, but what the OP is missing in their derivation. $\endgroup$
    – whuber
    Commented Oct 29, 2021 at 16:41
  • $\begingroup$ The first part of the question is: "I am stuck at computing $E[O_2 O_1^2]$" ... followed by a CLAIM of what the book states the answer to be, and the OPs uncertainty as to why his work does not match the book. In reply, (a) the book is correct, (b) your comment as to the author's error is verified. $\endgroup$
    – wolfies
    Commented Oct 29, 2021 at 16:51
  • $\begingroup$ Right: that makes your post a useful comment on the question, but it doesn't answer it. But I see that your remark "in the special case" is a sufficiently strong hint, +1. $\endgroup$
    – whuber
    Commented Oct 29, 2021 at 16:52
  • $\begingroup$ And thirdly, he asks: "Does this also mean the formula for the variance of the variance only holds for the assumption about the mean?? ... which is also resolved. $\endgroup$
    – wolfies
    Commented Oct 29, 2021 at 16:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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