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In "Serfling, R. J. (1980). Approximation theorems of mathematical statistics", we read

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In Theorem A, as one suspects, $k=1,2,...$, indicating the integer-moments, while $n$ is the sample size. As regards the $m_k$ and $\mu_k$ symbols, we have, drawing i.i.d from the distribution of some random variable $X$,

$$m_k = \frac 1n \sum_{i=1}^n (X_i-\bar X)^k, \;\;\;\mu_k = E[(X-E(X)]^k$$

Theorem A is concerned with each separate central sample moment, Theorem B is concerned with their joint limiting distribution.

Suppose that $k=3$. Then, from Theorem A we have

$$\text{AVar}(m_3) = \mu_6 - \mu_3^2 - 6\mu_2\mu_4 + 9\mu_2^3$$

But from Theorem B, setting $i=j=3$ we get

$$\text{AVar}(m_3) = \mu_8 - \mu_4^2 - 8\mu_3\mu_5 + 16\mu_3^2\mu_2$$

I don't see how the two can coincide, and some quick simulated checks show that they don't.

Question:is it possible that the expression in Theorem B is meant only for the off-diagonal covariance terms and not for the diagonal variance terms?

The author makes no such clarification.

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1 Answer 1

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Posting questions on CV has this magic effect many times: a little after you post them, you find the answer.

ANSWER: both expressions are correct, and the expression in Theorem B holds also for the diagonal variance terms. The catch?

...In the variance-covariance matrix expression, we realize that the numbering of the variables starts from $2$, while the $i,j$ indices start from $1$.

This means that when we want to obtain the variance of $m_3$ we must set $i=j=2$, and when we want to obtain, say, the covariance of $(m_3, m_4)$, we must set $i=2, j=3$.

In general, for $k,\ell$ moments, set $i=k-1, j=\ell-1$ in the variance-covariance expression.

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  • $\begingroup$ Re "must notice:" you don't seem to have supplied enough information to enable your readers to notice anything like that. $\endgroup$
    – whuber
    Commented Aug 11, 2020 at 19:55
  • $\begingroup$ @whuber The information is all in the statement of Theorem B: the 1st variable is $m_2$, not $m_1$, and no explicit numbering is given for the $i,j$ -in which case, one starts the counting from $1,...$. Strictly speaking, Serfling wrote everything correctly, but he should have anticipated the possible confusion here. $\endgroup$ Commented Aug 11, 2020 at 23:55
  • $\begingroup$ The principal problem is that in Theorem A, the meanings of the symbols--and especially the indexing conventions--are impossible to deduce from its mere statement. $\endgroup$
    – whuber
    Commented Aug 12, 2020 at 12:59
  • $\begingroup$ @whuber But immediately after the two theorems, I explicitly write what $m_k$ and what $\mu_k$ is, so I don't understand where the problem is. $\endgroup$ Commented Aug 12, 2020 at 16:05
  • $\begingroup$ I still cannot see anything in your post that gives any clue to the indexing convention in Theorem A. $\endgroup$
    – whuber
    Commented Aug 12, 2020 at 16:17

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