Unbiased estimators and moment of moments

Following section 7.4 of Rose and Smith "Mathematical Statistics with Mathematica" (book available online here), I'm trying to use the Fundamental Expectation Result (eq 7.15) and other building blocks given in section to derive by hand as an exercise the variance of the sample variance. (I know there are easier ways for this particular problem, I'm trying to understand the general methodology illustrated in the subsection)

An unbiased estimator of the population variance is the $$h_2$$ statistic

$$h_2 = \frac{n S_2 - S_1^2}{(n-1)n} = \frac{n}{n-1} m_2 = \frac{1}{n-1} \sum_{i=1}^n {\left( X_i - \bar{X} \right)}^2$$

where $$S_r = \sum_{i=1}^n X_i^r$$ is the $$r^{th}$$ power sum and $$E[h_2]=\mu_2$$, the second central population moment. The variance of $$h_2$$ is the second central moment of $$h_2$$ (shown here)

$$Var[h_2] = \frac{\mu_4}{n} - \frac{(n-3) \mu_2^2}{n(n-1)}$$

which simplifies to the well known result of $$\frac{2 \sigma^4}{n-1}$$ for the Normal Distribution as $$\{\mu_2,\mu_4\}=\{\sigma^2,3 \sigma^4\}$$

Progress:

First I solved for $$h_2$$, an unbiased estimate of $$\mu_2 = \mu'_2 - \mu'^2_1$$

$$h_2 = \mu'_2 - \mu'^2_1 \\ = \mu'_2 - \mu'_1 \mu'_1 \\ = \frac{A_{\{2\}}}{n} - \frac{A_{\{1,1\}}}{n(n-1)} \\ = \frac{S_2}{n} - \frac{S_1^2 - S_2}{n(n-1)} \\ = \frac{n S_2 - S_1^2}{(n-1)n}$$

Where used Fundamental Expectation result (eq 7.15 in ref) $$E[A_{\{a,b,c..\}}] = \mu'_a \mu'_b \mu'_c \cdots n (n-1) \cdots (n-\rho+1)$$ and the two facts: $$A_{\{2\}}=S_2$$ and $$A_{\{1,1\}}=S_1^2 - S_2$$

Problem

The second step is where I am having trouble: The second central moment of $$h_2$$. Do I use the augmented symmetrics, $$A_{\{a,b,c,\cdots\}}$$, again with the theorem? Additionally how to determine, from first principles, that $$A_{\{2\}}=S_2$$ and $$A_{\{1,1\}}=S_1^2 - S_2$$ ? I'm trying to understand the mathematics and methodology outside of the program.

The author even answered next-level question of what if the population values of $$\mu_4$$ and $$\mu^2_2$$ themselves were unknown, what is the unbiased estimate of $$Var[h_2]$$? answer That is outside the scope of this question, but hopefully with understanding how the general procedure all works that too can be done by hand as an exercise for myself.

Your chosen example illustrates that an algorithm that is completely general and works brilliantly with a computer can quickly become cumbersome for hand computation ... especially in this field. I think your chosen example quickly becomes too messy to solve by hand on a forum such as this (by this method). To illustrate the approach (which is completely general), we should choose a simpler example, such as:

Find: $$\text{Var}[\frac{s_2}{n}]$$

... where $$s_2 = \sum_{i=1}^n X_i^2$$. The step-by-step approach is exactly the same for finding $$\text{Var}[h_2]$$.

Step 1: $$\text{Var}[Z] = E[(Z-E[Z])^2] = E[\text{an expression}]$$ ... and hence can apply the Fundamental Expectation Result.

In this case, expand $$\left(\frac{s_2}{n}-\acute{\mu }_2\right)^2$$ as: $$\frac{s_2^2}{n^2}-\frac{2 s_2 \acute{\mu }_2}{n} + \acute{\mu }_2^2$$

Step 2: Express each term as an Augmented Symmetric function $$A_{}$$:

$$\frac{A_{\{2,2\}}+A_{\{4\}}}{n^2}-\frac{2 A_{\{2\}}\acute{\mu }_2}{n}+\acute{\mu }_2^2$$

Step 3: Apply the Fundamental Expectation Result (also Stuart and Ord (1994) - Section (12.5)):

$$E\left(A_{\{a,b,c,\ldots \}}\right)= \acute{\mu }_a \acute{\mu }_b \acute{\mu }_c \cdots \times n (n-1) \cdots (n-\rho +1)$$

... where given $$A_t$$, the symbol $$\rho$$ denotes the number of elements in the list $$t$$.

Doing so yields the following expectation for each term:

$$\frac{A_{\{2,2\}}}{n^2} \rightarrow \frac{(n-1) \acute{\mu }_2^2}{n}$$

$$\frac{A_{\{4\}}}{n^2} \rightarrow \frac{\acute{\mu }_4}{n}$$

$$\frac{2 A_{\{2\}}\acute{\mu }_2}{n} \rightarrow 2 \acute{\mu }_2^2$$

... which after simplifying yields:

$$\frac{\acute{\mu }_4-\acute{\mu }_2^2}{n}$$

• Was able to follow the given steps, along with answering the original question in the same manner - albeit with a lot more paper. Thank you! Commented Oct 11, 2023 at 17:24