How to compute exact confidence interval for the third moment of normal distribution $N(a, \sigma^2)$?

**Answer**

In order to find a confidence interval for this quantity you will need to form a pivotal quantity that uses the third raw moment as its only unknown parameter. It might not be possible to do this exactly, but you can usually get something that is an *approximately* pivotal quantity that can be used to form an approximate confidence interval. In order to do this, we will first find the form of the third raw moment that is being estimated, then construct a sample estimator of this moment, and then try to use this to construct a quasi-pivotal quantity and resulting confidence interval.

**What is the third raw moment of a normal distribution?** Take $X \sim \text{N}(\mu, \sigma^2)$ to be an arbitrary normal random variable and define $Y = X-\mu \sim \text{N}(0, \sigma^2)$. The third raw moment of $X$ is:

$$\begin{equation} \begin{aligned}

\mu_3 \equiv \mathbb{E}(X^3)

&= \mathbb{E}((\mu + Y)^3) \\[6pt]

&= \mathbb{E}(Y^3 + 3 \mu Y^2 + 3 \mu^2 Y + \mu^3) \\[6pt]

&= 0 + 3 \mu \sigma^2 + 0 + \mu^3 \\[6pt]

&= 3 \mu \sigma^2 + \mu^3. \\[6pt]

\end{aligned} \end{equation}$$

This is the parameter you are trying to estimate in your analysis.

**Unbiased estimator of the third raw moment:** Ordinarily we would estimate the mean parameter with the sample mean and the variance parameter with the sample variance, but in this case we want to estimate a function of these things, and substitution of these estimators is likely to lead to a biased estimator. We will start by trying to find an unbiased estimator of the third raw moment. To do this, we begin by noting that:

$$\begin{equation} \begin{aligned}

\mathbb{E}(\bar{X}_n^3)

&= \mathbb{E}((\mu + \bar{Y}_n)^3) \\[6pt]

&= \mathbb{E}(\bar{Y}_n^3 + 3 \mu \bar{Y}_n^2 + 3 \mu^2 \bar{Y}_n + \mu^3) \\[6pt]

&= 0 + 3 \mu \frac{\sigma^2}{n} + 0 + \mu^3 \\[6pt]

&= \frac{3}{n} \mu \sigma^2 + \mu^3. \\[6pt]

\end{aligned} \end{equation}$$

We know from Cochran’s theorem that the sample mean and sample variance from normal data are independent, and so we also have $\mathbb{E}(\bar{X}_n S_n^2) = \mathbb{E}(\bar{X}_n) \mathbb{E}(S_n^2) = \mu \sigma^2$. Hence, based on these results, we can form the **unbiased estimator**:

$$\hat{\mu}_3 = \frac{3(n-1)}{n} \cdot \bar{X}_n S^2 + \bar{X}_n^3.$$

**Variance of the estimator:** We know that the expected value of this estimator is equal to the third raw moment of the distribution (to see this, simple substitute the above expected value expressions), however the variance of the estimator is laborious to derive. As preliminary results we have:

$$\begin{equation} \begin{aligned}

\mathbb{V}(\bar{X}_n S^2)

&= \mathbb{V}(\bar{X}_n) \mathbb{V}(S^2) \\[6pt]

&= \frac{1}{n} \sigma^2 \cdot \frac{2}{n-1} \sigma^4 \\[6pt]

&= \frac{2}{n(n-1)} \sigma^6, \\[12pt]

\mathbb{V}(\bar{X}_n^3)

&= \mathbb{E}(\bar{X}_n^6) – \mathbb{E}(\bar{X}_n^3)^2 \\[6pt]

&= \Big( \frac{15}{n^3} \sigma^6 + \frac{45}{n^2} \mu^2 \sigma^4 + \frac{15}{n} \mu^4 \sigma^2 + \mu^6 \Big) – \Big( \frac{3}{n} \mu \sigma^2 + \mu^3 \Big)^2 \\[6pt]

&= \Big( \frac{15}{n^3} \sigma^6 + \frac{45}{n^2} \mu^2 \sigma^4 + \frac{15}{n} \mu^4 \sigma^2 + \mu^6 \Big) – \Big( \frac{9}{n^2} \mu^2 \sigma^4 + \frac{6}{n} \mu^4 \sigma^2 + \mu^6 \Big) \\[6pt]

&= \frac{15}{n^3} \sigma^6 + \frac{36}{n^2} \mu^2 \sigma^4 + \frac{9}{n} \mu^4 \sigma^2, \\[12pt]

\mathbb{C}(\bar{X}_n S^2, \bar{X}_n^3)

&= \mathbb{E}(\bar{X}_n^4 S^2) – \mathbb{E}(\bar{X}_n S^2) \mathbb{E}(\bar{X}_n^3) \\[6pt]

&= \mathbb{E}(\bar{X}_n^4) \mathbb{E}(S^2) – \mathbb{E}(\bar{X}_n) \mathbb{E}(\bar{X}_n^3) \mathbb{E}(S^2) \\[6pt]

&= \Big( \frac{3}{n^2} \sigma^4 + \frac{6}{n} \mu^2 \sigma^2 + \mu^4 \Big) \sigma^2 – \mu \Big( \frac{3}{n} \mu \sigma^2 + \mu^3 \Big) \sigma^2 \\[6pt]

&= \Big( \frac{3}{n^2} \sigma^4 + \frac{6}{n} \mu^2 \sigma^2 + \mu^4 \Big) \sigma^2 – \Big( \frac{3}{n} \mu^2 \sigma^2 + \mu^4 \Big) \sigma^2 \\[6pt]

&= \Big( \frac{3}{n^2} \sigma^4 + \frac{3}{n} \mu^2 \sigma^2 \Big) \sigma^2 \\[6pt]

&= \frac{3}{n^2} \sigma^6 + \frac{3}{n} \mu^2 \sigma^4. \\[6pt]

\end{aligned} \end{equation}$$

This gives us the variance:

$$\begin{equation} \begin{aligned}

\mathbb{V}(\hat{\mu}_3)

&= \mathbb{V} \Big( \frac{3(n-1)}{n} \cdot \bar{X}_n S^2 + \bar{X}_n^3 \Big) \\[6pt]

&= \frac{9(n-1)^2}{n^2} \cdot \mathbb{V}(\bar{X}_n S^2) + \mathbb{V}(\bar{X}_n^3) + \frac{3(n-1)}{n} \cdot \mathbb{C} (\bar{X}_n S^2, \bar{X}_n^3) \\[6pt]

&= \frac{18(n-1)}{n^3} \sigma^6 + \Big( \frac{15}{n^3} \sigma^6 + \frac{36}{n^2} \mu^2 \sigma^4 + \frac{9}{n} \mu^4 \sigma^2 \Big) + \Big( \frac{9(n-1)}{n^3} \sigma^6 + \frac{9(n-1)}{n^2} \mu^2 \sigma^4 \Big) \\[6pt]

&= \frac{27n-12}{n^3} \cdot \sigma^6 + \frac{9n+27}{n^2} \cdot \mu^2 \sigma^4 + \frac{9}{n} \cdot \mu^4 \sigma^2 \\[6pt]

&= \frac{3}{n^3} \Big[ (9n-4) \sigma^6 + (3n^2+9n) \mu^2 \sigma^4 + 3n^2 \mu^4 \sigma^2 \Big]. \\[6pt]

\end{aligned} \end{equation}$$

**Forming a confidence interval:** From the above results, we can obtain an unbiased estimator for the third raw moment, with known variance. The exact distribution of this estimator is complicated, and its density cannot be expressed in closed-form. It is possible to form a studentised quantity with this estimator, approximate its distribution, and treat it as a quasi-pivotal quantity to obtain an approximate confidence interval. However, this would not be an exact confidence interval.

**Attribution***Source : Link , Question Author : Lilith , Answer Author : Ben*