# asymptotic distribution of joint random variables

I am trying to understand the asymptotic distribution of the following expression under normality $${\hat \sigma \hat S - \sigma S}$$

Where $\sigma$ and $S$ are the population standard deviation and population skewness respectively and $\hat{\sigma}$ and $\hat{S}$ are the sample standard deviation and sample skewness respectively. Any bright ideas, references or suggestions are appreciated

I also need to understand the following covariance structure $${\mathop{\rm cov}} \left[ {\hat \mu - \mu ,\hat \sigma \hat S - \sigma S} \right]$$

Where $\mu$ is population mean and $\hat{\mu}$ is the sample mean

What I have is some standard results $$\frac{{\sqrt n \left( {\hat \mu - \mu } \right)}}{\sigma }\xrightarrow{d}{\cal N}\left( {0,1} \right)$$

$$\,\frac{{\sqrt n \left( {\hat \sigma - \sigma } \right)}}{\sigma }\xrightarrow{d}{\cal N}\left( {0,\frac{{\kappa - 1}}{4}} \right)$$

$$\sqrt {\frac{N}{6}} \hat S\xrightarrow{d}{\cal N}\left( {0,1} \right)$$

• Under normality you have a much stronger result than $\frac{{\sqrt n \left( {\hat \mu - \mu } \right)}}{\sigma }\xrightarrow{d}{\cal N}\left( {0,1} \right)$ -- since that is the distribution for any $n$ (i.e. you don't need to deal with asymptotic results for that). – Glen_b Oct 29 '14 at 0:40
• By 'skewness' do you mean the moment-skewness? -- are you just estimating that by $\hat{S}=m_3/s^3$ where $m_3=\tfrac{1}{n} \sum_{i=1}^n (x_i-\overline{x})^3$? – Glen_b Oct 29 '14 at 0:54
• That is definition of skewness I am using and a similar definition of kurtosis should you need it – Rohit Arora Oct 29 '14 at 1:06

I will use the established symbol for skewness, $\gamma_1$. The population Skewness is

$$\gamma_1 = \frac {\mu_3}{\sigma^3} \Rightarrow \sigma \gamma_1 = \frac {\mu_3}{\sigma^2}$$

where $\mu_3$ is the third central moment. For the normal distribution this is zero, so the product $\sigma\gamma_1$ will also be zero.

Assuming an i.i.d. normal sample of size $n$, using the sample analogues of the moments (no need to apply finite sample corrections since we are interested in the asymptotic behavior)

$$\hat \sigma \hat \gamma_1 = \frac {(1/n)\sum_{i=1}^n \left(X_i-\bar X\right)^3}{(1/n)\sum_{i=1}^n \left(X_i-\bar X\right)^2} = \frac {\hat \mu_3}{\hat \sigma^2} \xrightarrow{p} 0$$

Then (see for example Dasgupta 2008, Theorem 3.8, page 42) we have that

$$\sqrt n\hat \mu_3 \xrightarrow{d} {\cal N}(0, V_3)$$

$$V_3 = \mu_6 - \mu_3^2 -6\sigma^2\mu_4 + 9\sigma^6,\;\; \mu_k = E(X-\mu)^k$$

Using the values and relations between the central moments of the normal distribution we obtain $V_3 = 6\sigma^6$

Since moreover $\hat \sigma^2 \xrightarrow{p}\sigma^2$, we obtain

$$\sqrt n \hat \sigma \hat \gamma_1 = \frac {\sqrt n\hat \mu_3}{\hat \sigma^2} \xrightarrow{d} {\cal N}(0, 6\sigma^{2})$$

by Slutsky's lemma. Of course the above could be directly derived using again Slutsky's lemma from

$$\sqrt {\frac{n}{6}} \hat \gamma_1 \xrightarrow{d} {\cal N}\left( 0,1 \right) \Rightarrow \sqrt n \hat \sigma \hat \gamma_1 \xrightarrow{d} \sigma \sqrt 6 {\cal N}(0,1) = {\cal N}\left( {0,6\sigma^2}\right)$$

• Just to clarify the use of Slutsky's Lemma in the last line. Is the following correct ?$$\begin{array}{l} {{\hat \sigma }^2}\xrightarrow{p} {\sigma ^2} \Rightarrow \frac{\sigma }{{\hat \sigma }}\xrightarrow{p} 1\\ \sqrt {\frac{N}{6}} {{\hat \gamma }_1}\xrightarrow{d} {\cal N}\left( {0,1} \right) \Rightarrow \frac{{\sqrt {\frac{N}{6}} {{\hat \gamma }_1}}}{{\frac{\sigma }{{\hat \sigma }}}}\xrightarrow{d}{\cal N}\left( {0,1} \right) \end{array}$$ – Rohit Arora Oct 29 '14 at 3:55
• Also, how would I evaluate the covariance. $${\mathop{\rm cov}} \left[ {\hat \mu - \mu ,\hat \sigma \hat S - \sigma S} \right]$$. – Rohit Arora Oct 29 '14 at 4:00
• About Slutsky's lemma: the first relation is an application of the Continuous Mapping (Mann-Wald) Theorem - not of Slutsky's (Slutsky comes into play when we have two random variables). The second is Slutsky. About the covariance: what do you mean by "evaluate"? Finite-sample? Asymptotically? Why don't you use the usual expression of covariance and see where it gets you? – Alecos Papadopoulos Oct 29 '14 at 8:19
• It was easy to evaluate $${\mathop{\rm cov}} \left[ {\hat \mu - \mu ,\hat \sigma - \sigma } \right] = {\mathop{\rm cov}} \left[ {\hat \mu ,\hat \sigma } \right]$$ by using bivariate delta method and the joint distribution of mean and variance is well known but in this case I have no idea about any joint distributions. Hnce, can't go beyond $${\mathop{\rm cov}} \left[ {\hat \mu - \mu ,\hat \sigma \hat S - \sigma S} \right] = {\mathop{\rm cov}} \left[ {\hat \mu ,\hat \sigma \hat S} \right]$$ – Rohit Arora Oct 29 '14 at 11:33