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I am trying to estimate the asymptotic covariance between mean and standard deviation. I know the following $$\sqrt n \hat \mu \xrightarrow{d}N\left( {\mu ,{\sigma ^2}} \right),\sqrt n \hat \sigma \xrightarrow{d}N\left( {\sigma ,{\sigma ^2}\frac{{\kappa - 1}}{4}} \right),{\Sigma _{\mu ,{\sigma ^2}}} = \left[ {\begin{array}{*{20}{c}} {{\sigma ^2}}&{{\mu _3}} \\ {{\mu _3}}&{{\mu _4} - {\sigma ^4}} \end{array}} \right]$$

I need to estimate the following $${\text{cov}}\left[ {\sqrt n \left( {\hat \mu - \mu } \right),\sqrt n \left( {\hat \sigma - \sigma } \right)} \right] = n{\text{cov}}\left[ {\hat \mu ,\hat \sigma } \right]$$

I am using the multivariate delta method with $$g\left( {\mu ,\sigma } \right) = \mu + \sigma$$

$$\begin{array}{l} var\left[ {g\left( {\mu ,\sigma } \right)} \right] = {\left( {\nabla g} \right)^ \top }\Sigma \nabla g\\ =\left[ {\begin{array}{*{20}{c}} {\frac{{\partial \left( {\mu + \sqrt {{\sigma ^2}} } \right)}}{{\partial \mu }}}&{\frac{{\partial \left( {\mu + \sqrt {{\sigma ^2}} } \right)}}{{\partial {\sigma ^2}}}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{\sigma ^2}}&{{\mu _3}}\\ {{\mu _3}}&{{\mu _4} - {\sigma ^4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {\frac{{\partial \left( {\mu + \sqrt {{\sigma ^2}} } \right)}}{{\partial \mu }}}\\ {\frac{{\partial \left( {\mu + \sqrt {{\sigma ^2}} } \right)}}{{\partial {\sigma ^2}}}} \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} 1&{\frac{1}{{2\sigma }}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{\sigma ^2}}&{{\mu _3}}\\ {{\mu _3}}&{{\mu _4} - {\sigma ^4}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 1\\ {\frac{1}{{2\sigma }}} \end{array}} \right] = \left( {{\sigma ^2} + \frac{{{\mu _3}}}{\sigma } + \frac{1}{{4{\sigma ^2}}}\left( {{\mu _4} - {\sigma ^4}} \right)} \right) \end{array}$$

Now I have $${\text{nvar}}\left[ {\hat \mu } \right] = {\sigma ^2},n{\text{var}}\left[ {\hat \sigma } \right] = {\sigma ^2}\frac{{\kappa - 1}}{4},n{\text{var}}\left[ {\hat \mu + \hat \sigma } \right] = \left( {{\sigma ^2} + \frac{{{\mu _3}}}{\sigma } + \frac{1}{{4{\sigma ^2}}}\left( {{\mu _4} - {\sigma ^4}} \right)} \right)$$

Now I will try to recover the covariance from the known variances $$\begin{array}{*{20}{l}} {{\text{cov}}\left[ {\hat \mu ,\hat \sigma } \right] = \frac{{{\text{var}}\left[ {\hat \mu + \hat \sigma } \right] - {\text{var}}\left[ {\hat \mu } \right] - {\text{var}}\left[ {\hat \sigma } \right]}}{2}} \\ { = \frac{1}{2}\left( {{\sigma ^2} + \frac{{{\mu _3}}}{\sigma } + \frac{1}{{4{\sigma ^2}}}\left( {{\mu _4} - {\sigma ^4}} \right)} \right) - \frac{{{\sigma ^2}}}{2} - \frac{{{\sigma ^2}}}{2}\frac{{\kappa - 1}}{4}} \\ { = \frac{1}{2}\left( {\frac{{{\mu _3}}}{\sigma } + \frac{{3{\sigma ^2}}}{4}} \right) = \frac{{{\sigma ^2}}}{2}\left( {{\gamma _1} + \frac{3}{4}} \right)} \end{array}$$

Does this analysis look correct? Is there a better way to recover the covariance? Are there any known results to for calculating covariance matrix for functions of random variables whose large scale asymptotic covariance is known

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  • $\begingroup$ The initial distributional expressions are problematic, since they have zero variance. Since you are deriving the asymptotic covariance, you need to examine $${\mathop{\rm cov}} \left[ \sqrt n{\left( {\hat \mu - \mu } \right),\sqrt n\left( {\hat \sigma - \sigma } \right)} \right] $$ . Don't you think that the $n$ in the denominator in your last line is problematic? $\endgroup$ – Alecos Papadopoulos Nov 2 '14 at 2:09
  • $\begingroup$ Does that look better ? $\endgroup$ – Kumar Nov 2 '14 at 3:26
  • $\begingroup$ Just curious: what does $\kappa$ denote? Also, one can derive cov(sample mean, sample variance) exactly ... if you are happy to work with the sample variance rather than standard deviation. $\endgroup$ – wolfies Nov 2 '14 at 5:11
  • $\begingroup$ $\kappa$ is the kurtosis. Covariance between mean and standard deviation is $\mu_3$ and is already embedded in the asymptotic covariance matrix above. $\endgroup$ – Kumar Nov 2 '14 at 15:42
  • $\begingroup$ @user2142 wrote: "Covariance between mean and standard deviation is μ3". Well, Cov(sample mean, sample variance) = $\mu_3$, so your assertion that Cov(sample mean, standard deviation) = $\mu_3$ appears false. $\endgroup$ – wolfies Nov 6 '14 at 4:19

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