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the standard error of a standard deviation is given as : s/n^(1/2). Would the standard error of kurtosis and skewness follow the same idea? For example, se of kurtosis = kurtosis/n^(1/2)?

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Are you asking about the standard deviation of moment estimators? For a estimator of the mean, i.e. $$\hat{m} = \frac{1}{n} \sum_{k=1}^n X_k $$ the variance is $$ \mathbb{Var}(\hat{m}) = \mathbb{E}(\hat{m}^2) - \mathbb{E}(\hat{m})^2 = \frac{\mathbb{Var}(X)}{n}$$ which is what you claimed first. So, rephrasing the question, are you asking for variances of the skewness and kurtosis estimators? If so, you need to precise which estimators you have in mind, biased/unbiased, based on symmetric polynomials of the sample data, or on quantiles, etc. – Sasha Nov 3 '12 at 17:02
well, I already have a given data. I am trying to compute the standard error of the sample kurtosis of the data in R. So, I was wondering if the formula for Se = Kurtosis/n^(1/2) – Josh Nov 3 '12 at 17:13

The answer is no. Your "standard error of a standard deviation" is also incorrect (that's the standard error of the sample mean). You can calculate the distributions (and hence vasiance/standard errors) of higher sample moments assuming certain distributions, for example here for the Normal distribution--notice that it's pretty complicated. If you want to estimate the standard error of the sample kurtosis my recommendation would be to do bootstrapping.

Edit: Since you also mentioned skewness, you should note in particular that skewness can be negative, so the standard error according to your formula would end up being negative. That's not cool.

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