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Why has the Jarque-Bera test of normality two degrees of freedom?

My initial thought was that the sum of two squared standard normals (i.e., skewness and kurtosis in this test) should have n-1 = 2-1 = 1 df. So, why has it two?

Any help is very appreciated.

The test is described here: Wikipedia link

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  • $\begingroup$ You should note that the approach to bivariate normality and independence of the sample skewness and kurtosis is quite slow, necessitating small sample modification of the test for sample sizes below several hundred (at least). This issue was discussed in some detail by Bowman and Shenton, (1975). [i.e. five years before the paper by Jarque and Bera] $\endgroup$ – Glen_b -Reinstate Monica Dec 5 '19 at 12:34
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"$\chi^2$-distribution with $k$ degrees of freedom is the distribution of a sum of the squares of $k$ independent standard normal random variables" (Wikipedia: Chi-squared distribution)

So in the case of Jarque-Bera test the degrees of freedom must be 2, because - as you have stated - the test statistics is sum of two squared standard normally distributed values (skewness and kurtosis).

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