Calculating the variance and the central moments with a dumb calculator can be a pain. My question is if I have the standard deviation is there a quick way to calculate the skewness and the kurtosis
2 Answers
In order to calculate the standard deviation, you need to calculate $\sum x_i^2$.
To calculate the skewness, you need to calculate $\sum x_i^3$.
To calculate the kurtosis, you need to calculate $\sum x_i^4$.
Your calculator will calculate the standard deviation for you, but probably does not calculate higher order moments. These higher moments will need to be calculated if you want the skewness or kurtosis.
Calculate $s_0= \sum 1 , s_1= \sum x_i , s_2= \sum x_i^{2} , s_3= \sum x_i^{3} , s_4= \sum x_i^{4} $ then the first second, third & fourth moment quantities of interest , the mean, standard deviation , skewness & kurtosis can be calculated as follows \begin{eqnarray*} \mu &=& \frac{s_1}{s_0} \\ \sigma &=& \sqrt{\frac{s_2-s_1^2}{s_0}} \\ \gamma_1 &=& \frac{s_3-3 s_1 s_2 +2s_1^3}{s_0 \sigma^3} \\ \kappa &=& \frac{s_4-4s_1 s_3 +6 s_1^2 s_2 -3 s_1^4}{s_0 \sigma^4}. \end{eqnarray*}
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how do i finish after i got $\sum x_i^3$? – JDizzle Jun 04 '17 at 14:06
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Is there a source for these formulas? – Ben Jones Jan 15 '19 at 14:05
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@BenJones The first two formulae are standard, if I remember rightly, I derived the other two by hand. If you google Skewness & Kurtosis and look at the wiki pages for their definition, you will find it quite easy to derive these formulae. Send me another message if you need more help. $\ddot \smile$ – Donald Splutterwit Jan 15 '19 at 19:53
Skewness: $$skw(X) = E\left[\left(\frac{X - \mu}{\sigma}\right)^3\right]$$ $skew(X)$ can be expressed in terms of the first three moments of $X$: $$skw(X) = \frac{E\left(X^3\right) - 3 \mu E\left(X^2\right) + 2 \mu^3}{\sigma^3} = \frac{E\left(X^3\right) - 3 \mu \sigma^2 - \mu^3}{\sigma^3}$$ Kurtosis: $$kur(X) = E\left[\left(\frac{X - \mu}{\sigma}\right)^4\right]$$
$kurt(X)$ can be expressed in terms of the first four moments of $X$: $$kur(X) = \frac{E\left(X^4\right) - 4 \mu E\left(X^3\right) + 6 \mu^2 E\left(X^2\right) - 3 \mu^4}{\sigma^4} = \frac{E\left(X^4\right) - 4 \mu E\left(X^3\right) + 6 \mu^2 \sigma^2 + 3 \mu^4}{\sigma^4}$$
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Could you please tell me what is the distribution of $X$ you working on it? – Ahmed Jun 04 '17 at 14:27