The arithmetic mean and variance give us an information on variability (dispersion, spread) of a set of numbers. Therefore, in that way we get informations about the appearance of the distribution of the numbers. Also, they are known as the first two statistical moments.
If we want to know more about the shape of the distribution, we need to use certain measures which are called the third and fourth moments.
Definition and types of moments
Let $X:S \rightarrow X(S) = \{x_{1}, x_{2}, \cdots, x_{k}\} \subseteq \mathbf{R}$ be the numeric variable given on finite set with $N$ elements, with values $y_{1}, \cdots, y_{N}$. Furthermore, let $\{(x_{1}, f_{1}), \cdots, (x_{k}, f_{k})\}$ be its distribution and $\mu$ its arithmetic mean.
Moment of $X$ is the mean of deviations of values of $X$ from its arithmetic mean $\mu$ (central moment) or from some other value raised to a certain power $r \in \mathbf{N_{0}}$. In other words,
r – th central moment is defined as: $$\mu_{r} = \frac{\sum_{i = 1}^{k}f_{i}(x_{i} – \mu)^{r}}{\sum_{i = 1}^{k}f_{i}} = \frac{\sum_{i = 1}^{N}(y_{i} – \mu)^{r}}{N}.$$
Similarly, the r – th moment about the origin is defined as:
$$m_{r} = \frac{\sum_{i = 1}^{k} f_{i}x_{i}^{r}}{\sum_{i = 1}^{k}f_{i}} = \frac{\sum_{i = 1}^{N}y_{i}^{r}}{N}.$$
Obviously, $m_{1} = \mu$.
The first central moment is always equal to $0$, i.e. $\mu_{1} = 0$. This can be seen in the following:
$$\mu_{1} = \frac{f_{1}(x_{1} – \mu) + f_{2}(x_{2} – \mu) + \cdots + f_{k}(x_{k} – \mu)}{N} $$
$$= \frac{-\mu (f_{1} + f_{2} + \cdots + f_{k}) + f_{1}x_{1} + \cdots + f_{k}x_{k}}{N}$$
$$= \mu – \mu = 0$$
The second central moment is the variance, i.e. $\mu_{2} = \sigma^{2}$.
Moment $\mu_{3}$ represents the skewness of data in relation to the $\mu$, while $\mu_{4}$ represents the kurtosis of distribution.
Skewness and kurtosis
Skewness is a measure of the symmetry of the shape of a distribution. The value of skewness can be negative, zero or positive.
The skewness is equal to zero if a distribution is symmetric. In this case we say that the distribution is symmetrical.
It is negative if there is a long tail in the negative direction. Therefore, the distribution is negatively skewed.
Finally, it is positive if there is a long tail in the positive direction. In other words, the distribution is positively skewed.
Kurtosis is a measure of peakedness or flatness of a distribution. If a distribution is flat – looking, it is called ‘platykurtic’. Furthermore, if a distribution is peaked, it is called ‘leptokurtic’.
Examples
Example 1: Calculate the first four central moments for the following numbers: $3, 4, 8, 9, 11$.
Solution:
$$x_{1} = 3, x_{2} = 4, x_{3} = 8, x_{4} = 9, x_{5} = 11 \rightarrow \sum_{i = 1}^{5}f_{i} = 5$$
$$\mu = \frac{3 + 4 + 8 + 9 + 11}{5} = \frac{35}{5} = 7$$
$$\mu_{1} = \frac{(3 – 7) + (4 – 7) + (8 – 7) + (9 – 7) + (11 – 7)}{5} = 0$$
$$\mu _{2} = \frac{(3 – 7)^{2} + (4 – 7)^{2} + (8 – 7)^{2} + (9 – 7)^{2}+ (11 – 7)^{2}}{5} = \frac{46}{5} = 9.2$$
$$\mu_{3} = \frac{(3 – 7)^{3} + (4 – 7)^{3} + (8 – 7)^{3} + (9 – 7)^{3}+ (11 – 7)^{3}}{5} = -\frac{18}{5} = -3.6$$
$$\mu_{4} = \frac{(3 – 7)^{4} + (4 – 7)^{4} + (8 – 7)^{4} + (9 – 7)^{4}+ (11 – 7)^{4}}{5} = \frac{610}{5} = 122.$$
Example 2: In the following table marks distribution of $100$ students is given. Calculate the first four central moments.
Solution:
Class marks are $5, 15, 25, 35, 45, 55, 65$, respectively. Furthermore, $\sum_{i = 1}^{7}f_{i} = 100$ and
$$\mu = \frac{10 \cdot 5 + 14 \cdot 15 + 26 \cdot 25 + 9 \cdot 35 + 28 \cdot 45 + 4 \cdot 55 + 9 \cdot 65}{100} = \frac{3290}{100} = 32.9$$
$\mu_{1}$ is always equal to $0$.
$$\mu_{2} = \frac{10(5 – 32.9)^{2}+ 14(15 – 32.9)^{2} + 26(25 – 32.9)^{2}+ 9(35 – 32.9)^{2} + 28(45 – 32.9)^{2} + 4(55 – 32.9)^{2} + 9(65 – 32.9)^{2}}{100}$$
$$\mu_{2} = \frac{29259}{100} = 292.59$$
$$\mu_{3}= \frac{10(5 – 32.9)^{3}+ 14(15 – 32.9)^{3} + 26(25 – 32.9)^{3}+ 9(35 – 32.9)^{3} + 28(45 – 32.9)^{3} + 4(55 – 32.9)^{3} + 9(65 – 32.9)^{3}}{100}$$
$$\mu_{3} = \frac{80257.8}{100} = 802.578$$
$$\mu_{4} = \frac{10(5 – 32.9)^{4}+ 14(15 – 32.9)^{4} + 26(25 – 32.9)^{4}+ 9(35 – 32.9)^{4} + 28(45 – 32.9)^{4} + 4(55 – 32.9)^{4} + 9(65 – 32.9)^{4}}{100}$$
$$\mu_{4} = \frac{18708027.57}{100} = 187080.2757.$$
