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 $0$.

$$\mu_{2} = \frac{10(5 – 32.9)^{2}+ 14(15 – 32.9)^{2} + 26(25 – 32.9)^{2}+ 9(35 – 32.9)^{2} + 28(45 – 32.9)^{2} + 4(55 – 32.9)^{2} + 9(65 – 32.9)^{2}}{100}$$

$$\mu_{2} = \frac{29259}{100} = 292.59$$

$$\mu_{3}= \frac{10(5 – 32.9)^{3}+ 14(15 – 32.9)^{3} + 26(25 – 32.9)^{3}+ 9(35 – 32.9)^{3} + 28(45 – 32.9)^{3} + 4(55 – 32.9)^{3} + 9(65 – 32.9)^{3}}{100}$$

$$\mu_{3} = \frac{80257.8}{100} = 802.578$$

$$\mu_{4} = \frac{10(5 – 32.9)^{4}+ 14(15 – 32.9)^{4} + 26(25 – 32.9)^{4}+ 9(35 – 32.9)^{4} + 28(45 – 32.9)^{4} + 4(55 – 32.9)^{4} + 9(65 – 32.9)^{4}}{100}$$

$$\mu_{4} = \frac{18708027.57}{100} = 187080.2757.$$