# Moments

Can uniquely characterize a distribution

## Central Moments

Central Moments: $$\mu_r=\frac{\sum(x_i-\bar x)^r}{n}$$

• $$\mu_1=\frac{\sum(x_i-\bar x)}{n}=\frac{\sum x_i}{n}-\frac{n \bar x}{n}=\frac{n \bar x}{n}-\frac{n \bar x}{n}=0$$
• $$\mu_2=\frac{\sum(x_i-\bar x)^2}{n}=\sigma^2$$
• $$\mu_3=\frac{\sum(x_i-\bar x)^3}{n}$$
• $$\mu_4=\frac{\sum(x_i-\bar x)^4}{n}$$
• For grouped data: $$\mu_r=\frac{\sum f_i(x_i-\bar x)^3}{n}$$

## Raw Moments

$$\mu_r'=\frac{\sum(x_i-a)^r}{n}$$; a is arbitrary number

$$\mu_1'=\frac{\sum(x_i-a)}{n}=\frac{\sum x_i}{n}-\frac{na}{n}=\bar x-a$$

## Raw & Central Moments: Relationship

• $$\mu_2 = \mu_2'-\mu_1'^2$$
• $$\mu_3 = \mu_3'-3\mu_2'\mu_1'+2\mu_1'^3$$
• $$\mu_4 = \mu_4'-4\mu_3'\mu_1' + 6 \mu_2'\mu_1'^2-3\mu_1'^4$$

How to Remember??

$$\downarrow$$

## How to Remember (Skip)

$$(a-b)^3=a^3-3a^2b+3ab^2-b^3$$

$$(a-b)^4=a^4-4a^3b+6a^2b^2-4ab^3+b^4$$ (Pascal triangle can be used)

$$a^r=\mu_r'$$

$$b^r=\mu_1'^r$$

Now,

$$\mu_3 = \mu_3'-3\mu_2'\mu_1'+2\mu_1'^3$$

• Remove the penultimate term and let coefficient of it is k
• Use (-k) as the coefficient of the last term.

## Changing Origin of Moments

From $$\mu_r'(a)$$ to $$\mu_r'(k)$$

Assume, $$a^r = \mu_r'(a)$$, $$b = a - k$$

Binomial Formulae

1. $$(a+b)^1 = a + b$$
2. $$(a+b)^2 = a^2 + 2ab + b^2$$
3. $$(a+b)^3=a^3+3a^2b+3ab^2+b^3$$
4. $$(a+b)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4$$
1. $$\mu_1'(k) = \mu_r'(a) + b$$
1. $$\mu_2'(k) = \mu_2'(a) + 2 \mu_1'(a) + b^2$$
1. ?
1. ?

## Central Moments from Raws

Use $$k = \bar x$$

That’s it! DO NOT MEMORIZE!

## Skewness

Lack of symmetry

• -Ve Skew $$\rightarrow \bar X \lt Me \lt Mode$$
• +Ve Skew $$\rightarrow \bar X \gt Me \gt Mode$$
• No Skew $$\rightarrow \bar X = Me = Mode$$

## Kurtosis

Normal distribution $$\rightarrow$$

• Most students get average marks
• Higher and lower marks are obtained by lesser number of students

## Measures of SKewness

• Pearson’s Coefficient: $$SK_P=\frac{Mean-Mode}{\sigma}=\frac{3(Mean-Median)}{\sigma} ;(-3,3)$$

($$Mode=3Me-2\bar X$$)

• Bowley’s Coefficient: $$SK_B=\frac{Q_3+Q_1-2Me}{Q_3-Q_1}; (-1,1)$$
• Kelly’s Coefficient: $$SK_k=\frac{D_1+D_9-2Me}{D_9-D1}$$
• Method of Moments: $$\beta_1=\frac{\mu_3^2}{\mu_2^3}$$

Estimate Skewness

4, 23, 55, 70, 74, 78, 86, 89

## Measures of Kurtosis

• Pearsons’s Coefficient of Moments, $$\beta_2 = \frac{\mu_4}{\mu_2^2}$$
• Percentile Coefficient, $$K=\frac{\frac 1 2 (Q_3-Q_1)}{P_{90}-P_{10}}$$

$$\gamma_2=\beta_2-3$$

$$\gamma_2=0 \space or \space \beta_2=3) \rightarrow$$ Mesokurtik $$\gamma_2\gt0 \space or \space \beta_2 \gt 3) \rightarrow$$ Leptokurtik $$\gamma_2\lt0 \space or \space \beta_2\lt3) \rightarrow$$ Platykurtik

9, 7, 8, 6 $$\rightarrow$$ Find kurtosis

## Five Number Summary

• Minimum
• Maximum
• $$Q_1, Q_2, Q_3$$

## Problems

1. $$\bar X = 65, Me = 70, SK_P = -0.5$$; Mod = ? CV = ?
1. GM & HM of quartiles of a symmetrical distribution are 8 and 6.4, respectively. Find median.
1. Mean, SK(P), coefficient of skewness and CV of distribution are 50, 0.4, and 40%, respectively. Find SD, mode, & median.
1. Variance of a mesokurtik distribution is 4. Find $$\mu_4$$.

## Moments Conversion Problems

1. First three moments about 2 are 1, 16, and -40, , respectively. Find them about 0.
2. First four moments about 5 are 2, 20,40, and 50, respectively. Find the central moments.
3. First three moments about 2 are -1, 6, and 30, respectively. Find them about 5.
• 3, 24, 76
• 0, 16, -64, 162
• -4, 21, -78