###### Lognormal Distribution and Continuous ...

A random variable \(Y\) is lognormally distributed if its natural logarithm, In \(Y\),... **Read More**

Measures of dispersion are used to describe the variability or spread in a sample or population. They are usually used in conjunction with measures of central tendency such as the mean and the median. These are the range, variance, absolute deviation, and standard deviation. They are important because they give us an idea of how well the measures of central tendency represent the data. For example, if the standard deviation is large, there are large differences between individual data points. Consequently, the mean may not be representative of the data.

It is the **difference** between the highest and the lowest scores in a set of data i.e.

$$ \text{Range} = \text{maximum value} – \text{minimum value} $$

**Example 1**

Consider the following scores of 10 CFA Level 1 candidates:

78 56 67 51 43 89 57 67 78 50

$$ \text{Range} = 89 – 43 = 46 $$

It is a measure of dispersion representing the **average of the absolute values** of the deviations of individual observations from the arithmetic mean. Therefore:

$$ \text{MAD} \frac { \sum { |{ X }_{ i }-\bar { X } | } }{ n } $$

Remember that the sum of deviations from the arithmetic means is always zero, and that’s why we are using the **absolute** **values**.

**Example 2**

6 investment analysts attain the following returns on six different investments:

{12% 4% 23% 8% 9% 16%}

Calculate the mean absolute deviation and interpret it.

**Solution**

First, we have to calculate the arithmetic mean:

$$ X =\cfrac {(12 + 4 + 23 + 8 + 9 + 16)}{6} = 12\% $$

Next, we can now compute the MAD:

$$ \begin{align*} \text{MAD} & = \cfrac {\left\{ |12 – 12|+ |4 – 12| + |23 – 12| + |8 – 12| + |9 – 12| + |16 – 12| \right\}} {6} \\ & =\cfrac {30}{6} \\ & = 5\% \\ \end{align*} $$

Interpretation: it means that on average, an individual return deviates 5% from the mean return of 12%.

The population variance, denoted by σ^{2}, is the average of the squared deviations from the mean. Therefore:

$$ { \sigma }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }-\mu \right) }^{ 2 } } \right\} }{ N } $$

And the standard deviation is simply the square root of variance.

**Example 3**

Working with data from example 2 above, the variance will be calculated as follows:

$$ \begin{align*} { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 6 } \\ & = 37.67(\%^2) \\ & = 0.003767 \\ \end{align*} $$

Thus, the average variation from the mean (0.12) is 0.003767.

The standard deviation is 0.003767^{1/2} = 0.06137 or 6.14%.

Analysts use the standard deviation, instead of the variance, to interpret returns since it is much easier to comprehend.

The sample variance, S^{2}, is the measure of dispersion that applies when we are working with a sample as opposed to a population. (The two have been distinguished here)

$$ { S }^{ 2 }=\frac { \left\{ \sum { { \left( { X }_{ i }- \bar { X } \right) }^{ 2 } } \right\} }{ n-1 } $$

Note that we are dividing by n – 1. This is necessary so as to remove **bias**.

The sample standard deviation, S, is simply the square root of the sample variance.

**Example 4**

Assume that the returns realized in example 2 above were sampled from a population comprising 100 returns. Compute the sample mean and the corresponding sample variance.

**Solution**

The sample mean will still be 12%.

Hence,

$$ \begin{align*} { S }^{ 2 } & =\frac { \left\{ { \left( 12-12 \right) }^{ 2 }+{ \left( 4-12 \right) }^{ 2 }+{ \left( 23-12 \right) }^{ 2 }+{ \left( 8-12 \right) }^{ 2 }+{ \left( 9-12 \right) }^{ 2 }+{ \left( 16-12 \right) }^{ 2 } \right\} }{ 5 } \\ & = 45.20(\%^2) \\ & = 0.00452 \\ \end{align*} $$

Therefore,

$$ \begin{align*} S & = 0.00452^{\frac {1}{2}} \\ & = 0.0672 \end{align*} $$

QuestionYou have been given the following data:

{12 13 54 56 25}

Assuming this is complete data from a certain population, the population standard deviation is

closestto:

- 19.34.
- 374.
- 1,870.
The correct answer is A.

Working:

$$ \mu =\cfrac {(12 + 13 + \cdots +25)}{5} =\cfrac {160}{5} = 32 $$

Hence,

$$ \begin{align*} { \sigma }^{ 2 } & =\frac { \left\{ { \left( 12-32 \right) }^{ 2 }+{ \left( 13-32 \right) }^{ 2 }+{ \left( 54-32 \right) }^{ 2 }+{ \left( 56-32 \right) }^{ 2 }+{ \left( 25-32 \right) }^{ 2 } \right\} }{ 5 } \\ & =\cfrac {1870}{5} = 374 \\ \end{align*} $$

Therefore,

$$ \sigma = 19.34 $$