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When trying to estimate downside risk (i.e., returns below the mean), we can use the following measures:
The target semi deviation, \(s_{\text {Target }}\), is calculated as follows:
$$s_{\text {Target }}=\sqrt{ \sum_{\text {for all } X_{i} \leq B}^{n} \frac{\left(X_{i}-B\right)^{2}}{n-1}}$$
Where \(B\) is the target and \(n\) is the total number of sample observations.
Yearly returns of an equity mutual fund are provided as follows.
$$
\begin{array}{c|c}
\textbf { Month } & \textbf { Return % } \\
\hline 2010 & 36 \% \\
\hline 2011 & 29 \% \\
\hline 2012 & 10 \% \\
\hline 2013 & 52 \% \\
\hline 2014 & 41 \% \\
\hline 2015 & 16 \% \\
\hline 2016 & 10 \% \\
\hline 2017 & 23 \% \\
\hline 2018 & -10 \% \\
\hline 2019 & -19 \% \\
\hline 2020 & 2 \% \\
\end{array}
$$
What is the target downside deviation if the target return is 20%?
Solution
$$
\begin{array}{c|c|c|c|c}
\textbf { Month } & \begin{array}{c}
\textbf { Return } \\
\%
\end{array} & \begin{array}{c}
\textbf { Deviation } \\
\textbf { from the 20% } \\
\textbf { target }
\end{array} & \begin{array}{c}
\textbf { Deviation } \\
\textbf { below the } \\
\textbf { target }
\end{array} & \begin{array}{c}
\textbf { Squared } \\
\textbf { deviations } \\
\textbf { below the } \\
\textbf { target }
\end{array} \\
\hline 2010 & 36.00 & 16.00 & – & – \\
\hline 2011 & 29.00 & 9.00 & – & – \\
\hline 2012 & 10.00 & (10.00) & (10.00) & 100 \\
\hline 2013 & 52.00 & 32.00 & – & \\
\hline 2014 & 41.00 & 21.00 & – & \\
\hline 2015 & 16.00 & (4.00) & (4.00) & 16 \\
\hline 2016 & 10.00 & (10.00) & (10.00) & 100 \\
\hline 2017 & 23.00 & 3.00 & – & \\
\hline 2018 & (10.00) & (30.00) & (30.00) & 900 \\
\hline 2019 & (19.00) & (39.00) & (39.00) & 1,521 \\
\hline 2020 & 2.00 & (18.00) & (18.00) & 324 \\
\hline {\text { Sum }} & {}&{}&{}&{\textbf{2,961}}\\
\end{array}
$$
Here \(n = 11 – 1 = 10\) so that:
$$\text{Target semi-deviation} = \left(\frac{2961 }{10}\right)^{0.5} = 17.21\%$$
The coefficient of variation, \(CV\), is a measure of spread that describes the amount of variability of data relative to its mean. It has no units, so we can use it as an alternative to the standard deviation to compare the variability of data sets that have different means. The coefficient of variation is given by:
$$ \text{CV} = \cfrac {S}{\bar{X}} $$
Where:
\(S\) = The standard deviation of a sample.
\(\bar{X}\) = The mean of the sample.
Note: The formula can be replaced with \(\frac{σ}{μ}\) when dealing with a population.
Procedure to Follow While Calculating the Coefficient of Variation:
What is the relative variability for the samples 40, 46, 34, 35, and 45 of a population?
Solution
Step 1: Calculate the mean.
$$ \text{Mean} =\cfrac {(40 + 46 + 34 + 35 + 45)}{5} =\cfrac {200}{5} = 40 $$
Step 2: Calculate the sample standard deviation. (Start with the variance, \(S^2\).)
$$ \begin{align*} S^2 & =\cfrac {{(40 – 40)^2 + … + (45 – 40)^2 }}{4} \\ &=\cfrac {122}{4} \\ & = 30.5 \\ \end{align*} $$
Note: Since it is the sample standard deviation (not the population standard deviation), we use \(n – 1\) as the denominator.
Therefore,
$$ S = \sqrt{30.5} = 5.52268 $$
Step 3: Calculate the ratio.
$$ \text{Ratio} =\cfrac {5.52268}{40} = 0.13806 \text{ or } 13.81\% $$
In finance, the coefficient of variation is used to measure the risk per unit of return. For example, imagine that the mean monthly return on a T-Bill is 0.5% with a standard deviation of 0.58%. Suppose we have another investment, say, Y, with a 1.5% mean monthly return and standard deviation of 6%, then,
$$ \text{CV}_{\text T-\text {Bill}} =\cfrac {0.58}{0.5} = 1.16 $$
$$ \text{CV}_\text{Y} =\cfrac {6}{1.5} = 4 $$
Interpretation: The dispersion per unit monthly return of T-Bills is less than that of Y. Therefore, investment Y is riskier than an investment on T-Bills.
Question
If a security has a mean expected return of 10% and a standard deviation of 5%, its coefficient of variation is closest to:
- 0.005.
- 0.500.
- 2.000.
Solution
The correct answer is B.
$$ \text{CV} = \cfrac {S}{\text x̄} = \cfrac {0.05}{0.10} = 0.5$$
Where:
\(S\) = The standard deviation of the sample.
\(x̄\) = The mean of the sample.
A is incorrect. It assumes the following calculation.
$$CV=\frac{0.05}{10}=0.005$$
C is incorrect. It assumes the following calculation.
$$CV=\frac{10}{5}=2$$