Explain and calculate variance, standard deviation, and coefficient of variation

The variance of a discrete random variable is the sum of the square of all the values the variable can take times the probability of that value occurring minus the sum of all the values the variable can take times the probability of that value occurring squared as shown in the formula below:

$$ Var\left( X \right) =\sum { { x }^{ 2 }p\left( x \right) -{ \left[ \sum { { x }p\left( x \right) } \right] }^{ 2 } } $$

Or written in another way, as a function of \(E(X)\), then:

$$ Var\left( X \right) =E\left( { X }^{ 2 } \right) -E{ \left( X \right) }^{ 2 } $$

Often \(E(X)\) is written as \(\mu\) and therefore variance can also be shown as in the formula below:

$$ Var\left( X \right) =\sum { { \left( x-\mu \right) }^{ 2 }p\left( x \right) } $$

Example

Given the experiment of rolling a single die, calculate \(Var(X)\).

\( E(X) = 1 \ast ({1}/{6})+2 \ast ({1}/{6})+ 3 \ast ({1}/{6})+ 4 \ast ({1}/{6})+ 5 \ast ({1}/{6})+ 6 \ast ({1}/{6}) = 3.5 \)

\( E(X^2) = 1 \ast ({1}/{6})+2^2 \ast ({1}/{6})+ 3^2 \ast ({1}/{6})+ 4^2 \ast ({1}/{6})+ 5^2 \ast ({1}/{6})+ 6^2 \ast ({1}/{6}) = {91}/{6} \)

\( Var \left(X\right) = \left(91/6 \right) – \left(3.5 \right)^2 = {35}/{12} = 2.92 \)

We could also calculate \(Var(X)\) as:

\( E \left(X \right) = \mu =3.5 \)
\begin{align*}
Var \left(X \right) = & (1-3.5)^2 \ast (1/6) + (2-3.5)^2 \ast (1/6)+(3-3.5)^2 \ast (1/6)+(4-3.5)^2 \ast (1/6)+ \\
& (5-3.5)^2 \ast (1/6)+(6-3.5)^2 \ast (1/6) = 2.92 \\
\end{align*}

The variance of a continuous random variable is shown in the formula below:

$$ Var\left( X \right) =\int _{ -\infty }^{ \infty }{ { x }^{ 2 }f\left( x \right) dx } -{ \left[ \int _{ -\infty }^{ \infty }{ { x }f\left( x \right) dx } \right] }^{ 2 } $$

where \(f(x)\) is the probability density function of \(x\).

As in the discrete case, it can also be written as:

$$ Var\left( X \right) =E\left( { X }^{ 2 } \right) -E{ \left( X \right) }^{ 2 } $$

And also,

$$ Var\left( X \right) =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ 2 }f\left( x \right) dx } $$

Example

Given the following probability density function of a continuous random variable:

$$ f\left( x \right) =\begin{cases} \frac { x }{ 2 } , & 0 < x < 2 \\ 0, & otherwise \end{cases} $$

Calculate \(Var(X)\).

$$
\begin{align*}
& E\left( X \right) =\int _{ -\infty }^{ \infty }{ xf\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ x=0 }^{ x=2 }=\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \\
& E\left( X^2 \right) =\int _{ -\infty }^{ \infty }{ x^2 f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x^2 \ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 4 } }{ 8 } \right] }_{ x=0 }^{ x=2 }=2 \\
& Var \left(X \right) = 2 – {\left({4}/{3}\right)}^ 2 = {2}/{9} \\
\end{align*}
$$

Or,

$$ Var\left( X \right) =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ 2 }f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ { \left( x-\frac { 4 }{ 3 } \right) }^{ 2 }\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 4 } }{ 8 } -\frac { 4 }{ 9 } { x }^{ 3 }+\frac { 4 }{ 9 } { x }^{ 2 } \right] }_{ x=0 }^{ x=2 }=\frac { 2 }{ 9 } $$

The standard deviation, often written as \(\sigma\) , of either a discrete or continuous random variable can be defined as:

$$ S.D.\left( X \right) =\sigma =\sqrt { Var\left( X \right) } $$

The coefficient of variation of a random variable can be defined as the standard deviation divided by the mean (or expected value) of \(X\) as shown in the formula below:

$$ C.V.= {\frac {\sigma}{\mu}} $$

The variance of \(X\) is sometimes often referred to as the second moment of \(X\) about the mean.

The third moment of \(X\) is referred to as the skewness and the fourth moment is called kurtosis.

In general, the mth moment of \(X\) can be calculated from the following formula:

$$ mth\quad moment\left( X \right) =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ m }f\left( x \right)dx } $$

Example

Given the following probability density function of a continuous random variable:

$$ f\left( x \right) =\begin{cases} \frac { x }{ 2 } , & 0 < x < 2 \\ 0, & otherwise \end{cases} $$

Calculate the skewness.

\( \begin{align*}
\mu & =\int _{ -\infty }^{ \infty }{ xf\left( x \right) dx= } \int _{ 0 }^{ 2 }{ x\ast \frac { x }{ 2 } \ast dx } ={ \left[ \frac { { x }^{ 3 } }{ 6 } \right] }_{ x=0 }^{ x=2 }=\frac { 8 }{ 6 } =\frac { 4 }{ 3 } \\
Skew\left( X \right) & =\int _{ -\infty }^{ \infty }{ { \left( x-\mu \right) }^{ 3 }f\left( x \right) dx= } \int _{ 0 }^{ 2 }{ { \left( x-\frac { 4 }{ 3 } \right) }^{ 3 }\ast \frac { x }{ 2 } \ast dx } \\
& =\int _{ 0 }^{ 2 }{ \left( \frac { { x }^{ 4 } }{ 2 } -2{ x }^{ 3 }+\frac { 8{ x }^{ 2 } }{ 3 } -\frac { 32x }{ 27 } \right) dx={ \left[ \frac { { x }^{ 5 } }{ 10 } -\frac { { x }^{ 4 } }{ 2 } +\frac { 8{ x }^{ 3 } }{ 9 } -\frac { 16{ x }^{ 2 } }{ 27 } \right] }_{ x=0 }^{ x=2 }=-{ 8 }/{ 135 } }
\end{align*}
\)

Binomial (and Bernoulli)

An experiment is performed with probability of success, \(p\), and failure, \(1-p\), and the experiment is performed \(n\) times.

$$
\begin{align*}
& p\left( x \right) =\left( \begin{matrix} n \\ x \end{matrix} \right) { p }^{ X }{ \left( 1-p \right) }^{ n-x } \\
& E\left( X \right) =np \\
& Var\left( X \right) =np\left( 1-p \right) \\
\end{align*}
$$

A Bernoulli random variable is the special case of a binomial random variable where the experiment is performed only once.

Negative Binomial

Given an experiment that is performed \(X\) number of times until a total of \(r\) successes occur, then

$$
\begin{align*}
& p\left( x \right) =\left( \begin{matrix} x-1 \\ r-1 \end{matrix} \right) { \left( 1-p \right) }^{ x-r }{ p }^{ r } \\
& E\left( X \right) =\frac { r }{ p } \\
& Var\left( X \right) ={ r\left( 1-p \right) }/{ p^{ 2 } } \\
\end{align*}
$$

Geometric

Given an experiment that is performed \(X\) number of times until a success occurs, with the probability of a success on each trial being equal to \(p\), then

$$ \begin{align*}
& p\left(x \right)=\left(1-p \right)^\left(x-1\right) p \\
& E \left(X \right)={1}/{p} \\
& Var\left(X \right)={\left(1-p \right)}/{p^2} \\
\end{align*}
$$

Hypergeometric

$$ \begin{align*}
& p\left( x \right) =\frac { \left( \begin{matrix} M \\ x \end{matrix} \right) \left( \begin{matrix} N-M \\ n-x \end{matrix} \right) }{ \left( \begin{matrix} N \\ n \end{matrix} \right) } \\
& E\left(X \right)={\frac {nm}{N}} \\
& Var \left(X \right)={nm \left(N-M \right)\left(N-n\right)}/{N^2 \left(N-1 \right)} \\
\end{align*}
$$

Poisson

A Poisson random variable can be described as the number of events occurring in a fixed time period if the events occur at a known constant rate,\(\lambda\).

$$ \begin{align*}
& p \left(x \right)={\frac {{e}^{-\lambda} {\lambda}^{x}}{x!}} \\
& E(X) = \lambda \\
& Var \left(X \right) = \lambda \\
\end{align*}
$$

Uniform– Discrete

Given an experiment in which outcomes are all equally likely

$$ \begin{align*}
& p \left(x \right)={\frac {1}{b-a+1}} \\
& E \left(X \right)={\frac {b+a}{2}} \\
& Var \left(X \right)={\frac {\left(b-a+2 \right)\left(b-a\right)}{12}} \\
\end{align*}
$$

Uniform – Continuous

$$ \begin{align*}
& f\left(x \right)={\frac {1}{b-a}} \\
& E\left(X \right)={\frac {b+a}{2}} \\
& Var \left(X \right)={\frac {{\left(b-a \right)}^2}{12}} \\
\end{align*}
$$

Exponential

$$ \begin{align*}
& f\left(x \right)= {{\lambda}e}^{-{\lambda} x} \\
& E\left(X \right)={\frac {1}{\lambda}} \\
& Var \left(X \right)={\frac {1}{{\lambda}^2}} \\
\end{align*}
$$

Gamma

$$ \begin{align*}
& f\left( x \right) =\frac { {{\lambda}e}^{-{\lambda} x}{ \left( {\lambda} x \right) }^{ \alpha -1 } }{ \Gamma \left( \alpha \right) } \\
&          where\quad { \Gamma \left( \alpha \right) }=\int _{ 0 }^{ \infty }{ { e }^{ -y }{ y }^{ \alpha -1 }dy } \\
& E \left(X \right)={\frac {\alpha}{\lambda}} \\
& Var \left(X \right)={\frac {\alpha}{\lambda^2} } \\
\end{align*}
$$

Normal

$$ \begin{align*}
& f\left( x \right) =\frac { 1 }{ \sigma \sqrt { 2\pi } } { e }^{ -.5{ \left( \frac { x-\mu }{ \sigma } \right) }^{ 2 } }
\\
& E \left(X \right)= \mu \\
& Var \left(X\right)= \sigma \\
\end{align*}
$$

The standard normal distribution has \(E(X) = 0\) and \(Var(X) = 1\).

 

Learning Outcome

Topic 2.d: Univariate Random Variables – Explain and calculate variance, standard deviation, and coefficient of variation.


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