Question

Suppose that X has the normal distribution with mean\({\bf{\mu }}\)and variance\({{\bf{\sigma }}^{\bf{2}}}\). Express\({\bf{E}}\left( {{{\bf{X}}^{\bf{3}}}} \right)\)in terms of\({\bf{\mu }}\)and\({{\bf{\sigma }}^{\bf{2}}}\).

Short answer

\(E\left( {{X^3}} \right) = 3\mu {\sigma ^2} + {\mu ^3}\)

Step-by-step solution

Given information

Let X is a random variable that follows normal distribution that is \(X \sim N\left( {\mu ,\sigma } \right)\).

Calculating the first order and second order moment

The first order moment is expectation

\(E\left( X \right) = \mu \ldots \left( 1 \right)\)

The second order moment is

\(\begin{aligned}{}V\left( X \right)& = E\left( {{X^2}} \right) - {\left[ {E\left( X \right)} \right]^2}\\{\sigma ^2} &= E\left( {{X^2}} \right) - {\mu ^2}\\E\left( {{X^2}} \right) &= {\sigma ^2} + {\mu ^2} \ldots \left( 2 \right)\end{aligned}\)

Calculating the third order moment

Since X follows a normal distribution, the odd-order moments of X are 0.

Therefore,

\(\begin{array}{l}E\left( {{{\left( {X - \mu } \right)}^3}} \right) = 0\\E\left( {{X^3} - 3{X^2}\mu + 3{\mu ^2}X - {\mu ^3}} \right) = 0\end{array}\)

Using this derivation,

\(\begin{aligned}{}E\left( {{X^3}} \right) &= 3\mu E\left( {{X^2}} \right) - 3{\mu ^2}E\left( X \right) + {\mu ^3}\\ &= 3\mu \left( {{\mu ^2} + {\sigma ^2}} \right) - 3{\mu ^2} \times \mu + {\mu ^3} \ldots {\rm{From}}\,\,\left( {{\rm{1}}\,} \right)\,\,{\rm{and}}\,\,\left( {\rm{2}} \right)\\& = 3{\mu ^3} + 3\mu {\sigma ^2} - 3{\mu ^3} + {\mu ^3}\\& = \mu \left( {3{\sigma ^2} + {\mu ^2}} \right)\\ &= 3\mu {\sigma ^2} + {\mu ^3}\end{aligned}\)

Hence the final answer is\(3\mu {\sigma ^2} + {\mu ^3}\).