91Ó°ÊÓ

According to the example 4.4.3, X is a random variable with the following pdf:

\(f\left( x \right) = \left\{ {\begin{aligned}{*{20}{l}}{{e^{ - x}},\;{\rm{for}}\;x > 0}\\{0,\;{\rm{otherwise}}}\end{aligned}} \right.\)

\(\begin{aligned}{c}E\left( X \right) &= \int\limits_0^\infty {x{e^{ - x}}dx} \\ &= \int\limits_0^\infty {{x^{2 - 1}}{e^{ - 1 \cdot x}}dx} \\ = \frac{{\left| \!{\overline {\,

2 \,}} \right. }}{{{1^2}}}\,\,\left( {{\rm{using the gamma integral}}} \right)\\ &= 1.\end{aligned}\)

\(\begin{aligned}{c}E\left( {{X^2}} \right) &= \int\limits_0^\infty {{x^2}{e^{ - x}}dx} \\ = \int\limits_0^\infty {{x^{3 - 1}}{e^{ - 1 \cdot x}}dx} \\ &= \frac{{\left| \!{\overline {\,

3 \,}} \right. }}{{{1^3}}}\left( {{\rm{using the gamma integral}}} \right)\\ = 2.\end{aligned}\)

\(\begin{aligned}{c}Var\left( X \right) &= E\left( {{X^2}} \right) - {\left( {E\left( X \right)} \right)^2}\\ &= 2 - 1\\ &= 1.\end{aligned}\)

Then, the mean and the variance of X are:

\(\begin{aligned}{l}{\mu _X} &= 1.\\{\sigma _X} = 1.\end{aligned}\)

Now,

\(\begin{aligned}{c}E\left( {{X^3}} \right) &= \int\limits_0^\infty {{x^3}{e^{ - x}}dx} \\ &= \int\limits_0^\infty {{x^{4 - 1}}{e^{ - x}}dx} \\ &= \left| \!{\overline {\,

4 \,}} \right. \;\left( {{\rm{Using the gamma integral}}} \right)\\ = 6.\end{aligned}\)

The skewness of X is defined as:

\(\begin{aligned}{c}Skewness\left( X \right) = \frac{{E\left( {{{\left( {X - {\mu _X}} \right)}^3}} \right)}}{{{\sigma _X}^3}}\\ &= \frac{{E\left( {{X^3} - 3{\mu _X}{X^2} + 3{\mu _X}^2X - {\mu _X}^3} \right)}}{{{\sigma _X}^3}}\\ = \frac{{E\left( {{X^3}} \right) - 3{\mu _X}E\left( {{X^2}} \right) + 3{\mu _X}^2E\left( X \right) - {\mu _X}^3}}{{{\sigma _X}^3}}\\ &= \frac{{6 - 3 \times 1 \times 2 + 3 \times 1 \times 1 - 1}}{1}\\ = 2.\end{aligned}\)

Therefore, the skewness of X is 2.