Proof: Skewness of the exponential distribution
Theorem: Let $X$ be a random variable following an exponential distribution:
\[\label{eq:exp} X \sim \mathrm{Exp}(\lambda) \; .\]Then the skewness of $X$ is
\[\label{eq:exp-skew} \mathrm{Skew}(X) = 2 \; .\]Proof:
To compute the skewness of $X$, we partition the skewness into expected values:
\[\label{eq:skew-mean} \mathrm{Skew}(X) = \frac{\mathrm{E}(X^3)-3\mu\sigma^2-\mu^3}{\sigma^3} \; ,\]where $\mu$ and $\sigma$ are the mean and standard deviation of $X$, respectively. Since $X$ follows an exponential distribution, the mean of $X$ is given by
\[\label{eq:exp-mean} \mu = \mathrm{E}(X) = \frac{1}{\lambda}\]and the standard deviation of $X$ is given by
\[\label{eq:exp-var} \sigma = \sqrt{\mathrm{Var}(X)} = \sqrt{\frac{1}{\lambda^2}} = \frac{1}{\lambda} \; .\]Substituting \eqref{eq:exp-mean} and \eqref{eq:exp-var} into \eqref{eq:skew-mean} gives:
\[\label{eq:skew-mean-alt} \begin{split} \mathrm{Skew}(X) &= \frac{\mathrm{E}(X^3)-3\mu\sigma^2-\mu^3}{\sigma^3}\\ &= \frac{\mathrm{E}(X^3)}{\sigma^3}-\frac{3\mu\sigma^2+\mu^3}{\sigma^3}\\ &= \frac{\mathrm{E}(X^3)}{\left(\frac{1}{\lambda}\right)^3}-\frac{3\left(\frac{1}{\lambda}\right)\left(\frac{1}{\lambda}\right)^2 + \left(\frac{1}{\lambda}\right)^3}{\left(\frac{1}{\lambda}\right)^3}\\ &= \lambda^3\cdot\mathrm{E}(X^3)-\frac{\frac{3}{\lambda^3}+\frac{1}{\lambda^3}}{\frac{1}{\lambda^3}}\\ &= \lambda^3\cdot\mathrm{E}(X^3)-4 \; . \end{split}\]Thus, the remaining work is to compute $\mathrm{E}(X^3)$. To do this, we use the moment-generating function of the exponential distribution to calculate
\[\label{eq:exp-moment} \mathrm{E}(X^3) = M_X'''(0)\]based on the relationship between raw moment and moment-generating function.
First, we differentiate the moment-generating function of the exponential distribution
\[\label{eq:exp-mgf} M_X(t) = \frac{\lambda}{\lambda-t} = \lambda(\lambda-t)^{-1}\]with respect to $t$. Using the chain rule gives:
\[\label{eq:exp-skew-s1} \begin{split} M_X'(t) &= -1\cdot \lambda(\lambda-t)^{-2}\cdot (-1) \\ &= \lambda(\lambda-t)^{-2} \; . \end{split}\]We continue using the chain rule to obtain the second derivative:
\[\label{eq:exp-skew-s2} \begin{split} M_X''(t) &= -2\cdot \lambda(\lambda-t)^{-3}\cdot (-1) \\ &= 2\lambda(\lambda-t)^{-3} \; . \end{split}\]Finally, one more application of the chain rule gives us the third derivative:
\[\label{eq:exp-skew-s3} \begin{split} M_X'''(t) &= -3\cdot 2\lambda(\lambda-t)^{-4}\cdot (-1)\\ &= 6\lambda(\lambda-t)^{-4} \\ &= \frac{6\lambda}{(\lambda-t)^4} \; . \end{split}\]Applying \eqref{eq:exp-moment}, together with \eqref{eq:exp-skew-s3}, yields
\[\label{eq:exp-skew-s4} \begin{split} \mathrm{E}(X^3) &= M_X'''(0)\\ &= \frac{6\lambda}{(\lambda-0)^4}\\ &= \frac{6\lambda}{\lambda^4}\\ &= \frac{6}{\lambda^3} \; . \end{split}\]We now substitute \eqref{eq:exp-skew-s4} into \eqref{eq:skew-mean-alt}, giving
\[\label{eq:exp-skew-s5} \begin{split} \mathrm{Skew}(X) &= \lambda^3\cdot\mathrm{E}(X^3)-4 \\ &= \lambda^3\cdot \left(\frac{6}{\lambda^3}\right)-4\\ &= 6-4\\ &= 2 \; . \end{split}\]This completes the proof of \eqref{eq:exp-skew}.
Metadata: ID: P409 | shortcut: exp-skew | author: tomfaulkenberry | date: 2023-04-24, 12:00.