Skewness of Erlang Distribution

Theorem

Let $k$ be a strictly positive integer.

Let $\lambda$ be a strictly positive real number.

Let $X$ be a continuous random variable with an Erlang distribution with parameters $k$ and $\lambda$.

Then the skewness $\gamma_1$ of $X$ is given by:

$\gamma_1 = \dfrac 2 {\sqrt k}$

Proof

From Skewness in terms of Non-Central Moments, we have:

$\gamma_1 = \dfrac {\expect {X^3} - 3 \mu \sigma^2 - \mu^3} {\sigma^3}$

where:

$\mu$ is the expectation of $X$.

$\sigma$ is the standard deviation of $X$.

By Expectation of Erlang Distribution we have:

$\mu = \dfrac k \lambda$

By Variance of Erlang Distribution we have:

$\sigma = \dfrac {\sqrt k} \lambda$

We also have:

\(\ds \expect {X^3}\)

\(=\)

\(\ds \frac 1 {\lambda^3} \prod_{m \mathop = 0}^2 \paren {k + m}\)

\(\ds \)

\(=\)

\(\ds \frac {k \paren {k + 1} \paren {k + 2} } {\lambda^3}\)

\(\ds \)

\(=\)

\(\ds \frac {k^3 + 3 k^2 + 2 k} {\lambda^3}\)

So:

\(\ds \gamma_1\)

\(=\)

\(\ds \frac {\lambda^3} {k^{3 / 2} } \paren {\frac {k^3 + 3 k^2 + 2 k} {\lambda^3} - \frac {3 k^2} {\lambda^3} - \frac {k^3} {\lambda^3} }\)

\(\ds \)

\(=\)

\(\ds \frac {\lambda^3} {k^{3 / 2} } \cdot \frac {2 k} {\lambda^3}\)

\(\ds \)

\(=\)

\(\ds \frac 2 {\sqrt k}\)

$\blacksquare$