Variance 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 variance of $X$ is given by:

$\var X = \dfrac k {\lambda^2}$

Proof

By Variance as Expectation of Square minus Square of Expectation, we have:

$\var X = \expect {X^2} - \paren {\expect X}^2$

By Expectation of Erlang Distribution, we have:

$\expect X = \dfrac k \lambda$

We also have:

\(\ds \expect {X^2}\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

So:

\(\ds \var X\)

\(=\)

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

\(\ds \)

\(=\)

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

\(\ds \)

\(=\)

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

$\blacksquare$