Variance of the geometric distribution

Stub grade: A

This page is a stub

This page is a stub, so it contains little or minimal information and is on a to-do list for being expanded.The message provided is:

There's work to do, not just writing out the entire proof Alec (talk) 15:04, 16 January 2018 (UTC)

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\newcommand{\E}[1]{ {\mathbb{E}{\left[{#1}\right]} } }$$\newcommand{\Mdm}[1]{\text{Mdm}{\left({#1}\right) } }$$\newcommand{\Var}[1]{\text{Var}{\left({#1}\right) } }$$\newcommand{\ncr}[2]{ \vphantom{C}^{#1}\!C_{#2} }$$\newcommand{\d}[0]{\mathrm{d} }$$\newcommand{\ddq}[1]{ \frac{\d}{\d q}{\left[{#1}\middle]\right\vert_q} }$

Statement

Let $X\sim$$\text{Geo}$

$$(p)$$ - as defined on the geometric distribution page - then the variance of $X$ is:

• $$\Var{X}\eq \frac{1-p}{p^2}$$ for $p\in(0,1)$ with easy extension (as per expectation of the geometric distribution) to $p\in(0,1]$

Proof

Workings so far

Final steps

Recall $q:\eq 1-p$

Computing $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q$$

We leave the bottom of the paper workings with:

• $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q$$

  $$\eq\frac{\d}{\d q}\left[\frac{1}{(1-q)^2}-1-2q\middle]\right\vert_q$$

  $$\eq -2+\ddq{(1-q)^{-2} }$$

  $$\eq -2+(-2)(1-q)^{-3}\cdot\ddq{1-q}$$

  $$\eq -2+\frac{-2}{(1-q)^3}\cdot (-1)$$

  $$\eq\ 2\left(\frac{1}{(1-q)^3}-1\right)$$

• We may now substitute $q\eq 1-p$ (as $q:\eq 1-p$ so $p\eq 1-q$ follows)
  • This yields:
    • $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q \eq 2\left(\frac{1}{p^3}-1\right)$$

Computing $\E{X^2}$

Recall:

• $q:\eq 1-p$
• $\alpha:\eq \P{X\eq 1}+4\P{X\eq 2}$
• $\beta:\eq \P{X\eq 1}+2\P{X\eq 2}$

The previous step yielded:

• $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q \eq\ 2\left(\frac{1}{p^3}-1\right)$$

and we got as far as:

• $$\E{X^2}\eq \alpha-\beta+\E{X}+pq\left(\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3}q^k\middle]\right\vert_q\right)$$

So:

• $$pq\left(\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3}q^k\middle]\right\vert_q\right)$$

  $$\eq 2pq\left(\frac{1}{p^3}-1\right)$$

  $$\eq 2q\left(\frac{1}{p^2}-p\right)$$

  $$\eq 2(1-p)\left(\frac{1}{p^2}-p\right)$$

Now we substitute this all in to $\E{X^2}\eq \alpha-\beta+\E{X}+pq\left(\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3}q^k\middle]\right\vert_q\right)$ and:

• $$\E{X^2}\eq \P{X\eq 1}+4\P{X\eq 2}-\P{X\eq 1}-2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-p\right)$$

  $$\eq 2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-\frac{p^3}{p^2}\right)$$

  $$\eq 2(1-p)p+\frac{1}{p}+2(1-p)\frac{1-p^3}{p^2}$$

  $$\eq \frac{1}{p}+2(1-p)\left(p+\frac{1-p^3}{p^2}\right)$$

  $$\eq \frac{1}{p}+2(1-p)\left(\frac{p^3}{p^2}+\frac{1-p^3}{p^2}\right)$$

  $$\eq \frac{1}{p}+2(1-p)\frac{1}{p^2}$$

  $$\eq \frac{1}{p} +\frac{2}{p^2} - \frac{2}{p}$$

  $$\eq \frac{2}{p^2}-\frac{1}{p}$$

  $$\eq \frac{2}{p^2}-\frac{p}{p^2}$$ - it's probably easier in this form

Given we'll need to subtract $\frac{1}{p^2}$ there's no point in proceeding any further

Computing $\Var{X}$

Lastly:

• $\Var{X}\eq\E{X^2}-(\E{X})^2$, so
  • $$\Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2}$$

    $$\eq\frac{1}{p^2}-\frac{p}{p^2}$$

    $$\eq\frac{1-p}{p^2}$$

Thus

• $$\Var{X}\eq\frac{1-p}{p^2}$$