Variance of the geometric distribution

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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[Note 1] - then the variance of X is:

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}

Notes

  1. Jump up So using our convention, to say it explicitly

References