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)

[ilmath]\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} }[/ilmath]

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

Notes

Workings so far

The variance is [math]\frac{1-p}{p^2} [/math]

Final steps

Recall [ilmath]q:\eq 1-p[/ilmath]

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

We leave the bottom of the paper workings with:

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

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

  [math]\eq -2+\ddq{(1-q)^{-2} } [/math]

  [math]\eq -2+(-2)(1-q)^{-3}\cdot\ddq{1-q} [/math]

  [math]\eq -2+\frac{-2}{(1-q)^3}\cdot (-1)[/math]

  [math]\eq\ 2\left(\frac{1}{(1-q)^3}-1\right)[/math]

• We may now substitute [ilmath]q\eq 1-p[/ilmath] (as [ilmath]q:\eq 1-p[/ilmath] so [ilmath]p\eq 1-q[/ilmath] follows)
  • This yields:
    • [math]\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)[/math]

Computing [ilmath]\E{X^2} [/ilmath]

Recall:

• [ilmath]q:\eq 1-p[/ilmath]
• [ilmath]\alpha:\eq \P{X\eq 1}+4\P{X\eq 2} [/ilmath]
• [ilmath]\beta:\eq \P{X\eq 1}+2\P{X\eq 2} [/ilmath]

The previous step yielded:

• [math]\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)[/math]

and we got as far as:

• [math]\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)[/math]

So:

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

  [math]\eq 2pq\left(\frac{1}{p^3}-1\right)[/math]

  [math]\eq 2q\left(\frac{1}{p^2}-p\right)[/math]

  [math]\eq 2(1-p)\left(\frac{1}{p^2}-p\right)[/math]

Now we substitute this all in to [ilmath]\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)[/ilmath] and:

• [math]\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)[/math]

  [math]\eq 2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-\frac{p^3}{p^2}\right)[/math]

  [math]\eq 2(1-p)p+\frac{1}{p}+2(1-p)\frac{1-p^3}{p^2} [/math]

  [math]\eq \frac{1}{p}+2(1-p)\left(p+\frac{1-p^3}{p^2}\right)[/math]

  [math]\eq \frac{1}{p}+2(1-p)\left(\frac{p^3}{p^2}+\frac{1-p^3}{p^2}\right)[/math]

  [math]\eq \frac{1}{p}+2(1-p)\frac{1}{p^2} [/math]

  [math]\eq \frac{1}{p} +\frac{2}{p^2} - \frac{2}{p} [/math]

  [math]\eq \frac{2}{p^2}-\frac{1}{p} [/math]

  [math]\eq \frac{2}{p^2}-\frac{p}{p^2} [/math] - it's probably easier in this form

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

Lastly:

• [ilmath]\Var{X}\eq\E{X^2}-(\E{X})^2[/ilmath], so
  • [math]\Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2} [/math]

    [math]\eq\frac{1}{p^2}-\frac{p}{p^2} [/math]

    [math]\eq\frac{1-p}{p^2} [/math]

Thus

• [math]\Var{X}\eq\frac{1-p}{p^2} [/math]