Variance of the geometric distribution

[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

It's [math]\frac{1-p}{p^2} [/math] I believe - I did this quickly on a sticky note starting from the warning below. My "formal attempt" (here) though I messed up the final step. I'll fix it later today. DO NOT USE RESULT unless you complete the proof, everything is fine until you hit the red and highly visible WARNING marker below.

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^2}{p^2}\right)[/math] - Warning:Error is here, the [math]\frac{p^2}{p^2} [/math] should be [math]\frac{p^3}{p^2} [/math] instead! - I'm just saving my work, I scrutinised everything and the error was here!

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

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

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

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