Variance of the geometric distribution
From Maths
[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]Contents
Notes
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]
- This yields:
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]