Difference between revisions of "Variance of the geometric distribution"

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(Completed the initial workings, most of the proof is on paper, marked as stub.)
m (Adding definition note and confirming the convention.)
 
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{{Stub page|grade=A|msg=There's work to do, not just writing out the entire proof [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 15:04, 16 January 2018 (UTC) }}
 
{{Stub page|grade=A|msg=There's work to do, not just writing out the entire proof [[User:Alec|Alec]] ([[User talk:Alec|talk]]) 15:04, 16 January 2018 (UTC) }}
 
{{ProbMacros}}{{M|\newcommand{\d}[0]{\mathrm{d} } }}{{M|\newcommand{\ddq}[1]{ \frac{\d}{\d q}{\left[{#1}\middle]\right\vert_q} }  }}__TOC__
 
{{ProbMacros}}{{M|\newcommand{\d}[0]{\mathrm{d} } }}{{M|\newcommand{\ddq}[1]{ \frac{\d}{\d q}{\left[{#1}\middle]\right\vert_q} }  }}__TOC__
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==Notes==
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==Statement==
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[[File:MostOfGeometricDistributionVarianceComputations.JPG|thumbnail|Workings so far]]The variance is {{MM|\frac{1-p}{p^2} }}
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Let {{M|X\sim}}[[Geometric distribution|{{M|\text{Geo} }}]]{{MM|(p) }} - as defined on the [[geometric distribution]] page<ref group="Note">So using ''our'' convention, to say it explicitly</ref> - then the [[variance]] of {{M|X}} is:
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==Final steps==
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* {{MM|\Var{X}\eq \frac{1-p}{p^2} }} for {{M|p\in(0,1)}} with easy extension (as per ''[[expectation of the geometric distribution]]'') to {{M|p\in(0,1]}}  
 +
==Proof==
 +
[[File:MostOfGeometricDistributionVarianceComputations.JPG|thumbnail|Workings so far]]
 +
===Final steps===
 
Recall {{M|q:\eq 1-p}}
 
Recall {{M|q:\eq 1-p}}
−
===Computing {{MM|\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q}}===
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====Computing {{MM|\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:
 
We leave the bottom of the paper workings with:
 
* {{MM|\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q}}
 
* {{MM|\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q}}
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** This yields:
 
** This yields:
 
*** {{MM|\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)}}
 
*** {{MM|\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)}}
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===Computing {{M|\E{X^2} }}===
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====Computing {{M|\E{X^2} }}====
 
Recall:
 
Recall:
 
* {{M|q:\eq 1-p}}
 
* {{M|q:\eq 1-p}}
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Given we'll need to subtract {{M|\frac{1}{p^2} }} there's no point in proceeding any further
 
Given we'll need to subtract {{M|\frac{1}{p^2} }} there's no point in proceeding any further
  
−
 
+
====Computing {{M|\Var{X} }}====
 
Lastly:
 
Lastly:
 
* {{M|\Var{X}\eq\E{X^2}-(\E{X})^2}}, so
 
* {{M|\Var{X}\eq\E{X^2}-(\E{X})^2}}, so

Latest revision as of 15:08, 16 January 2018

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)
[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]

Statement

Let [ilmath]X\sim[/ilmath][ilmath]\text{Geo} [/ilmath][math](p) [/math] - as defined on the geometric distribution page[Note 1] - then the variance of [ilmath]X[/ilmath] is:

Proof

Workings so far

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

Computing [ilmath]\Var{X} [/ilmath]

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]

Notes

  1. ↑ So using our convention, to say it explicitly

References

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