Difference between revisions of "Variance of the geometric distribution"
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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) }} | ||
{{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__ | ||
− | == | + | ==Statement== |
− | [[ | + | 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: |
− | ==Final steps== | + | * {{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}}=== | + | ====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)}} | ||
− | ===Computing {{M|\E{X^2} }}=== | + | ====Computing {{M|\E{X^2} }}==== |
Recall: | Recall: | ||
* {{M|q:\eq 1-p}} | * {{M|q:\eq 1-p}} | ||
Line 34: | Line 38: | ||
Now we substitute this all in to {{M|\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: | Now we substitute this all in to {{M|\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: | ||
* {{MM|\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)}} | * {{MM|\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)}} | ||
− | *: {{MM|\eq 2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-\frac{p^ | + | *: {{MM|\eq 2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-\frac{p^3}{p^2}\right)}} |
− | *: {{MM|\eq | + | *: {{MM|\eq 2(1-p)p+\frac{1}{p}+2(1-p)\frac{1-p^3}{p^2} }} |
− | *: {{MM|\eq \frac{1}{p}+2(1-p)\left(p+\frac{1-p^ | + | *: {{MM|\eq \frac{1}{p}+2(1-p)\left(p+\frac{1-p^3}{p^2}\right)}} |
− | *: {{MM|\eq \frac{1}{p}+2(1-p)\ | + | *: {{MM|\eq \frac{1}{p}+2(1-p)\left(\frac{p^3}{p^2}+\frac{1-p^3}{p^2}\right)}} |
− | * | + | *: {{MM|\eq \frac{1}{p}+2(1-p)\frac{1}{p^2} }} |
+ | *: {{MM|\eq \frac{1}{p} +\frac{2}{p^2} - \frac{2}{p} }} | ||
+ | *: {{MM|\eq \frac{2}{p^2}-\frac{1}{p} }} | ||
+ | *: {{MM|\eq \frac{2}{p^2}-\frac{p}{p^2} }} - it's probably easier in this form | ||
+ | Given we'll need to subtract {{M|\frac{1}{p^2} }} there's no point in proceeding any further | ||
+ | |||
+ | ====Computing {{M|\Var{X} }}==== | ||
+ | Lastly: | ||
+ | * {{M|\Var{X}\eq\E{X^2}-(\E{X})^2}}, so | ||
+ | ** {{MM|\Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2} }} | ||
+ | **: {{MM|\eq\frac{1}{p^2}-\frac{p}{p^2} }} | ||
+ | **: {{MM|\eq\frac{1-p}{p^2} }} | ||
+ | Thus | ||
+ | * {{MM|\Var{X}\eq\frac{1-p}{p^2} }} | ||
+ | ==Notes== | ||
+ | <references group="Note"/> | ||
+ | ==References== | ||
+ | <references/> | ||
+ | {{Theorem Of|Probability|Statistics|Elementary Probability}} | ||
+ | [[Category:Variance Calculations]] |
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:
[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
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:
- [math]\Var{X}\eq \frac{1-p}{p^2} [/math] for [ilmath]p\in(0,1)[/ilmath] with easy extension (as per expectation of the geometric distribution) to [ilmath]p\in(0,1][/ilmath]
Proof
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^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]
- [math]\Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2} [/math]
Thus
- [math]\Var{X}\eq\frac{1-p}{p^2} [/math]
Notes
- ↑ So using our convention, to say it explicitly
References
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