Difference between revisions of "Variance of the geometric distribution"

Revision as of 15:04, 16 January 2018 (view source) Alec (Talk | contribs) (Completed the initial workings, most of the proof is on paper, marked as stub.) ← Older edit		Latest revision as of 15:08, 16 January 2018 (view source) Alec (Talk | contribs) 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__
−	==Notes==	+	==Statement==
−	[[File:MostOfGeometricDistributionVarianceComputations.JPG|thumbnail|Workings so far]]The variance is {{MM|\frac{1-p}{p^2} }}	+	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}}
Line 16:		Line 19:
	** 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 45:		Line 48:
	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

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

Statement

Let $X\sim$$\text{Geo}$

$$(p)$$ - as defined on the geometric distribution page^{[Note 1]} - then the variance of $X$ is:

• $$\Var{X}\eq \frac{1-p}{p^2}$$ for $p\in(0,1)$ with easy extension (as per expectation of the geometric distribution) to $p\in(0,1]$

Proof

Final steps

Recall $q:\eq 1-p$

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

We leave the bottom of the paper workings with:

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

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

  $$\eq -2+\ddq{(1-q)^{-2} }$$

  $$\eq -2+(-2)(1-q)^{-3}\cdot\ddq{1-q}$$

  $$\eq -2+\frac{-2}{(1-q)^3}\cdot (-1)$$

  $$\eq\ 2\left(\frac{1}{(1-q)^3}-1\right)$$

• We may now substitute $q\eq 1-p$ (as $q:\eq 1-p$ so $p\eq 1-q$ follows)
  • This yields:
    • $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q \eq 2\left(\frac{1}{p^3}-1\right)$$

Computing $\E{X^2}$

Recall:

• $q:\eq 1-p$
• $\alpha:\eq \P{X\eq 1}+4\P{X\eq 2}$
• $\beta:\eq \P{X\eq 1}+2\P{X\eq 2}$

The previous step yielded:

• $$\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3} q^k\middle]\right\vert_q \eq\ 2\left(\frac{1}{p^3}-1\right)$$

and we got as far as:

• $$\E{X^2}\eq \alpha-\beta+\E{X}+pq\left(\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3}q^k\middle]\right\vert_q\right)$$

So:

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

  $$\eq 2pq\left(\frac{1}{p^3}-1\right)$$

  $$\eq 2q\left(\frac{1}{p^2}-p\right)$$

  $$\eq 2(1-p)\left(\frac{1}{p^2}-p\right)$$

Now we substitute this all in to $\E{X^2}\eq \alpha-\beta+\E{X}+pq\left(\frac{\d^2}{\d q^2}\left[\sum^\infty_{k\eq 3}q^k\middle]\right\vert_q\right)$ and:

• $$\E{X^2}\eq \P{X\eq 1}+4\P{X\eq 2}-\P{X\eq 1}-2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-p\right)$$

  $$\eq 2\P{X\eq 2}+\frac{1}{p}+2(1-p)\left(\frac{1}{p^2}-\frac{p^3}{p^2}\right)$$

  $$\eq 2(1-p)p+\frac{1}{p}+2(1-p)\frac{1-p^3}{p^2}$$

  $$\eq \frac{1}{p}+2(1-p)\left(p+\frac{1-p^3}{p^2}\right)$$

  $$\eq \frac{1}{p}+2(1-p)\left(\frac{p^3}{p^2}+\frac{1-p^3}{p^2}\right)$$

  $$\eq \frac{1}{p}+2(1-p)\frac{1}{p^2}$$

  $$\eq \frac{1}{p} +\frac{2}{p^2} - \frac{2}{p}$$

  $$\eq \frac{2}{p^2}-\frac{1}{p}$$

  $$\eq \frac{2}{p^2}-\frac{p}{p^2}$$ - it's probably easier in this form

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

Computing $\Var{X}$

Lastly:

• $\Var{X}\eq\E{X^2}-(\E{X})^2$, so
  • $$\Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2}$$

    $$\eq\frac{1}{p^2}-\frac{p}{p^2}$$

    $$\eq\frac{1-p}{p^2}$$

Thus

• $$\Var{X}\eq\frac{1-p}{p^2}$$

Notes

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

References