Difference between revisions of "Variance of the geometric distribution"

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Revision as of 06:41, 16 January 2018

[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

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]