Variance of the geometric distribution
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\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} } Contents
[hide]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)
- This yields:
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}
- \Var{X}\eq \frac{2}{p^2}-\frac{p}{p^2} - \frac{1}{p^2}
Thus
- \Var{X}\eq\frac{1-p}{p^2}
Notes
- Jump up ↑ So using our convention, to say it explicitly
References
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