Difference between revisions of "Expectation of the geometric distribution"

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Revision as of 23:27, 30 November 2017

[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]

Statement

Let [ilmath]X\sim[/ilmath][ilmath]\text{Geo} [/ilmath][ilmath](p)[/ilmath] where [ilmath]p[/ilmath] is the probability of any trial being a success, and each trial is i.i.d as [ilmath]X_i\sim[/ilmath][ilmath]\text{Borv} [/ilmath][ilmath](p)[/ilmath], from this we have:

• For [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath] that [ilmath]\P{X\eq k}\eq p(1-p)^{k-1} [/ilmath]

We now define [ilmath]q:\eq 1-p[/ilmath] as this will simplify calculations further on, meaning that now:

• For [ilmath]k\in\mathbb{N}_{\ge 1} [/ilmath] that [ilmath]\P{X\eq k}\eq pq^{k-1} [/ilmath]

Expectation

• The expectation of [ilmath]X[/ilmath] is:
  • [math]\sum^\infty_{k\eq 1}k\P{X\eq k} [/math] which of course is actually a limit of a series, [math]\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}k\P{X\eq k}\right)[/math]

We claim that that [ilmath]\E{X}\eq\frac{1}{p} [/ilmath] for [ilmath]p\in[/ilmath][ilmath](0,1][/ilmath][ilmath]\subseteq\mathbb{R} [/ilmath] and undefined for [ilmath]p\eq 0[/ilmath]

To do so we will consider the 3 cases, [ilmath]p\eq 0[/ilmath], [ilmath]p\in (0,1)\subseteq\mathbb{R} [/ilmath] and [ilmath]p\eq 1[/ilmath] separately and in reverse of this order.

See also

• Variance of the geometric distribution
• Mdm of the geometric distribution

Proof

We introduce the following for short.

1. [math]S'_n:\eq\sum^n_{k\eq 1}kpq^{k-1} [/math] - this forms the sequence used in the limit - which is a series.
   • Thus [math]\E{X}\eq\lim_{n\rightarrow\infty}\Big(S'_n\Big)[/math]
2. [math]S_n:\eq\sum^n_{k\eq 1}kq^{k-1} [/math]
   • This comes from the sequence inside the limit, [math]\sum^n_{k\eq 1}k\P{X\eq k}\eq\sum^n_{k\eq 1}kpq^{k-1}\eq p\sum^n_{k\eq 1} kq^{k-1} \eq pS_n[/math], so:
     • [math]\E{X}\eq\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}k\P{X\eq k}\right)\eq\lim_{n\rightarrow\infty}\Big(pS_n\Big)[/math]

Notice that [ilmath]S'_n\eq pS_n[/ilmath] - introduced purely to save typing.

Case 1: [ilmath]p\eq 1[/ilmath]

Notice that in this case, [ilmath]q\eq 1-p\eq 0[/ilmath].

We now consider the [ilmath]S'_n[/ilmath] terms:

• [math]S'_n\eq pS_n\eq p\left(\sum^n_{k\eq 1}kq^{k-1}\right)[/math] - [ilmath]0^0[/ilmath] comes up here

Case 2: [ilmath]p\in (0,1)\subseteq\mathbb{R} [/ilmath]

Here we use:

• [math]\frac{\mathrm{d} }{\mathrm{d}q}\Big[q^k\Big]\Bigg\vert_q\eq kq^{k-1} [/math] and then the [math]\sum^n_{k\eq 1}q^k[/math] is a geometric series - starting at [ilmath]q[/ilmath] though not [ilmath]1[/ilmath]

Notes