Expectation of the geometric distribution

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:

Finish in morning

$\newcommand{\P}[2][]{\mathbb{P}#1{\left[{#2}\right]} } \newcommand{\Pcond}[3][]{\mathbb{P}#1{\left[{#2}\!\ \middle\vert\!\ {#3}\right]} } \newcommand{\Plcond}[3][]{\Pcond[#1]{#2}{#3} } \newcommand{\Prcond}[3][]{\Pcond[#1]{#2}{#3} }$

$\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} }$

Statement

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

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

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

• For $k\in\mathbb{N}_{\ge 1}$ that $\P{X\eq k}\eq pq^{k-1}$

Expectation

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

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

To do so we will consider the 3 cases, $p\eq 0$, $p\in (0,1)\subseteq\mathbb{R}$ and $p\eq 1$ 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. $$S'_n:\eq\sum^n_{k\eq 1}kpq^{k-1}$$ - this forms the sequence used in the limit - which is a series.
   • Thus $$\E{X}\eq\lim_{n\rightarrow\infty}\Big(S'_n\Big)$$
2. $$S_n:\eq\sum^n_{k\eq 1}kq^{k-1}$$
   • This comes from the sequence inside the limit, $$\sum^n_{k\eq 1}k\P{X\eq k}\eq\sum^n_{k\eq 1}kpq^{k-1}\eq p\sum^n_{k\eq 1} kq^{k-1} \eq pS_n$$ , so:
     • $$\E{X}\eq\lim_{n\rightarrow\infty}\left(\sum^n_{k\eq 1}k\P{X\eq k}\right)\eq\lim_{n\rightarrow\infty}\Big(pS_n\Big)$$

Notice that $S'_n\eq pS_n$ - introduced purely to save typing.

Case 1: $p\eq 1$

Notice that in this case, $q\eq 1-p\eq 0$.

We now consider the $S'_n$ terms:

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

Case 2: $p\in (0,1)\subseteq\mathbb{R}$

Here we use:

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