Difference between revisions of "Geometric distribution"

Revision as of 03:17, 16 January 2018

Definition

Consider a potentially infinite sequence of $\text{Borv}$ variables, $({ X_i })_{ i = 1 }^{ n }$, each independent and identically distributed (i.i.d) with $X_i\sim$$\text{Borv}$$(p)$, so $p$ is the probability of any particular trial being a "success".

The geometric distribution models the probability that the first success occurs on the $k^\text{th}$ trial, for $k\in\mathbb{N}_{\ge 1}$.

As such:

• $\P{X\eq k} :\eq (1-p)^{k-1}p$ - pmf / pdf - Claim 1 below
• $\mathbb{P}[X\le k]\eq 1-(1-p)^k$ - cdf - Claim 2 below
  • $\mathbb{P}[X\ge k]\eq (1-p)^{k-1}$ - an obvious extension.

Convention notes

Properties

For $p\in[0,1]\subseteq\mathbb{R}$ and $X\sim\text{Geo}(p)$ we have the following results about the geometric distribution:

• $\E{X}\eq\frac{1}{p}$ for $p\in(0,1]$ and is undefined or tentatively defined as $+\infty$ if $p\eq 0$
  • Proof: Expectation of the geometric distribution

To do:

1. Variance of the geometric distribution
2. Mdm of the geometric distribution

Proof of claims

Claim 1: $\P{X\eq k}\eq (1-p)^{k-1} p$

• $\P{X\eq k} :\eq (1-p)^{k-1}p$ - which is derived as folllows:
  • $\P{X\eq k} :\eq \Big(\P{X_1\eq 0}\times\Pcond{X_2\eq 0}{X_1\eq 0}\times\cdots\times \Pcond{X_{k-1}\eq 0}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k-2}\eq 0}\Big)\times\Pcond{X_k\eq 1}{X_1\eq 0\cap X_2\eq 0\cap\cdots\cap X_{k-1}\eq 0}$
    • Using that the $X_i$ are independent random variables we see:
      • $$\P{X\eq k}\eq \left(\prod^{k-1}_{i\eq 1}\P{X_i\eq 0}\right)\times \P{X_k\eq 1}$$

        $$\eq (1-p)^{k-1} p$$ as they all have the same distribution, namely $X_i\sim\text{Borv}(p)$

Claim 2: $\mathbb{P}[X\le k]\eq 1-(1-p)^k$

See also

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

Distributions

• Binomial distribution
• Exponential distribution
  • Obtaining the exponential distribution from the geometric distribution

Notes

1. Jump up ↑ Do we make this distinction for cumulative distributions?
2. Jump up ↑ Due to different conventions on the definition of geometric (for example $X':\eq X-1$ for my $X$ and another's $X'\sim\text{Geo}(p)$) or even differing by using $1-p$ in place of $p$ in the $X$ and $X'$ just mentioned - I cannot be sure without working it out that it's $$\frac{1-p}{p^2}$$ - I record this value only for a record of what was once there with the correct expectation - DO NOT USE THIS EXPRESSION

References

1. Jump up ↑ See Expectation of the geometric distribution