Notes:Geometric probability distribution
From Maths
What I recall
It's been nearly 5 years since I did this, I hope I haven't slipped too far.
- Let [ilmath]X\sim\text{Geometrically}(p)[/ilmath] where [ilmath]p[/ilmath] is the probability of an event happening.
- For an [ilmath]n\in\mathbb{N} [/ilmath]:
- [ilmath]\mathbb{P}\left[X=n\right]=(1-p)^{n-1}p[/ilmath] - that is the probability of an event not happing [ilmath]n-1[/ilmath] times, then happening on the [ilmath]n^\text{th} [/ilmath] trial. So
- [ilmath]\mathbb{P}\left[X=n\right][/ilmath] is the probability that [ilmath]X[/ilmath] first occurs on the [ilmath]n^\text{th} [/ilmath] trial.
- [ilmath]\mathbb{P}\left[X=n\right]=(1-p)^{n-1}p[/ilmath] - that is the probability of an event not happing [ilmath]n-1[/ilmath] times, then happening on the [ilmath]n^\text{th} [/ilmath] trial. So
Expected value
- [math]\sum^\infty_{i=1}i\mathbb{P}[X=i]=\sum_{i=1}^\infty ip(1-p)^{i-1}=p\sum^\infty_{i=1}i(1-p)^{i-1}\frac{p}{1-p}\sum^\infty_{i=1}i(1-p)^i[/math]
- Notice [ilmath]\frac{p}{1-p} [/ilmath] is a simple scalar value.
- Define [ilmath]s_n:=\sum^n_{i=1}i(1-p)^i[/ilmath] then...
Anyway! Just by applying "logic", the expected value is (unsurprisingly) [ilmath]\frac{1}{p} [/ilmath]
