Geometric distribution

Consider a sequence of independent Bernoulli trials, each with same success probability $p$, where $0<p<1$. We continue performing trials until the first success.

Let $X$ be the number of failures before the first success. Then $X$ has the Geometric distribution with parameter $p$. The textbook writes $X\sim \text{Geom}(p)$.

PMF

The PMF of $X$ for a support $k$ is given by

$$P(X=k)=q^{k}p$$

Where $k\in [0,1,2,...]$ and $q=1-p$. Intuitively, this says that we fail $k$ times, each with probability $1-p$ (since $p$ is success rate), and then we finally succeed once.

Expectation

Given that $X\sim \text{Geom}(p)$, then $\mathbb{E}(X)=\frac{1-p}{p}$. The proof is a bit gnarly (and there are like 5 of them) so we’ll prove it later in the semester.

First Success distribution

Some people define the Geometric distribution to include the first success within the r.v. $X$. The textbook doesn’t, but instead defines a new distribution.

If $X\sim \text{Geom}(p)$, then the First Success distribution is given by $Y=X+1$, and the textbook writes this as $Y\sim \text{FS}(p)$ with the same success rate $p$ as $X$.

Expectation

Furthermore, the First Success expectation is then given by

$$\mathbb{E}(Y)=\mathbb{E}(X+1)=\mathbb{E}(X)+1=\frac{1-p}{p}+1=\frac{1}{p}$$