Geometric Distribution

Discrete
Used to calculate the probability of a success on trial number $x + 1$ on independent trials
$p$ is the probability of a success
$1 - p$ is the probability of a failure
To observe a success on trial $x + 1$
$\begin{matrix} (1 - p) & (1 - p) & (1 - p) & \dots & (1 - p) & | & p \\ F & F & F & \dots & F & | & S \\ 1 & 2 & 3 & \dots & x & | & x + 1 \end{matrix}$
$x$ trials are failures, then $x + 1$ is the successful trial
Probability Mass Function:
- $P (X = x) = p (1 - p)^{x}$
- Support: $x = 0, 1, 2, \dots$
- Properties
  - Mean: $μ = E (X) = \frac{1 - p}{p}$
  - Variance: $σ^{2} = v a r (x) = \frac{1 - p}{p^{2}}$
    - Does this imply that $\frac{1 - p}{p^{3}}$ is the third moment?
  - MGF is on formula sheet
  - #tk practice getting the MGF through the formulas through the formula sheet
Cumulative Distribution Function:
- #tk try deriving it from the Probability Mass Function
- $P (X \leq x) = 1 - (1 - p)^{x + 1}$
- $P (X > x) = (1 - p)^{x + 1}$
- $P (X < x) = 1 - (1 - p)^{x}$
  - Take the complement
- $P (X \leq x) = 1 - (1 - p)^{x + 1}$
  - Add one more trial
- #tk check the examples
Example: STA256 Lecture 20#^9495ff