Geometric Distribution

For a Bernoulli Trial, let $X$ be the number of tests when we first meet 1. Then $X$ is said to have a geometric distribution. Another definition lets $X$ be the number of failures before the first success, then the PMF is $p{q}^n$.

• Parameter
  • $p\in [0,1]$
• CDF
  • $F(n)=1-q^n$
• PMF
  • $p(n)=p{q}^{n-1}$
• Mean
  • $1/p$
• Variance
  • $q/p^2$
• MGF
  • $\frac{p{e}^t}{1-q{e}^t}$, $e^t<1/q$

Memoryless

An geometric random variable is memoryless:

$$P(X\ge n+k|X>n)=\frac{q^{n+k-1}}{q^n}=q^{k-1}=P(x\ge k).$$

This gives an alternative method to calculate the expectation of geometric distribution using Conditional Expectation. For the mean, we have

$$\mathbb{E}[X]=\mathbb{E}[X\mid X=1]P(X=1)+\mathbb{E}[X\mid X>1]P(X>1)=1\cdot p+\left(1+\mathbb{E}[X]\right)\cdot q,$$

which implies $\mathbb{E}[X]=(p+q)/p=1/p$. Similarly, for the second moment, we have

$$\mathbb{E}[X^2]=1\cdot p+(\mathbb{E}[(X+1{)}^2])\cdot q,$$

which implies $\mathbb{E}[X^2]=(2-p)/p^2$.