Geometric Distribution

Definition

$X \sim \text{Geo}(p)$ counts the number of Bernoulli trials until the first success, where each trial is independent with success probability $p$.

PMF

$P(X = n) = (1-p)^{n-1} p, \quad n = 1, 2, 3, \ldots$

Why this formula? To have $X = n$:

• First $n-1$ trials must fail: probability $(1-p)^{n-1}$
• Trial $n$ succeeds: probability $p$

Range

$X \in \{1, 2, 3, \ldots\}$ (positive integers, starting at 1)

Expectation

$E[X] = \sum_{n=1}^{\infty} n \cdot (1-p)^{n-1} p$

Using $\sum_{n=1}^{\infty} n r^{n-1} = \frac{1}{(1-r)^2}$ for $|r| < 1$:

$E[X] = p \cdot \frac{1}{p^2} = \frac{1}{p}$

Intuition: If success has probability $p$, on average you need $1/p$ trials.

Example

For a fair coin ($p = 0.5$), expected number of flips until heads is $E[X] = 2$.

If $p = 0.1$, expected trials until success is $E[X] = 10$.

Variance

$\text{Var}(X) = \frac{1-p}{p^2}$

Derivation: Using $E[X^2] = \frac{2-p}{p^2}$, we get $\text{Var}(X) = E[X^2] - E[X]^2 = \frac{2-p}{p^2} - \frac{1}{p^2} = \frac{1-p}{p^2}$.

Memoryless Property

The geometric distribution is the only discrete distribution that is memoryless:

$P(X > m + n \mid X > m) = P(X > n)$

Interpretation: If you’ve already waited $m$ trials without success, the probability of waiting at least $n$ more trials is the same as starting fresh.

Proof

$P(X > n) = (1-p)^n$

$P(X > m + n \mid X > m) = \frac{P(X > m + n)}{P(X > m)} = \frac{(1-p)^{m+n}}{(1-p)^m} = (1-p)^n = P(X > n)$

Example

Flip a coin until heads appears. If you’ve flipped 10 tails already, the probability of flipping at least 5 more is the same as if you just started fresh. The coin has no memory.

CDF

$F(n) = P(X \leq n) = 1 - (1-p)^n$

This is the probability of at least one success in $n$ trials.

Examples

Example 1: Coin Flips

Flip a fair coin until heads. $X \sim \text{Geo}(0.5)$.

• $P(X = 1) = 0.5$
• $P(X = 2) = 0.5 \times 0.5 = 0.25$
• $P(X = 3) = 0.5^2 \times 0.5 = 0.125$
• $E[X] = 2$

Example 2: Bit Generation

Generate random bits until the first 1, where each bit is 1 with probability $p = 0.3$.

$E[X] = \frac{1}{0.3} \approx 3.33$

$P(X \leq 5) = 1 - (0.7)^5 = 1 - 0.168 = 0.832$

Example 3: Network Packets

A network sends packets until acknowledgment received. Packet loss rate is 5%.

Expected packets sent until acknowledgment:

$E[X] = \frac{1}{0.95} \approx 1.053$

Alternative Convention

Some textbooks define geometric distribution starting at 0:

$Y = X - 1 \sim \text{Geo}_0(p)$

This counts failures before first success. Then $Y \in \{0, 1, 2, \ldots\}$ with:

$E[Y] = \frac{1-p}{p}, \quad \text{Var}(Y) = \frac{1-p}{p^2}$

Always check which convention is being used.

Relation to Other Distributions

• Bernoulli: Each trial is Bernoulli$(p)$
• Binomial: Counts successes in $n$ trials; geometric counts trials until first success
• Negative Binomial: Counts trials until $r$th success; geometric is the special case $r = 1$

Summary

Property	Value
PMF	$P(X = n) = (1-p)^{n-1} p$
Range	$\{1, 2, 3, \ldots\}$
Mean	$\frac{1}{p}$
Variance	$\frac{1-p}{p^2}$
Memoryless	$P(X > m + n \mid X > m) = P(X > n)$
CDF	$F(n) = 1 - (1-p)^n$