Geometric distribution

The geometric distribution $𝑋 \sim G e o m (𝑝)$ describes the number of failures of independent Bernoulli trials with success probability $𝑝$ before the first success. #m/def/prob It has the probability mass function

ℙ (𝑋 = 𝑥) = (1 - 𝑝) 𝑥 𝑝 = 𝑞 𝑥 𝑝

for $𝑥 \in ℕ 0$ .

Proof

Let $𝑌 𝑖 \sim B e r n (𝑝)$ be independent for $𝑖 \in ℕ$ . Then

ℙ (𝑋 = 𝑥) = ℙ ({𝑋 𝑥 + 1 = 1} \cap 𝑥 ⋂ 𝑖 = 1 {𝑋 𝑖 = 0}) = ℙ (𝑋 𝑥 + 1 = 1) 𝑥 \prod 𝑖 = 1 ℙ (𝑋 𝑖 = 0) = 𝑝 (1 - 𝑝) 𝑥

as claimed.

This is related to the Negative binomial distribution, which is the sum of i.i.d. geometric variables.

Properties

Let $𝑋 \sim G e o m (𝑝)$ and $𝑞 = 1 - 𝑝$

Expectation: $𝔼 [𝑋] = 𝑞 𝑝$
Variance: $V a r [𝑋] = 𝑞 𝑝 2$
Moment-generating function:

𝑀 𝑋 : (- \infty, - l n 𝑞) \to ℝ : 𝑡 \mapsto 𝑝 1 - 𝑞 e 𝑡

Proof of 1

Invoking the expansion for a geometric series

𝔼 [𝑋] = \infty \sum 𝑥 = 0 𝑥 ℙ (𝑥 = 𝑥) = \infty \sum 𝑥 = 0 𝑥 (1 - 𝑝) 𝑥 𝑝 = (1 - 𝑝) 𝑝 \infty \sum 𝑥 = 0 𝑥 (1 - 𝑝) 𝑥 - 1 = (1 - 𝑝) 𝑝 \infty \sum 𝑥 = 0 - 𝑑 𝑑 𝑝 [(1 - 𝑝) 𝑥] = (𝑝 - 1) 𝑝 𝑑 𝑑 𝑝 \infty \sum 𝑥 = 0 (1 - 𝑝) 𝑥 = (𝑝 - 1) 𝑝 𝑑 𝑑 𝑝 1 1 (1 - 𝑝) = (𝑝 - 1) 𝑝 𝑑 𝑑 𝑝 𝑝 - 1 = (1 - 𝑝) 𝑝 - 1

as claimed, proving ^P1.

Alternately we may invoke conditional expected value. Let $𝑆$ denote the event that the first trial is successful. Then noting that $(𝑋 ∣ 𝑆) \sim 𝑋 + 1$ , we have

𝔼 [𝑋] = 𝔼 [𝑋 ∣ 𝑆] 𝑝 + 𝔼 [𝑋 ∣ 𝑆 𝑐] 𝑞 = 𝔼 [1 + 𝑋] 𝑞 = 𝑞 (1 + 𝔼 [𝑋])

whence

𝔼 [𝑋] = 𝑞 𝑝

proving ^P1.

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