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Topics Covered

Definition PMF Mean & Variance MGF/CF/CGF/PGF Skewness & Kurtosis Lack of Memory Recurrence
On this page
  1. 1. Definition & PMF
  2. 2. MGF, Mean & Variance
  3. 3. Other Generating Functions
  4. 4. Additive Property
  5. 5. Skewness & Kurtosis
  6. 6. Lack-of-Memory Property (Memoryless Property)
  7. 7. Recurrence Relation
  8. 8. Worked Examples
  9. Key Take-aways from Unit 4

1. Definition & PMF

DEFINITION

Let \(X\) = number of failures preceding the first success in a sequence of independent Bernoulli\((p)\) trials. Then \(X\) follows a Geometric distribution with parameter \(p\) (\(0 < p < 1\)) with PMF

\[ P(X = x) \;=\; q^x p, \quad x = 0, 1, 2, \ldots, \quad q = 1 - p. \]

This is the special case NB\((1, p)\) of Negative Binomial.

Alternative form (number of trials including the success)

Some texts define \(Y =\) trial number on which the first success occurs:

\[ P(Y = y) \;=\; q^{y-1} p, \quad y = 1, 2, 3, \ldots \]

The two forms differ by 1: \(Y = X + 1\). Means & variances differ accordingly.

Validity: \(\sum_{x=0}^{\infty} q^x p = p \cdot \dfrac{1}{1-q} = p \cdot \dfrac{1}{p} = 1\) ✓.

2. MGF, Mean & Variance

MGF \[ M_X(t) \;=\; \sum_{x=0}^{\infty} e^{tx} q^x p \;=\; \dfrac{p}{1 - q e^t}, \quad q e^t < 1. \]

Differentiating:

\(M'(0) = q/p = E(X)\).

\(M''(0) = (2q^2 + qp)/p^2;\;\) hence \(E(X^2) = (q + q^2)/p^2\).

\[ E(X) = \dfrac{q}{p}, \qquad \text{Var}(X) = \dfrac{q}{p^2}. \]

(For the alternative form \(Y\): \(E(Y) = 1/p\); same variance.)

3. Other Generating Functions

\[ \phi_X(t) = \dfrac{p}{1 - q e^{it}}, \qquad K_X(t) = \ln p - \ln(1 - q e^t), \] \[ P_X(s) = \dfrac{p}{1 - qs}, \quad |s| < 1/q. \]

4. Additive Property

The sum of \(r\) independent Geometric\((p)\) variables is NB\((r, p)\) — i.e., Geometric is not closed under addition, but Negative Binomial is. So the geometric does not have an additive property in the usual sense.

5. Skewness & Kurtosis

\[ \mu_3 = \dfrac{q(1+q)}{p^3}, \qquad \mu_4 = \dfrac{q(p^2 + 6q + 3q)}{p^4} = \dfrac{q(p^2 + 9q)}{p^4}. \] \[ \gamma_1 = \dfrac{1 + q}{\sqrt{q}}, \qquad \beta_2 = 3 + \dfrac{p^2 + 6q}{q}. \]

Strongly right-skewed and leptokurtic for all \(p < 1\).

0 1 2 3 4 5 6 7 8 9 Geometric (p = 0.4) — strictly decreasing
Fig 4.1 — Each successive probability is q times the previous one

6. Lack-of-Memory Property (Memoryless Property)

STATEMENT

The Geometric distribution is the only discrete distribution with the lack-of-memory property:

\[ P(X \ge m + n \mid X \ge m) \;=\; P(X \ge n), \quad m, n \in \{0, 1, 2, \ldots\}. \]

Proof. Here \(X\) counts the number of failures before the first success, so \(P(X = k) = q^{k}p\) with \(q = 1-p\).

  1. Tail probability. The event \(\{X \ge k\}\) means the first \(k\) trials are all failures. Summing the geometric series, \[ P(X \ge k) = \sum_{j=k}^{\infty} q^{j}p = p\,q^{k}\sum_{i=0}^{\infty} q^{i} = p\,q^{k}\cdot\dfrac{1}{1-q} = p\,q^{k}\cdot\dfrac{1}{p} = q^{k}. \]
  2. Apply the definition of conditional probability. Since \(\{X \ge m+n\}\subseteq\{X \ge m\}\), the intersection is just \(\{X \ge m+n\}\), so \[ P(X \ge m+n \mid X \ge m) = \dfrac{P(X \ge m+n)}{P(X \ge m)}. \]
  3. Substitute the tail probabilities from step 1. \[ P(X \ge m+n \mid X \ge m) = \dfrac{q^{m+n}}{q^{m}} = q^{n}. \]
  4. Recognise the result. By step 1, \(q^{n} = P(X \ge n)\). Therefore \(P(X \ge m+n \mid X \ge m) = P(X \ge n)\), which is the lack-of-memory property. \(\blacksquare\)

Interpretation: the past does not influence the future — given that the first \(m\) trials are failures, the conditional distribution of the remaining wait is the same as the original.

7. Recurrence Relation

\[ P(X = x + 1) \;=\; q \cdot P(X = x), \qquad P(X = 0) = p. \]

Geometric probabilities form a geometric sequence with ratio \(q\) — hence the name.

8. Worked Examples

EXAMPLE 1

The probability that a die shows 6 is 1/6. Find the probability that the first 6 occurs on the 4th throw.

Using \(Y\) form (trial of first success): \(P(Y=4) = (5/6)^3 (1/6) = 125/1296 = 0.0965\).

Equivalently, \(X = 3\) (3 failures before the success), \(P(X=3) = (5/6)^3 (1/6)\).

EXAMPLE 2 (Lack of memory)

A telephone exchange has the property that for each minute, an outage occurs with probability 0.1 (independently of past).

Given that no outage has happened in the last 5 minutes, what is the probability of no outage in the next 5 minutes?

By memoryless property: same as \(P(\text{no outage for 5 min}) = (0.9)^5 \approx 0.5905\).

Key Take-aways from Unit 4

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