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Geometric distribution

The number of failures before the first success in repeated independent trials.

Parameters and support

$p$: probability of success on one trial - $0 < p \leq 1$
Support: $x \in {0, 1, 2, \dots}$

The formulas

p(x): $p (x) = (1 - p)^{x} p$
F(x): $F (x) = 1 - (1 - p)^{⌊ x ⌋ + 1}$
Mean: $E [X] = \frac{1 - p}{p}$
Variance: $Var (X) = \frac{1 - p}{p ^{2}}$
MGF: $M (t) = \frac{p}{1 - ( 1 - p ) e ^{t}}, t < - ln (1 - p)$

Memory hook. Decide FIRST which version you are in. Counting failures: mean (1−p)/p. Counting trials: mean 1/p. The variance is (1−p)/p² either way - shifting by one changes the mean, never the variance.

Where the moments come from

$P (X = x) = (1 - p)^{x} p$ : $x$ failures then a success, by independence.
$E [X] = \sum_{x \geq 0} x (1 - p)^{x} p$ . Differentiating the geometric series $\sum q^{x} = \frac{1}{1 - q}$ gives $\sum x q^{x} = \frac{q}{( 1 - q ) ^{2}}$ with $q = 1 - p$ .
So $E [X] = p \cdot \frac{1 - p}{p ^{2}} = \frac{1 - p}{p}$ .
The same trick on the second derivative gives $E [X^{2}]$ , and $Var (X) = \frac{1 - p}{p ^{2}}$ .
The geometric is the only discrete distribution that is MEMORYLESS: $P (X > m + n ∣ X > m) = P (X > n)$ .

Worked example

A claims adjuster reviews files one at a time; each is independently misfiled with probability 0.2. Find the probability that the first misfiled file is the fourth one reviewed.

'The fourth file is the first success' means 3 failures then a success, so $X = 3$ under the failures-before-first-success parameterisation.
$P (X = 3) = (1 - 0.2)^{3} (0.2) = (0.8)^{3} (0.2)$ .
$(0.8)^{3} = 0.512$ , so $P (X = 3) = 0.512 \times 0.2 = 0.1024$ .

Answer: 0.1024

The mean, variance, CDF and moment generating function above are re-derived numerically from this distribution’s own density on every test run - summed over the support for a discrete distribution, integrated by quadrature for a continuous one - and compared with the closed forms printed here. A typo on this page fails the build.

Traps

Mixing the two parameterisations mid-problem - the trials version is the failures version plus 1.
Applying memorylessness to a non-geometric discrete distribution; it is the only one that has it.

Drill this: the Exam P question bank has original questions on this distribution, and the recall trainer builds its prompts from exactly the formulas above.

Geometric distribution

Parameters and support

The formulas

Where the moments come from

Worked example

Traps

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