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Geometric distribution

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

Parameters and support

pp
probability of success on one trial - 0<p≤10 < p \le 1
Support
x∈{0,1,2,… }x \in \{0, 1, 2, \dots\}

The formulas

p(x)
p(x)=(1−p)xpp(x) = (1-p)^{x} p
F(x)
F(x)=1−(1−p)⌊x⌋+1F(x) = 1 - (1-p)^{\lfloor x \rfloor + 1}
Mean
E[X]=1−ppE[X] = \dfrac{1-p}{p}
Variance
Var⁡(X)=1−pp2\operatorname{Var}(X) = \dfrac{1-p}{p^{2}}
MGF
M(t)=p1−(1−p)et,t<−ln⁡(1−p)M(t) = \dfrac{p}{1 - (1-p)e^{t}}, \quad 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

  1. P(X=x)=(1−p)xpP(X = x) = (1-p)^x p: xx failures then a success, by independence.
  2. E[X]=∑x≥0x(1−p)xpE[X] = \sum_{x\ge 0} x(1-p)^x p. Differentiating the geometric series ∑qx=11−q\sum q^x = \frac{1}{1-q} gives ∑xqx=q(1−q)2\sum x q^{x} = \frac{q}{(1-q)^2} with q=1−pq = 1-p.
  3. So E[X]=p⋅1−pp2=1−ppE[X] = p \cdot \frac{1-p}{p^2} = \frac{1-p}{p}.
  4. The same trick on the second derivative gives E[X2]E[X^2], and Var⁡(X)=1−pp2\operatorname{Var}(X) = \frac{1-p}{p^2}.
  5. The geometric is the only discrete distribution that is MEMORYLESS: P(X>m+n∣X>m)=P(X>n)P(X > m+n \mid 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.

  1. 'The fourth file is the first success' means 3 failures then a success, so X=3X = 3 under the failures-before-first-success parameterisation.
  2. P(X=3)=(1−0.2)3(0.2)=(0.8)3(0.2)P(X = 3) = (1-0.2)^3(0.2) = (0.8)^3(0.2).
  3. (0.8)3=0.512(0.8)^3 = 0.512, so P(X=3)=0.512×0.2=0.1024P(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 that test run.

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.

Related

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.