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Geometric Distribution[edit]

Probability mass function
Geometric pmf.svg
Cumulative distribution function
Geometric cdf.svg
Parameters 0< p \leq 1 success probability (real)
Support k \in \{1,2,3,\dots\}\!
PMF (1 - p)^{k-1}\,p\!
CDF 1-(1 - p)^k\!
Mean \frac{1}{p}\!
Median \left\lceil \frac{-1}{\log_2(1-p)} \right\rceil\! (not unique if -1/\log_2(1-p) is an integer)
Mode 1
Variance \frac{1-p}{p^2}\!
Skewness \frac{2-p}{\sqrt{1-p}}\!
Ex. kurtosis 6+\frac{p^2}{1-p}\!
Entropy \tfrac{-(1-p)\log_2 (1-p) - p \log_2 p}{p}\!
MGF \frac{pe^t}{1-(1-p) e^t}\!,
for t<-\ln(1-p)\!
CF \frac{pe^{it}}{1-(1-p)\,e^{it}}\!

There are two similar distributions with the name "Geometric Distribution".

  • The probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set { 1, 2, 3, ...}
  • The probability distribution of the number Y = X − 1 of failures before the first success, supported on the set { 0, 1, 2, 3, ... }

These two different geometric distributions should not be confused with each other. Often, the name shifted geometric distribution is adopted for the former one. We will use X and Y to refer to distinguish the two.


The shifted Geometric Distribution refers to the probability of the number of times needed to do something until getting a desired result. For example:

  • How many times will I throw a coin until it lands on heads?
  • How many children will I have until I get a girl?
  • How many cards will I draw from a pack until I get a Joker?

Just like the Bernoulli Distribution, the Geometric distribution has one controlling parameter: The probability of success in any independent test.

If a random variable X is distributed with a Geometric Distribution with a parameter p we write its probability mass function as:

P\left( X=i \right) =p\left( 1-p\right)^{i-1}

With a Geometric Distribution it is also pretty easy to calculate the probability of a "more than n times" case. The probability of failing to achieve the wanted result is \left( 1-p\right)^k.

Example: a student comes home from a party in the forest, in which interesting substances were consumed. The student is trying to find the key to his front door, out of a keychain with 10 different keys. What is the probability of the student succeeding in finding the right key in the 4th attempt?

P\left( X=4 \right) =\frac{1}{10}\left( 1-\frac{1}{10}\right)^{4-1}=\frac{1}{10}\left( \frac{9}{10}\right)^{3}=0.0729


The probability mass function is defined as:

f(x) = p(1-p)^x \, for x \in \{0, 1, 2, \}


\operatorname{E}[X] = \sum_i f(x_i) x_i = \sum_0^{\infin} p (1-p )^x x

Let q=1-p

\operatorname{E}[X] = \sum_0^{\infin}(1-q) q^x x
\operatorname{E}[X] = \sum_0^{\infin}(1-q)q q^{x-1} x
\operatorname{E}[X] = (1-q)q\sum_0^{\infin} q^{x-1} x
\operatorname{E}[X] = (1-q)q\sum_0^{\infin} \frac{d}{dq}q^x

We can now interchange the derivative and the sum.

\operatorname{E}[X] = (1-q)q\frac{d}{dq}\sum_0^{\infin} q^x
\operatorname{E}[X] = (1-q)q\frac{d}{dq}{1 \over 1-q}
\operatorname{E}[X] = (1-q)q{1 \over (1-q)^2}
\operatorname{E}[X] = q{1 \over (1-q)}
\operatorname{E}[X] = {(1-p ) \over p }


We derive the variance using the following formula:

\operatorname{Var}[X] = \operatorname{E}[X^2] - (\operatorname{E}[X])^2

We have already calculated E[X] above, so now we will calculate E[X2] and then return to this variance formula:

\operatorname{E}[X^2] = \sum_i f(x_i) \cdot x^2
\operatorname{E}[X^2] = \sum_0^{\infin} p(1-p)^x x^2

Let q=1-p

\operatorname{E}[X^2] = \sum_0^{\infin} (1-q)q^x x^2

We now manipulate x2 so that we get forms that are easy to handle by the technique used when deriving the mean.

\operatorname{E}[X^2] = (1-q)\sum_0^{\infin} q^x [(x^2-x)+x]
\operatorname{E}[X^2] = (1-q)\left[\sum_0^{\infin} q^x (x^2-x)+\sum_0^{\infin}q^x x\right]
\operatorname{E}[X^2] = (1-q)\left[q^2\sum_0^{\infin} q^{x-2} x(x-1)+q\sum_0^{\infin}q^{x-1} x\right]
\operatorname{E}[X^2] = (1-q)q\left[q\sum_0^{\infin} \frac{d^2}{(dq)^2}q^x+\sum_0^{\infin}\frac{d}{dq}q^x\right]
\operatorname{E}[X^2] = (1-q)q\left[q\frac{d^2}{(dq)^2}\sum_0^{\infin} q^x+\frac{d}{dq}\sum_0^{\infin}q^x\right]
\operatorname{E}[X^2] = (1-q)q\left[q\frac{d^2}{(dq)^2}{1 \over 1-q}+\frac{d}{dq}{1 \over 1-q}\right]
\operatorname{E}[X^2] = (1-q)q\left[q{2 \over (1-q)^3}+{1 \over (1-q)^2}\right]
\operatorname{E}[X^2] = {2q^2 \over (1-q)^2}+{q \over (1-q)}
\operatorname{E}[X^2] = {2q^2 +q(1-q) \over (1-q)^2}
\operatorname{E}[X^2] = {q(q+1) \over (1-q)^2}
\operatorname{E}[X^2] = {(1-p)(2-p) \over p^2}

We then return to the variance formula

\operatorname{Var}[X] = \left[{(1-p)(2-p) \over p^2}\right] - \left({1-p \over p}\right)^2
\operatorname{Var}[X] = {(1-p) \over p^2}

External links[edit]