Bernoulli Process as Geometric Distribution

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Theorem

Let $\sequence {X_i}$ be a Bernoulli process with parameter $p$.

Let $\EE$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a failure occurs, and then stop.

Let $k$ be the number of successes before a failure is encountered.

Then $k$ is modelled by a geometric distribution with parameter $p$.


Shifted Geometric Distribution

Let $\sequence {Y_i}$ be a Bernoulli process with parameter $p$.

Let $\EE$ be the experiment which consists of performing the Bernoulli trial $Y_i$ as many times as it takes to achieve a success, and then stop.

Let $k$ be the number of Bernoulli trials to achieve a success.

Then $k$ is modelled by a shifted geometric distribution with parameter $p$.


Proof

Follows directly from the definition of geometric distribution.

Let $X$ be the discrete random variable defined as the number of successes before a failure is encountered.

Thus the last trial (and the last trial only) will be a failure, and the others will be successes.

The probability that $k$ successes are followed by a failure is:

$\map \Pr {X = k} = p^k \paren {1 - p}$

Hence the result.

$\blacksquare$


Also presented as

This proof uses as its model the first formulation of the geometric distribution.

However, if the second formulation is used, $k$ is then the number of failures before a success is encountered.

Hence this bears similarity to the shifted geometric distribution.


Sources