Bernoulli Process as Geometric Distribution

Theorem

Let $\sequence {X_i}$ be a binomial experiment 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 $p$.

Let $\sequence {Y_i}$ be a binomial experiment 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 $p$.

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.

$\blacksquare$

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.

1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): geometric distribution
2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): geometric distribution