Bernoulli Process as Geometric Distribution

Geometric Distribution
Let $$\left \langle{X_i}\right \rangle$$ be a Bernoulli process with parameter $p$.

Let $$\mathcal E$$ 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 $$\left \langle{Y_i}\right \rangle$$ be a Bernoulli process with parameter $p$.

Let $$\mathcal E$$ 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 for Geometric Distribution
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:
 * $$P \left({X = k}\right) = p^k \left({1 - p}\right)$$

Hence the result.

Proof for Shifted Geometric Distribution
Follows directly from the definition of shifted geometric distribution.

Let $$Y$$ be the discrete random variable defined as the number of trials for the first success to be achieved.

Thus the last trial (and the last trial only) will be a success, and the others will be failures.

The probability that $$k-1$$ failures are followed by a success is:
 * $$P \left({Y = k}\right) = \left({1 - p}\right)^{k-1} p$$

Hence the result.