Bernoulli Process as Negative Binomial Distribution

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

First Form
Let $\mathcal E$ be the experiment which consists of performing the Bernoulli trial $X_i$ until a total of $n$ failures have been encountered.

Let $X$ be the discrete random variable defining the number of successes before $n$ failures have been encountered.

Then $X$ is modeled by a negative binomial distribution of the first form.

Second Form
Let $\mathcal E$ be the experiment which consists of performing the Bernoulli trial $X_i$ as many times as it takes to achieve a total of $n$ successes, and then stops.

Let $Y$ be the discrete random variable defining the number of trials before $n$ successes have been achieved.

Then $X$ is modeled by a negative binomial distribution of the second form.

Proof for First Form
The number of Bernoulli trials may be as few as $0$, so the image is correct: $\operatorname{Im} \left({X}\right) = \left\{{0, 1, 2, \ldots}\right\}$.

If $X$ takes the value $x$, then there must have been $n + x$ trials altogether.

So, after $n + x - 1$ trials, there must have been $n - 1$ failures, as (from the description of the experiment) the last trial is a failure.

So the probability of the occurrence of the event $\left[{X = x}\right]$ is given by the binomial distribution, as follows:
 * $\displaystyle p_X \left({x}\right) = \binom {n + x - 1} {n - 1} p^x \left({1-p}\right)^n$

where $x \in \left\{{0, 1, 2, \ldots}\right\}$

Hence the result, by definition of first form of the negative binomial distribution.

Proof for Second Form
First note that the number of Bernoulli trials has to be at least $n$, so the image is correct: $\operatorname{Im} \left({X}\right) = \left\{{n, n+1, n+2, \ldots}\right\}$.

Now, note that if $X$ takes the value $x$, then in the first $x-1$ trials there must have been $n-1$ successes.

Hence there must have been $x-n$ failures, and so a success happens at trial number $x$.

So the probability of the occurrence of the event $\left[{X = x}\right]$ is given by the binomial distribution, as follows:
 * $\displaystyle p_X \left({x}\right) = \binom {x - 1} {n - 1} \left({1-p}\right)^{x - n} p^n$

where $x \in \left\{{k, k+1, k+2, \ldots}\right\}$

Hence the result, by definition of second form of the negative binomial distribution.