Bernoulli Process as Binomial Distribution

Theorem
Let $\sequence {X_i}$ be a finite Bernoulli process of length $n$ such that each of the $X_i$ in the sequence is a Bernoulli trial with parameter $p$.

Then the number of successes in $\sequence {X_i}$ is modelled by a binomial distribution with parameters $n$ and $p$.

Hence it can be seen that:
 * $X \sim \Binomial 1 p$ is the same thing as $X \sim \Bernoulli p$

Proof
Consider the sample space $\Omega$ of all sequences $\sequence {X_i}$ of length $n$.

The $i$th entry of any such sequence is the result of the $i$th trial.

We have that $\Omega$ is finite.

Let us take the event space $\Sigma$ to be the power set of $\Omega$.

As the elements of $\Omega$ are independent, by definition of the Bernoulli process, we have that:
 * $\forall \omega \in \Omega: \map \Pr \omega = p^{\map s \omega} \paren {1 - p}^{n - \map s \omega}$

where $\map s \omega$ is the number of successes in $\omega$.

In the same way:
 * $\ds \forall A \in \Sigma: \map \Pr A = \sum_{\omega \mathop \in A} \map \Pr \omega$

Now, let us define the discrete random variable $Y_i$ as follows:
 * $\map {Y_i} \omega = \begin{cases}

1 & : \text {$\omega_i$ is a success} \\ 0 & : \text {$\omega_i$ is a failure} \end{cases}$ where $\omega_i$ is the $i$th element of $\omega$.

Thus, each $Y_i$ has image $\set {0, 1}$ and a probability mass function:
 * $\map \Pr {Y_i = 0} = \map \Pr {\set {\omega \in \Omega: \text {$\omega_i$ is a success} } }$

Thus we have:

Then:
 * $\map \Pr {Y_i = 0} = 1 - \map \Pr {Y_i = 1} = 1 - p$

So (by a roundabout route) we have confirmed that $Y_i$ has the Bernoulli distribution with parameter $p$.

Now, let us define the random variable:
 * $\ds \map {S_n} \omega = \sum_{i \mathop = 1}^n \map {Y_i} \omega$

By definition:
 * $\map {S_n} \omega$ is the number of successes in $\omega$
 * $S_n$ takes values in $\set {0, 1, 2, \ldots, n}$ (as each $Y_i$ can be $0$ or $1$).

Also, we have that:

Hence the result.