Bernoulli Process as Binomial Distribution

Theorem
Let $$\left \langle{X_i}\right \rangle$$ 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 $$\left \langle{X_i}\right \rangle$$ is modelled by a binomial distribution with parameters $n$ and $p$.

Hence it can be seen that:
 * $$X \sim \operatorname{B} \left({1, p}\right)$$ is the same thing as $$X \sim \operatorname{Bern} \left({n, p}\right)$$

Proof
Consider the sample space $$\Omega$$ of all sequences $$\left \langle{X_i}\right \rangle$$ 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 $$\Sigma$$ to be the power set of $$\Omega$$.

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

where $$s \left({\omega}\right)$$ is the number of successes in $$\omega$$.

In the same way:
 * $$\forall A \in \Sigma: \Pr \left({A}\right) = \sum_{\omega \in A} \Pr \left({\omega}\right)$$

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

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

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

Thus we have:

$$ $$ $$ $$ $$

Then:
 * $$\Pr \left({Y_i = 1}\right) = 1 - \Pr \left({Y_i = 0}\right) = p$$

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

Now, if we let $$S_n \left({\omega}\right) = \sum_{i=1}^n Y_i \left({\omega}\right)$$

It is clear that:
 * $$S_n \left({\omega}\right)$$ is the number of successes in $$\omega$$;
 * $$S_n$$ takes values in $$\left\{{0, 1, 2, \ldots, n}\right\}$$ (as each $$Y_i$$ can be $$0$$ or $$1$$).

Also, we have that:

$$ $$ $$

Hence the result.