Bernoulli Process as Binomial Distribution

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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:

\(\ds \map \Pr {Y_i = 1}\) \(=\) \(\ds \sum_{\omega: \text {$\omega_i$ success} } p^{\map s \omega} \paren {1 - p}^{n - \map s \omega}\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^n \sum_{\substack {\omega: \text {$\omega_i$ success} \\ \map s \omega = r} } p^r \paren {1 - p}^{n - r}\)
\(\ds \) \(=\) \(\ds \sum_{r \mathop = 1}^n \binom {n - 1} {r - 1} p^r \paren {1 - p}^{n - r}\) as we already know the position of one success (namely $i$)
\(\ds \) \(=\) \(\ds p \sum_{r \mathop = 0}^{n - 1} \binom {n - 1} r p^r \paren {1 - p}^{\paren {n - 1} - r}\) switching summation index
\(\ds \) \(=\) \(\ds p \paren {p + \paren {1 - p} }^{n - 1}\) Binomial Theorem
\(\ds \) \(=\) \(\ds p\)


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:

\(\ds \map \Pr {S_n = k}\) \(=\) \(\ds \map \Pr {\set {\omega \in \Omega: \map s \omega = k} }\)
\(\ds \) \(=\) \(\ds \sum_{\omega: \map s \omega \mathop = k} \map \Pr \omega\)
\(\ds \) \(=\) \(\ds \sum_{\omega: \map s \omega \mathop = k} p^k \paren {1 - p}^{n - k}\)
\(\ds \) \(=\) \(\ds \binom n k p^k \paren {1 - p}^{n - k}\)

Hence the result.

$\blacksquare$


Sources

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