Binomial to Poisson Distribution

Given that \(n\) is the number of Bernoulli trials and \(p\) is the probability of success, we have:

\[ P(\mathbf{X} = 1) = p \\ P(\mathbf{X} = 0) = q = 1 - p \]

If \(n \to \infty\), \(p \to 0\), and \(n \cdot p\) or \(n \cdot q\) remains constant, the distribution approaches the Poisson Distribution.

\[ \mathbf{X} \sim \text{Binomial}(n, p) \rightarrow \mathbf{X} \sim \text{Poisson}(\lambda) \]

where \(\lambda = n \cdot p\) is the parameter for the Poisson Distribution.

Let \(\lambda = np\), \(p = \frac{\lambda}{n}\), and \(\mathbf{X} \sim \text{Binomial}(n, p)\).

\[ \begin{align*} P(\mathbf{X} = k) &= \binom{n}{k} p^k (1 - p)^{n - k} \\ &= \frac{n!}{k!(n - k)!} \left( \frac{\lambda}{n} \right)^k \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\ &= \frac{\lambda^k}{k!} \frac{n!}{(n - k)!} \left( \frac{\lambda}{n} \right)^k \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\ &= \frac{\lambda^k}{k!} \frac{n(n - 1)(n - 2) \cdots (n - k + 1)}{n^k} \left( 1 - \frac{\lambda}{n} \right)^{n - k} \end{align*} \]

As \(n\) approaches a large number of trials, \(n \to \infty\), we have:

\[ \lim_{n \to \infty} \left( 1 - \frac{\lambda}{n} \right)^n = e^{-\lambda} \]

\[ \lim_{n \to \infty} \left( 1 - \frac{1}{n} \right) \left( 1 - \frac{2}{n} \right) \cdots \left( 1 - \frac{k + 1}{n} \right) = 1 \]

\[ \lim_{n \to \infty} \left( 1 - \frac{\lambda}{n} \right)^{-k} = 1 \]

Thus, we obtain the Poisson distribution:

\[ P(\mathbf{X} = k) = \frac{\lambda^k e^{-\lambda}}{k!} \]