Answer 3.1 (a)

_posts/2019-02-14-pset2.md

@@ -8,6 +8,27 @@ title: Problem Set 2
 
 Derive equation (3.16) from the binomial distribution and Stirling’s approximation.
 
+Fix any $$\lambda$$ between 0 and 1. For any $$n$$, the binomial probability mass function for $$n$$
+trials with probability of success $$p = \lambda / n$$ is
+
+$$
+\begin{align*}
+p(k; n, \lambda/n)
+&= {n \choose k} \left( \frac{\lambda}{n} \right)^k \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{n (n - 1) \cdots (n - k + 1)}{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{n^k + \mathcal{O}\left( n^{k - 1} \right) }{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{\lambda^k}{k!} \left( 1 - \frac{\lambda}{n} \right)^{n - k} +
+   \frac{\mathcal{O}\left( n^{k - 1} \right) }{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k}
+\end{align*}
+$$
+
+As $$n$$ approaches infinity, the second term goes to zero because the $$n^{-k}$$ dominates
+$$\mathcal{O}\left( n^{k - 1} \right)$$ (and the last factor is always between 0 and 1). The first
+term approaches $$\lambda^k e^{-\lambda}/k!$$ which is the Poisson distribution.
+
 ### (b)
 
 Use it to derive equation (3.18).