Reading the textbook Mathematical Statistics and Data Analysis 3rd ed, by Rice. I've come up on an example that I'm trying to extend beyond the text:
So I am trying to obtain one of the stated Poisson probabilities, but using the binomial distribution instead. I'm not sure if I am interpreting things right to get my stated goal. For instance let's take trying to get $\text{Number of Deaths} = 0$. From the Poisson Probability this is given as $0.543$.
With the given information I am able to calculate a "probability" but I'm not sure what it means:
$$np = \lambda \\ \Rightarrow p = \frac{\lambda}{n}$$
So we know that $n = 200$ and $\lambda = 0.61$, meaning
$$p = \frac{0.61}{200} = 0.00305$$
I took this as meaning the "probability of dying from horse kick". Here is where I get stuck trying to convert the problem into a binomial distribution problem. I could see framing things in terms of deaths -no deaths and that may possibly look like:
$$\binom{200}{109}(0.00305)^{109}(0.99695)^{91}$$
But how would I go about things if I wanted to get 1 death, 2 deaths,…etc? How could I frame things to get the same (or close to) Poisson probabilities stated but with a binomial distribution instead ?
Best Answer
The random variable to which Bortkiewicz attributes the Poisson distribution with expected value $0.61$ is the number of such deaths in each corps in each year. Thus if $n$ is the number of soldiers in each corps and $p$ is the probability that a soldier is killed in this way during a year, then $np=\lambda = 0.61.$ So let $X$ be the number of such deaths in a particular corps in one year. Then we have \begin{align} & \Pr(X=3) = \binom n 3 p^3(1-p)^{n-3} \\[10pt] = {} & \frac{n(n-1)(n-2)}{3!} p^3 (1-p)^{n-3} \\[10pt] = {} & \frac{\big(np\big)^3 }{3!}\cdot {} \underbrace{ \frac{n(n-1)(n-2)}{n^3} \cdot \left( 1-\frac\lambda n \right)^{-3} }_\text{These approach 1 as $n\,\to\,\infty$} {} \cdot \left( 1 - \frac \lambda n \right)^n \\[12pt] \to {} & \frac{\lambda^3}{3!}\cdot 1 \cdot 1 \cdot e^{-\lambda} = \frac{0.61^3 e^{-0.61}}{3\cdot2\cdot1} \quad \text{as } n\to\infty. \end{align}