SST 204 • UNIT 8 • Statistics

Poisson Distribution And Binomial Approximation

Topics covered: Poisson distribution, Approximating binomial probabilities

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1. Core Concept Overview & Intuition

The Poisson distribution is the canonical probability model for counting the number of occurrences of a rare event in a fixed interval of time, length, area, or volume. It arises whenever we observe a large number of independent trials, each with a very small probability of "success," and we count the total number of successes. The classical motivating setting is the binomial experiment with parameters $n$ and $p$ in which $n$ is large and $p$ is small; in that regime the binomial probability mass function is closely reproduced by a single-parameter Poisson mass function indexed only by the mean $\lambda = np$.

The intuition is that of a "law of rare events." If a process generates events independently and at a constant average rate, so that the chance of two events occurring in the same instant is negligible, then the count of events in a fixed window is Poisson distributed. The parameter $\lambda$ is simultaneously the mean and the variance of the count, so the model is fully determined once the average rate is known.

The second subtopic, approximating binomial probabilities, is the practical bridge between the two models. When $n$ is large and $p$ is small, direct evaluation of binomial coefficients becomes computationally prohibitive and numerically unstable, whereas the Poisson mass function involves only powers and a single factorial. The approximation replaces the binomial probability of $x$ successes by the Poisson probability with mean $\lambda = np$, and the approximation improves as $n \to \infty$ while $np$ remains fixed.

2. Mathematical Formalization & Core Definitions

> Definition (Poisson random variable). A discrete random variable $X$ is said to have a Poisson distribution with parameter $\lambda > 0$ if its probability mass function is

$$
f(x) = P(X = x) = \frac{e^{-\lambda}\lambda^{x}}{x!}, \qquad x = 0, 1, 2, \dots
$$

> and $f(x) = 0$ otherwise. The parameter $\lambda$ is the mean number of occurrences in the interval under study.

The mass function satisfies the two conditions required of any discrete probability distribution. Non-negativity is immediate because $e^{-\lambda} > 0$, $\lambda^{x} > 0$, and $x! > 0$. The total mass equals one because

$$
\sum_{x=0}^{\infty} \frac{e^{-\lambda}\lambda^{x}}{x!}
= e^{-\lambda}\sum_{x=0}^{\infty}\frac{\lambda^{x}}{x!}
= e^{-\lambda}e^{\lambda}
= 1.
$$

> Definition (Binomial random variable). A discrete random variable $X$ has a binomial distribution with parameters $n$ and $p$ if

$$
P(X = x) = \binom{n}{x}p^{x}(1-p)^{n-x}, \qquad x = 0, 1, \dots, n.
$$

The binomial mean is $np$ and its variance is $np(1-p)$. The Poisson model is obtained as a limiting case in which $n \to \infty$, $p \to 0$, and $np \to \lambda$.

> Definition (Poisson approximation to the binomial). For large $n$ and small $p$, the binomial probability is approximated by

$$
P(X = x) \approx \frac{e^{-np}(np)^{x}}{x!}, \qquad x = 0, 1, 2, \dots
$$

> where $\lambda = np$ is the assumed mean of the count.

The approximation is justified by the limiting argument. Write $p = \lambda/n$ and expand the binomial coefficient:

$$
\binom{n}{x}p^{x}(1-p)^{n-x}
= \frac{n(n-1)\cdots(n-x+1)}{x!}\left(\frac{\lambda}{n}\right)^{x}\left(1-\frac{\lambda}{n}\right)^{n-x}.
$$

As $n \to \infty$ with $x$ fixed, the falling factorial ratio satisfies

$$
\frac{n(n-1)\cdots(n-x+1)}{n^{x}} \to 1,
$$

the factor $\left(1-\lambda/n\right)^{n} \to e^{-\lambda}$, and $\left(1-\lambda/n\right)^{-x} \to 1$. Hence the binomial probability converges to $e^{-\lambda}\lambda^{x}/x!$, which is the Poisson mass function with mean $\lambda$.

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Question 4 (12 marks)

(a) A random variable $X$ is said to have a Poisson distribution with parameter $\mu>0$ if its probability mass function is

$$
f(x)=\frac{e^{-\mu}\mu^{x}}{x!},\qquad x=0,1,2,\dots
$$

(i) State what the parameter $\mu$ represents in this distribution. (2 marks)
(ii) Using the probability mass function, show that $E(X)=\mu$. (5 marks)
(b) In a certain county, the average number of days that schools are closed owing to floods during the month of April is $4$. Assume that the number of such closure days follows a Poisson distribution with probability mass function

$$
f(x)=\begin{cases}
\dfrac{e^{-4}4^{x}}{x!}, & x=0,1,2,\dots,
0, & \text{otherwise}.
\end{cases}
$$

Find the probability that schools in the county will be closed for exactly $6$ days during floods in April. (5 marks)

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Question 2 (10 marks)

A manufacturing process produces items of which $3\%$ are defective. A random sample of $200$ items is taken. Using the Poisson approximation to the binomial distribution, find the probability that the sample contains:

(i) no defective items; (3 marks)
(ii) exactly three defective items. (2 marks)

If $X$ is a Poisson random variable such that $P(X=2)=P(X=3)$, determine the value of $E(X)$. (5 marks)

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