IB Mathematics Applications & Interpretation AHL 4.17—poisson Distribution Notes

Why One Number Controls The Whole Distribution Of Rare Events

1. Many real situations involve counting how many times something happens in a fixed stretch of time or space, such as calls to a helpline in an hour or flaws along a metre of cable.
2. These counts have no natural upper limit; in principle any whole number $0, 1, 2, \dots$ could occur, even if large counts are unlikely.
3. The Poisson distribution models exactly this kind of count, provided the events happen at a steady average rate and do not influence one another.
4. Its defining feature is efficiency: a single number, the mean $m$, fixes the entire distribution, including its spread.

The Conditions That Let You Reach For Poisson

1. A random variable follows a Poisson distribution when it counts events that satisfy three conditions, and all three must hold before the model is valid.
   1. First, the events occur in a fixed interval of time or space, such as one hour, one page, or one square metre.
   2. Second, they happen at a constant average rate, written $m$, that does not drift across the interval.
   3. Third, the events are independent, so one occurrence never changes the chance of another.
2. A useful consequence is that counts in non-overlapping intervals are independent of each other.
3. The count itself is discrete and unbounded above, taking the values $0, 1, 2, \dots$ with no fixed maximum.

The Probability Mass Function

1. The probability of observing exactly $x$ events in the interval is given by the Poisson probability mass function. $$P(X = x) = \dfrac{m^x e^{-m}}{x!}$$
   1. $m > 0$ is the mean number of events in the interval,
   2. $e$ is Euler's number,
   3. $x!$ is the factorial of the count $x$.
2. Because $m$ appears both as a power and inside the exponential, the whole shape of the distribution is locked in once $m$ is known.

Why The Mean Equals The Variance: Poisson As A Binomial Limit

1. The Poisson distribution is the limiting case of a binomial distribution when there are very many chances for an event but each one is very unlikely.
2. Start with $X \sim B(n, p)$, the number of successes in $n$ independent trials, and hold the mean $np = m$ fixed while $n$ grows large and $p$ shrinks.
   1. The binomial mean is $np$, which by construction is exactly $m$.
   2. The binomial variance is $np(1 - p)$, and since $np = m$ this equals $m(1 - p)$.
   3. As $p$ tends to $0$ the factor $1 - p$ tends to $1$, so the variance tends to $m$ as well.
3. The mean and the variance therefore meet at the same value, a property called equidispersion. $$E(X) = m \quad\text{and}\quad \operatorname{Var}(X) = m$$
4. The formula booklet gives both results under Topic 4 (AHL), so you may quote $E(X) = m$ and $\operatorname{Var}(X) = m$ directly.
5. In the same limit the binomial pmf $\binom{n}{x} p^x (1 - p)^{n - x}$ collapses to $\dfrac{m^x e^{-m}}{x!}$, which is where the Poisson formula comes from.

Rescaling The Mean And Reading Probabilities From The GDC

1. The mean $m$ is tied to the length of the interval, so changing the interval changes $m$ in direct proportion.
2. If events occur at a rate of $m$ per unit interval, then over an interval of length $t$ the mean becomes $mt$.
3. For example, a rate of $4$ calls per $10$ minutes gives a mean of $10$ calls over a $25$ minute period, found by scaling $4$ by $\tfrac{25}{10}$.

Using The GDC For Poisson Probabilities

1. AI provides a calculator on every paper, so Poisson probabilities are read from technology once the correct $m$ is set.
2. Use the Poisson probability density function $\texttt{poissonpdf}$, with inputs $m$ and $x$, to find $P(X = x)$.
3. Use the Poisson cumulative density function $\texttt{poissoncdf}$ to find $P(X \leq x)$ for at most problems.
4. Write the distribution and the probability statement, such as $X \sim \text{Po}(m)$ and $P(X \geq 3)$, before quoting the calculator value, because that set-up earns the method mark.
5. For at least problems use the complement $P(X \geq x) = 1 - P(X \leq x - 1)$, which avoids summing an infinite tail.

Adding Independent Poisson Processes

1. When two independent Poisson variables are added, the result is again Poisson, with the two means simply added together.
2. If $X_1 \sim \text{Po}(m_1)$ and $X_2 \sim \text{Po}(m_2)$ are independent, then their sum follows a single Poisson distribution. $$X_1 + X_2 \sim \text{Po}(m_1 + m_2)$$
3. Intuitively the combined process still has a constant overall rate, namely $m_1 + m_2$, so it satisfies the Poisson conditions.
4. The rule extends to any number of independent Poisson variables, so several separate sources can be pooled into one distribution.

Choosing Between Binomial, Poisson And Normal

1. Many contexts could be modelled in more than one way, so the exam skill is matching the situation to the right distribution.
2. Choose Poisson for a count of events in a fixed interval at a constant average rate, with no fixed maximum count.
3. Choose binomial for a fixed number $n$ of independent trials, each with the same success probability $p$.
4. Choose normal for a continuous measurement that is roughly symmetric about its mean.
5. When $n$ is large and $p$ is small, the binomial and Poisson give almost identical answers, so either model is acceptable.