FS1 2.2, 2.3 - Means, variances and the Poisson approximation | Edexcel A-Level Further Mathematics

A binomial and a Poisson model can have the same mean while predicting different spreads and probabilities. You will use their mean and variance formulae (FS1 2.2), recover binomial parameters and justify when a Poisson distribution is a useful approximation (FS1 2.3).

Matching the mean does not match the spread

Two experiments both expect three successes: $X\sim\mathrm B(300,0.01)$ and $W\sim\mathrm B(6,0.5)$ . Should the same $\operatorname{Po}(3)$ model approximate both? No: one has rare successes across many trials, while the other has only six trials and success probability one half.

Their exact zero-success probabilities are approximately $0.0490$ and $0.0156$ , more than threefold apart despite the equal means. By the end you will see why the Poisson approximation belongs to the first experiment.

You already know

From A-level Mathematics, $X\sim\mathrm B(n,p)$ counts successes in $n$ independent trials, each with the same success probability $p$ , and has values $0,1,\ldots,n$ . From discrete expectation, a mean describes centre and a variance describes squared spread; matching one does not establish a model's suitability.

The mean and variance of these two distributions are supplied results. You do not need to derive them from their probability formulae.

Binomial and Poisson moments

If $X\sim\mathrm B(n,p)$ , where $n$ is a positive integer and $0\le p\le1$ , then

\mathrm E(X)=np,\qquad \operatorname{Var}(X)=np(1-p)

If $Y\sim\operatorname{Po}(\lambda)$ , where $\lambda>0$ , then

\mathrm E(Y)=\lambda,\qquad \operatorname{Var}(Y)=\lambda

These are variances, not standard deviations.

In the formula booklet.

For the opening experiments,

\begin{array}{c|ccc} \text{Model}&\mathrm B(300,0.01)&\mathrm B(6,0.5)&\operatorname{Po}(3)\\\hline \text{Mean}&3&3&3\\ \text{Variance}&2.97&1.5&3 \end{array}

The first binomial's variance is close to the Poisson variance; the second's is only half as large. In general, a binomial variance is at most its mean because $0\le1-p\le1$ . A Poisson variance equals its mean exactly under the model.

Observed sample means and variances will fluctuate, so a small difference in a dataset does not by itself disprove a Poisson model. A substantial difference, together with a contextual reason such as event clustering, can indicate that the model is unsuitable. The exact equalities above concern the model's population moments.

Key idea

For a binomial model, $\operatorname{Var}(X)=\mathrm E(X)(1-p)$ . As $p$ becomes small with the mean held fixed, its variance approaches the equal-mean Poisson variance.

The word “population” makes the following an exact parameter question, rather than a judgement about sampling variation.

Recovering the number of trials and success probability

When a binomial mean and variance are both known, divide the variance by the mean to remove the unknown number of trials.

For a positive mean $\mu=np$ , the ratio is $np(1-p)/(np)=1-p$ . Once $p$ is known, the mean gives $n$ . Keep the parameter conditions in view: a computed probability must lie between zero and one, and the number of trials must be an integer.

Binomial parameters from moments

If a binomial variable has mean $\mu>0$ , variance $v$ and $p>0$ , then

p=1-\frac v\mu,\qquad n=\frac\mu p

The resulting $n$ must be a positive integer and $0<p\le1$ . If the proposed moments violate these conditions, they do not define such a binomial distribution.

Not in the formula booklet: learn it.

These rearrangements come from the supplied mean and variance formulae. A positive mean with variance zero gives the boundary case $p=1$ , provided the mean itself is a possible integer trial count.

Worked example 1: a variance noticeably below the mean

Question. A binomial variable has mean $6$ and variance $4.8$ . Find $n$ and $p$ .

Plan. Use the ratio of variance to mean to find $p$ , then substitute into $np=6$ .

\begin{aligned} np&=6,\qquad np(1-p)=4.8 &&\text{state the two moment equations}\\ 1-p&=\frac{4.8}{6}=0.8 &&\text{divide the equations}\\ p&=0.2 &&\text{solve for the probability}\\ n&=\frac6{0.2}=30 &&\text{use the mean} \end{aligned}

Why this step? Division removes $n$ entirely. Solving for $n$ first would leave a less convenient expression involving $p$ in both equations.

Answer. $n=30$ , $p=0.2$ .

Check. $30(0.2)=6$ and $30(0.2)(0.8)=4.8$ . The trial count is a positive integer and the probability is valid.

Worked example 2 changes only the variance, from $4.8$ to $5.94$ , keeping the mean at $6$ . A variance closer to the mean will force a smaller success probability and more trials.

Worked example 2: many trials with rare successes

Question. A binomial variable has mean $6$ and variance $5.94$ . Find $n$ and $p$ .

Plan. Repeat the ratio calculation, retaining the exact decimal values before division.

\begin{aligned} 1-p&=\frac{5.94}{6}=0.99 &&\text{variance divided by mean}\\ p&=0.01 &&\text{small success probability}\\ n&=\frac6{0.01}=600 &&\text{preserve the given mean} \end{aligned}

Why this step? The mean remains $6$ , so reducing $p$ requires a compensating increase in $n$ . A mean alone could not distinguish the two experiments.

Answer. $n=600$ , $p=0.01$ .

Check. $600(0.01)=6$ and $600(0.01)(0.99)=5.94$ . Both supplied moments are recovered.

There is no reason to round an exact trial count arising from exact moment data. If it is not an integer, the proposed exact moments are inconsistent with a binomial model.

Watch out

Wrong: $p=v/\mu$ . That ratio is $1-p$ . Right: subtract the ratio from $1$ , then check both the original mean and variance.

Use a new pair of moments to distinguish the success probability from its complement.

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FS1 2.2, 2.3 - Means, variances and the Poisson approximation

Matching the mean does not match the spread

Recovering the number of trials and success probability

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