FS1 3.2, 3.3 - Means and variances of both distributions | Edexcel A-Level Further Mathematics

Waiting times can have a much larger spread than their most likely value suggests. You will use the geometric mean and variance (FS1 3.2), use the corresponding negative binomial results (FS1 3.3), recover parameters from moments, and translate requirements about an average wait into conditions on the success probability.

The centre and spread of a first-success wait

How can one trial be the most likely wait, while the average wait is twenty trials?

For a geometric variable with $0<p<1$ , consecutive probabilities have ratio $1-p<1$ , so trial $1$ is the most likely individual outcome. But there are infinitely many longer waits, and their combined contribution to the average can be substantial.

You already know

Geometric counts the trial of the first success, including that trial. For any discrete variable, $\mathrm E(X)$ is its weighted mean and $\operatorname{Var}(X)=\mathrm E(X^2)-(\mathrm E(X))^2$ . Variance is not standard deviation.

The formula booklet supplies the moments directly, so you do not need to sum infinitely many weighted probabilities each time.

Geometric mean and variance

For $X\sim\operatorname{Geo}(p)$ , with the trial-count convention and $0<p\le1$ ,

\mathrm E(X)=\frac1p,\qquad \operatorname{Var}(X)=\frac{1-p}{p^2}

At $p=1$ , $X=1$ certainly and the variance is zero. In the formula booklet.

A lower success probability increases both the mean wait and its variance. The denominator of the variance is $p^2$ , which makes spread grow particularly quickly when successes are rare.

Worked example 1: moments of a first-success wait

Question. Independent trials succeed with probability $1/4$ . Let $X$ be the trial of the first success. Find its mean, variance and standard deviation.

Plan. State the geometric model, substitute into the two supplied formulae, then take the non-negative square root.

\begin{aligned} X&\sim\operatorname{Geo}(1/4) && \text{first-success trial} \\ \mathrm E(X)&=\frac1{1/4}=4 && \text{mean formula} \\ \operatorname{Var}(X)&=\frac{3/4}{(1/4)^2}=12 && \text{variance formula} \\ \sigma&=\sqrt{12}=2\sqrt3 && \text{standard deviation} \end{aligned}

Why this step? Squaring the probability in the denominator is essential; dividing $3/4$ by $1/4$ would not give the variance.

Answer. Mean $4$ trials, variance $12$ , standard deviation $2\sqrt3$ trials.

Check. The mean is above the minimum wait $1$ , and $\sigma^2=12$ .

Worked example 2 changes only the success probability, from $1/4$ to $1/5$ .

Worked example 2: a rarer success

Question. Find the same moments when $X\sim\operatorname{Geo}(1/5)$ .

Plan. Recalculate both moments with the new probability.

\begin{aligned} \mathrm E(X)&=\frac1{1/5}=5 && \text{new mean} \\ \operatorname{Var}(X)&=\frac{4/5}{(1/5)^2}=20 && \text{new squared denominator} \\ \sigma&=\sqrt{20}=2\sqrt5 && \text{non-negative root} \end{aligned}

Why this step? Reducing the success probability makes long waits more common, so neither moment should be copied unchanged.

Answer. Mean $5$ , variance $20$ , standard deviation $2\sqrt5$ .

Check. Both mean and variance exceed their values for $p=1/4$ .

If a question counts failures instead, use $F=X-1$ . The coding rules give $\mathrm E(F)=(1-p)/p$ , while $\operatorname{Var}(F)=\operatorname{Var}(X)$ . These failure-count consequences are not printed as separate results; derive them from the trial-count formulae.

Watch out

Wrong: report $(1-p)/p$ as the mean trial number. Right: that is the mean number of failures. Pearson's waiting-time variable includes its final success and has mean $1/p$ .

Change the success probability again.

The centre and spread of a later-success wait

Waiting for more successes increases the average trial count and the variance, with both growing in proportion to the number of required successes.

For a fixed $p$ , each additional requested success adds $1/p$ to the mean given by the booklet. Its variance formula also scales by $r$ , not $r^2$ . This is a stopping rule across fresh trials, rather than multiplication of one realised wait by $r$ .

Negative binomial mean and variance

If $T\sim\operatorname{NB}(r,p)$ counts the trial of success number $r$ , with positive integer $r$ and $0<p\le1$ ,

\mathrm E(T)=\frac rp,\qquad \operatorname{Var}(T)=\frac{r(1-p)}{p^2}

At $p=1$ , $T=r$ certainly. The case $r=1$ gives the geometric moments. In the formula booklet.

The mean is at least $r$ , as it must be: success number $r$ cannot occur before trial $r$ .

Worked example 3: waiting for three successes

Question. Trials succeed independently with probability $1/4$ . Let $T$ count the trial of the third success. Find its mean, variance and standard deviation.

Plan. Use $r=3,p=1/4$ in the negative binomial moment formulae.

\begin{aligned} T&\sim\operatorname{NB}(3,1/4) && \text{third-success trial} \\ \mathrm E(T)&=\frac3{1/4}=12 && \text{mean} \\ \operatorname{Var}(T)&=\frac{3(3/4)}{(1/4)^2}=36 && \text{variance} \\ \sigma_T&=\sqrt{36}=6 && \text{standard deviation} \end{aligned}

Why this step? The target number of successes is the first parameter; it is not the number of failures or a fixed trial limit.

Answer. Mean $12$ , variance $36$ , standard deviation $6$ .

Check. Compared with $r=1$ , the mean and variance are both tripled, matching the supplied formulae.

Worked example 4 changes only the success target, from $3$ to $6$ .

Worked example 4: doubling the success target

Question. Find the moments when $T\sim\operatorname{NB}(6,1/4)$ .

Plan. Keep $p$ fixed and update $r$ .

\begin{aligned} \mathrm E(T)&=\frac6{1/4}=24 && \text{twice the preceding mean} \\ \operatorname{Var}(T)&=\frac{6(3/4)}{(1/4)^2}=72 && \text{twice the preceding variance} \\ \sigma_T&=\sqrt{72}=6\sqrt2 && \text{not twice the preceding deviation} \end{aligned}

Why this step? Variance doubles with $r$ , so standard deviation grows by $\sqrt2$ , not $2$ .

Answer. Mean $24$ , variance $72$ , standard deviation $6\sqrt2$ .

Check. $r=6$ is twice $r=3$ , and the moment ratios agree with the linear dependence on $r$ .

For failures $F=T-r$ , the mean is $r(1-p)/p$ and the variance remains $r(1-p)/p^2$ , by the earlier coding rule. These are derived consequences, not a different Pearson convention.

Watch out

Wrong: replace the stopping time by $rX$ , where $X$ is one observed first-success wait. That would give variance $r^2\operatorname{Var}(X)$ . Right: use the supplied negative binomial formula, whose variance is proportional to $r$ .

Change the target once more.

Recovering the success probability and target

Two known moments can reveal both negative binomial parameters, if the resulting success target is a valid integer.

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FS1 3.2, 3.3 - Means and variances of both distributions

The centre and spread of a first-success wait

The centre and spread of a later-success wait

Recovering the success probability and target

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