HookCounting the silences between the clicks
In 1910 Ernest Rutherford and Hans Geiger sat in a darkened Manchester laboratory counting flashes on a zinc sulphide screen — one flash per alpha particle spat out by a polonium source. Over 2,608 intervals of 7.5 seconds each they logged 10,097 scintillations, a mean of 3.87 per interval, and the counts matched a Poisson distribution almost perfectly. That dataset became the standard demonstration that radioactive decay is genuinely random: no nucleus 'knows' when its neighbours decayed, and none remembers how long it has already waited. But hidden inside the same experiment is a second distribution the pair never tabulated — not the number of clicks in a window, but the silences between the clicks. Those waiting times follow the distribution this section is about.
The exponential distribution is the continuous shadow of the Poisson: whenever events occur singly, independently and at a constant average rate \(\lambda\), the number of events in a fixed window is Poisson and the gap until the next event is exponential. Its density \(f(x)=\lambda e^{-\lambda x}\) starts at its highest point and decays forever — short waits are always the most likely, long waits never quite impossible. And it carries one property no other continuous distribution has: it is memoryless. A nucleus that has survived an hour is statistically identical to one created a second ago. You will define the model, compute its probabilities two ways, prove its mean and variance from first principles, and derive the Poisson link that makes it the mathematics of waiting.
ModelThe model — one parameter, two functions
The conditions are exactly the Poisson conditions read sideways: events occur singly (never two at the same instant), independently, and at a constant average rate \(\lambda\) per unit time. Let \(X\) be the waiting time until the next event. Then \(X\) is exponentially distributed with probability density function
\[f(x)=\lambda e^{-\lambda x}\quad(x\geq 0),\qquad f(x)=0\ \text{otherwise},\]
and cumulative distribution function
\[F(x)=P(X\leq x)=1-e^{-\lambda x}\quad(x\geq 0).\]
Check the anatomy before using it. The density starts at height \(f(0)=\lambda\) — the larger the rate, the taller and steeper the curve — and decays exponentially towards the axis, so the mode is always at zero: the single most likely waiting time is 'immediately'. It integrates to one, as any density must: \(\int_0^{\infty}\lambda e^{-\lambda x}\,dx=\big[-e^{-\lambda x}\big]_0^{\infty}=0-(-1)=1\), an improper integral of exactly the kind section E trained you for. The CDF is obtained by the same integration with a variable upper limit, and differentiating \(F\) recovers \(f\) — the two functions are one object wearing two outfits, and AQA expects you to move between them without being told which to use.
MechanismProbabilities — the CDF does the work
Every exponential probability reduces to three templates. Left tail: \(P(X\leq x)=F(x)=1-e^{-\lambda x}\). Right tail: \(P(X>x)=1-F(x)=e^{-\lambda x}\) — worth memorising on its own because it is the cleanest of the three and the engine of the Poisson link later. Interval: \(P(a<X<b)=F(b)-F(a)=e^{-\lambda a}-e^{-\lambda b}\). Note the order in that last expression: the smaller limit gives the larger survival term, so the difference comes out positive — writing it the other way round is the fastest route to an impossible negative probability. Because \(X\) is continuous, \(P(X=x)=0\) for any exact value, so strict and non-strict inequalities give identical answers; agonising over \(<\) versus \(\leq\) is wasted time here. The specification says 'using \(F(x)\) or by integrating \(f(x)\)' — the two are the same calculation, since integrating \(\lambda e^{-\lambda x}\) between limits just rebuilds the CDF, but you must be able to show the integration when the question demands it.
Calls arrive at a helpline at a constant average rate of 2 per minute, singly and independently, so the waiting time \(T\) minutes until the next call satisfies \(T\sim\text{Exp}(2)\). Probability the next call arrives within 30 seconds: \(P(T\leq 0.5)=1-e^{-2\times 0.5}=1-e^{-1}=0.632\) (3 s.f.). Probability of a silence longer than a minute: \(P(T>1)=e^{-2}=0.135\). Probability the call lands between 30 seconds and a minute: \(P(0.5<T<1)=e^{-1}-e^{-2}=0.368-0.135=0.233\). For the median, solve \(F(m)=0.5\): \(1-e^{-2m}=0.5\Rightarrow e^{-2m}=0.5\Rightarrow m=\frac{\ln 2}{2}=0.347\) minutes, about 20.8 seconds. The mean wait (next block) is 0.5 minutes — mean above median, the signature of a right-skewed distribution: many short gaps, a long tail of rare long ones.
ModelProving the mean and variance — integration by parts, twice
AQA asks you to prove, not quote, that \(\mathrm{E}(X)=\frac{1}{\lambda}\), \(\mathrm{Var}(X)=\frac{1}{\lambda^2}\) and hence the standard deviation is \(\frac{1}{\lambda}\). Both proofs are improper integrals attacked by parts, and both lean on the limit result from the pure core, \(\lim_{x\to\infty}x^k e^{-x}=0\) for \(k>0\): the exponential crushes any power of \(x\), so every boundary term at infinity vanishes. Notice what the results say. The mean waiting time is the reciprocal of the rate — events at 2 per minute mean an average gap of half a minute, which is common sense wearing a formula. And the standard deviation equals the mean: exponential waiting times are exactly as spread out as they are long on average, one reason a single long silence from a Geiger counter tells you almost nothing about the source.
Mean: \(\mathrm{E}(X)=\int_0^{\infty}x\,\lambda e^{-\lambda x}\,dx\). Integrate by parts with \(u=x\), \(dv=\lambda e^{-\lambda x}dx\), so \(du=dx\), \(v=-e^{-\lambda x}\): \(\mathrm{E}(X)=\big[-xe^{-\lambda x}\big]_0^{\infty}+\int_0^{\infty}e^{-\lambda x}\,dx\). The boundary term is zero at both ends — at infinity because \(xe^{-\lambda x}\to 0\), at zero because of the factor \(x\) — leaving \(\int_0^{\infty}e^{-\lambda x}\,dx=\big[-\tfrac{1}{\lambda}e^{-\lambda x}\big]_0^{\infty}=\frac{1}{\lambda}\). Variance: first \(\mathrm{E}(X^2)=\int_0^{\infty}x^2\lambda e^{-\lambda x}\,dx\). Parts again with \(u=x^2\), \(v=-e^{-\lambda x}\): \(\mathrm{E}(X^2)=\big[-x^2e^{-\lambda x}\big]_0^{\infty}+2\int_0^{\infty}xe^{-\lambda x}\,dx=0+\frac{2}{\lambda}\int_0^{\infty}x\lambda e^{-\lambda x}\,dx=\frac{2}{\lambda}\cdot\frac{1}{\lambda}=\frac{2}{\lambda^2}\) — the second integral is \(\frac{1}{\lambda}\times\mathrm{E}(X)\) after slipping a \(\lambda\) inside, a recycling move worth showing explicitly. Then \(\mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}\), and the standard deviation is \(\sqrt{\tfrac{1}{\lambda^2}}=\frac{1}{\lambda}\).
MechanismThe Poisson connection — why waiting times must be exponential
The link the specification names is a one-line derivation, and deriving it is worth more than remembering it. Suppose events follow a Poisson process at rate \(\lambda\) per unit time, and let \(T\) be the wait until the next event. The key observation: the wait exceeds \(t\) precisely when the window \([0,t]\) contains no events. The number of events in that window is \(\mathrm{Po}(\lambda t)\), so
\[P(T>t)=P(\text{0 events in }[0,t])=\frac{e^{-\lambda t}(\lambda t)^0}{0!}=e^{-\lambda t},\]
which is exactly the exponential survival function: \(T\sim\text{Exp}(\lambda)\). The two distributions are one random process described from two angles — Poisson counts events per window, exponential clocks the gaps — and they share the parameter with reciprocal meanings: Poisson mean \(\lambda t\) events, exponential mean \(\frac{1}{\lambda}\) time units. In Rutherford and Geiger's data, 3.87 decays per 7.5-second interval is a rate of \(\lambda=0.516\) per second, so the silences averaged \(\frac{1}{0.516}\approx 1.94\) seconds — a number their published table implies but never states. Exam questions exploit the dictionary in both directions: given a Poisson rate, find waiting-time probabilities; given a mean gap, recover \(\lambda\) and count events.
CaseMemorylessness — the property that picks the model for you
Take the survival function and condition on having already waited \(s\): \(P(X>s+t\mid X>s)=\frac{P(X>s+t)}{P(X>s)}=\frac{e^{-\lambda(s+t)}}{e^{-\lambda s}}=e^{-\lambda t}=P(X>t)\). The past wait cancels out entirely. This is memorylessness, and among continuous distributions the exponential is the only one that has it — a component whose failure times are exponential does not age, and a nucleus that has survived a million years is no closer to decaying than a fresh one. That is why the model fits radioactive decay, call arrivals, and random breakdowns of electronics beautifully.
It is also why the model fails where things wear out. Human lifetimes are emphatically not exponential — a 90-year-old and a newborn do not face equal hazards — and nor are machine parts that fatigue, batteries that drain, or tyres that wear. The evaluation move examiners reward is to test the constant-rate assumption in context: bus arrivals on a timetable are not a Poisson process (gaps cluster around the scheduled interval), calls to a helpline surge at lunchtime (rate not constant), aftershocks cluster after earthquakes (not independent). One sentence naming the broken condition — singly, independently, constant rate — turns a modelling question from an assertion into an answer.
VocabularyKey terms the mark scheme pays for
TrapsMisconceptions that cost marks
ExamWhat examiners want
State the model before you compute: '\(T\sim\text{Exp}(\lambda)\) where \(\lambda=\dots\), assuming calls occur singly, independently, at constant average rate' earns the modelling mark that a bare calculation forfeits. Watch the units of \(\lambda\) like a hawk — a rate of 3 per hour used with a time in minutes is the single commonest arithmetic wreck in this topic; convert everything to one time unit before touching the formula. For probabilities, quote the template (\(F(x)\), \(1-F(x)\), or \(F(b)-F(a)\)), substitute, and give at least 3 significant figures; if the question says 'by integration', show \(\int\lambda e^{-\lambda x}dx\) with limits and the evaluated bracket, not just the CDF answer.
The proofs of the mean and variance are bookwork that AQA sets essentially verbatim, so rehearse them until the structure is automatic: parts, boundary term killed by \(\lim_{x\to\infty}x^k e^{-x}=0\) (say so — the examiner wants the limit acknowledged, not assumed), then for the variance the recycling step \(\mathrm{E}(X^2)=\frac{2}{\lambda}\mathrm{E}(X)\). On Poisson-link questions, write the sentence 'the wait exceeds \(t\) if and only if there are no events in \([0,t]\)' before any algebra — that sentence is the method mark. And when asked whether the exponential is a suitable model, name a specific condition and test it against the context; 'it might not be constant' scores nothing without saying what varies and why.