Mean and variance, and link to Poisson processesAQA A-Level Further Maths: Revision notes

Section 1

Mean, variance and standard deviation

For $X$ with an exponential distribution with parameter $\lambda$ : $\mathrm{E}(X)=\frac{1}{\lambda},\qquad \mathrm{Var}(X)=\frac{1}{\lambda^2},\qquad \sigma=\frac{1}{\lambda}.$ The mean and standard deviation are equal. Example: $\lambda=0.05$ per day gives mean $20$ days, variance $400$ and standard deviation $20$ days. A higher rate $\lambda$ means shorter gaps on average.

Key termsmeanstandard deviation

Common mistake

Giving $\frac{1}{\lambda}$ as the variance. The variance is $\frac{1}{\lambda^2}$ ; $\frac{1}{\lambda}$ is the mean and the standard deviation.

Section 2

Proof of the mean

$\mathrm{E}(X)=\int_0^\infty x\lambda e^{-\lambda x}\,dx.$ Integrate by parts with $u=x$ and $\frac{dv}{dx}=\lambda e^{-\lambda x}$ , so $v=-e^{-\lambda x}$ : $\left[-xe^{-\lambda x}\right]_0^\infty+\int_0^\infty e^{-\lambda x}\,dx=0+\left[-\frac{1}{\lambda}e^{-\lambda x}\right]_0^\infty=\frac{1}{\lambda}.$ The boundary term is $0$ because $xe^{-\lambda x}\to0$ as $x\to\infty$ .

Key termsintegration by parts

Exam tip

State that the boundary term is zero because $e^{-\lambda x}$ decays faster than $x$ grows. The mark depends on saying so.

Section 3

Proof of the variance

First, by parts again: $\mathrm{E}(X^2)=\int_0^\infty x^2\lambda e^{-\lambda x}\,dx=\left[-x^2e^{-\lambda x}\right]_0^\infty+\int_0^\infty2xe^{-\lambda x}\,dx=\frac{2}{\lambda}\int_0^\infty x\lambda e^{-\lambda x}\,dx=\frac{2}{\lambda}\times\frac{1}{\lambda}=\frac{2}{\lambda^2}.$ Then $\mathrm{Var}(X)=\mathrm{E}(X^2)-[\mathrm{E}(X)]^2=\frac{2}{\lambda^2}-\frac{1}{\lambda^2}=\frac{1}{\lambda^2}.$ The remaining integral is $\frac{2}{\lambda}\mathrm{E}(X)$ , which reuses the previous result.

Section 4

The link to Poisson processes

Events occurring at random, independently, at a constant average rate $\lambda$ per unit time form a Poisson process. Then:

the number of events in a time $t$ has a Poisson distribution with mean $\lambda t$ ,
the time between successive events has an exponential distribution with parameter $\lambda$ . The mean gap is $\frac{1}{\lambda}$ . Example: 12 strikes per hour means $\lambda=0.2$ per minute, mean gap 5 minutes.

Key termsPoisson process

Common mistake

Mixing units: a rate of 12 per hour gives a mean gap of $\frac{1}{12}$ hour (5 minutes), not 12 minutes.

Section 5

Linking the two distributions

A gap longer than $t$ is the same event as no events in a time $t$ . Both have probability $e^{-\lambda t}$ : $\mathrm{P}(X>t)=\mathrm{P}(\text{Poisson with mean }\lambda t\text{ equals }0)=e^{-\lambda t}$ . Example: $\lambda=0.8$ faults per km, so $\mathrm{P}(\text{gap}>2\text{ km})=\mathrm{P}(\text{no faults in 2 km})=e^{-1.6}=0.202$ . If the rate varies, for instance faults are more common near joints, neither model applies.

Exam tip

Use the Poisson count for 'how many events in a period', and the exponential for 'how long until the next event'.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Mean and variance, and link to Poisson processes

The lifetime, $X$ days, of a type of battery is modelled by an exponential distribution with parameter $\lambda=0.05$ .
Find the probability that a battery lasts longer than its mean lifetime.2 marks
Particles strike a detector at random, independently of each other, at a constant average rate of 12 per hour, so the number of strikes follows a Poisson process. Let $T$ minutes be the time between two successive strikes.
Find the standard deviation of $T$ in minutes.2 marks
The continuous random variable $X$ has an exponential distribution with parameter $\lambda>0$ , so that $f(x)=\lambda e^{-\lambda x}$ for $x\ge0$ . You may use the fact that $xe^{-\lambda x}\to0$ and $x^2e^{-\lambda x}\to0$ as $x\to\infty$ .
Prove that $\mathrm{E}(X)=\frac{1}{\lambda}$ .3 marks

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Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).