Exponential Distribution Revision Notes for AQA A-Level Further Maths

Exponential Distribution

What is the exponential distribution?

The exponential distribution is a continuous probability distribution used to model the time between randomly occurring events. It is particularly useful for representing:

Waiting times between random events
Time until radioactive disintegrations occur
Failure times of electronic components
Duration of telephone calls
Any scenario involving "memoryless" waiting times

The distribution has a positive skew and is only defined for positive values of the random variable.

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The exponential distribution is unique because it is memoryless: the probability of an event occurring in the next time interval does not depend on how long you've already waited. This makes it ideal for modeling truly random processes where the past does not influence the future.

Probability density function (PDF)

Definition: If $X$ is a random variable with an exponential distribution with parameter $\lambda$ (where $\lambda > 0$ ), then its probability density function is given by:

$f(x) = \lambda e^{-\lambda x}, \quad x > 0$

Key properties:

The parameter $\lambda$ is called the rate parameter
The PDF is defined only for $x > 0$ (non-negative values)
The function decreases exponentially as $x$ increases
Different values of $\lambda$ produce different decay rates

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The exponential distribution is only defined for positive values: $x > 0$ . Always include this condition when stating the PDF or CDF.

The graph shows how the PDF changes for different values of $\lambda$ . Notice that:

Larger values of $\lambda$ result in steeper initial decline
Smaller values of $\lambda$ produce more gradual decay
All curves start at different heights on the y-axis (equal to $\lambda$ )
All curves approach zero as $x$ increases

Cumulative distribution function (CDF)

The cumulative distribution function gives the probability that the random variable $X$ takes a value less than or equal to $x$ . We find it by integrating the PDF:

$F(x) = P(X < x) = \int_0^x f(t) \, dt$

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Derivation of the CDF:

Starting with the integral: $F(x) = \int_0^x \lambda e^{-\lambda t} \, dt$

$= \left[ \lambda \cdot \frac{e^{-\lambda t}}{-\lambda} \right]_0^x$

$= \left[ -e^{-\lambda t} \right]_0^x$

$= -e^{-\lambda x} - (-e^0)$

$= 1 - e^{-\lambda x}$

Key result: If $X$ follows an exponential distribution with parameter $\lambda$ , then:

$F(x) = 1 - e^{-\lambda x}, \quad x > 0$

This formula allows us to calculate probabilities directly without integration, making it extremely useful for problem-solving.

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A useful alternative form is $P(X > x) = e^{-\lambda x}$ , which is often quicker to use than calculating $1 - F(x)$ .

Mean and variance

Expected value (mean)

The mean of the exponential distribution is found using integration by parts.

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Derivation of the Mean:

$E(X) = \int_0^\infty x f(x) \, dx = \int_0^\infty x \lambda e^{-\lambda x} \, dx$

Using integration by parts with $u = x$ and $dv = \lambda e^{-\lambda x} dx$ :

$E(X) = \lambda \left[ x \cdot \frac{e^{-\lambda x}}{-\lambda} \right]_0^\infty - \int_0^\infty \frac{e^{-\lambda x}}{-\lambda} \, dx$

After evaluating the boundary terms and simplifying:

$E(X) = \frac{1}{\lambda}$

Variance

Similarly, we calculate $E(X^2)$ using integration by parts twice:

$E(X^2) = \int_0^\infty x^2 \lambda e^{-\lambda x} \, dx = \frac{2}{\lambda^2}$

Then using the variance formula:

$\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - \left(\frac{1}{\lambda}\right)^2 = \frac{1}{\lambda^2}$

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Key result: If $X$ has probability density function $f(x) = \lambda e^{-\lambda x}$ , $x > 0$ , then:

$E(X) = \frac{1}{\lambda} \quad \text{and} \quad \text{Var}(X) = \frac{1}{\lambda^2}$

The standard deviation is: $\text{SD}(X) = \sqrt{\text{Var}(X)} = \frac{1}{\lambda}$

Notice that the mean and standard deviation are equal for the exponential distribution.

Connection to the Poisson process

The exponential distribution has a special relationship with the Poisson distribution, making it particularly useful for modeling real-world random processes.

Poisson process: If random events occur at a mean rate of $\mu$ per unit time, and $X$ represents the number of occurrences in a unit time interval, then $X \sim \text{Poisson}(\mu)$ .

Waiting times: Let $T$ be the time between successive events in a Poisson process. Then:

$P(T < t) = 1 - P(T > t) = 1 - P(\text{no occurrences in interval } t)$

Since the number of occurrences follows a Poisson distribution: $P(\text{no occurrences in interval } t) = e^{-\mu t}$

Therefore: $P(T < t) = 1 - e^{-\mu t}$

This is the CDF of an exponential distribution with parameter $\mu$ .

chatImportant

Key result: If $T$ is the time between successive random events that occur at a mean rate of $\mu$ per unit time, then $T$ follows an exponential distribution with parameter $\mu$ , where:

$\text{Mean} = \frac{1}{\mu} \quad \text{and} \quad \text{Variance} = \frac{1}{\mu^2}$

The parameter $\mu$ represents the rate of the underlying Poisson process, while $\frac{1}{\mu}$ represents the mean waiting time.

Worked examples

Example 1: Basic probability calculations

A random variable $X$ has an exponential distribution with parameter $\lambda = \frac{1}{4}$ .

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Part (a): Write down the cumulative distribution function and find:

i) $P(X < 1)$
ii) $P(X > 3)$
iii) $P(1 < X < 3)$

Solution:

The CDF is: $F(x) = 1 - e^{-\frac{1}{4}x}$

i) $P(X < 1) = F(1) = 1 - e^{-\frac{1}{4} \times 1} = 1 - e^{-0.25} = 0.221$ (3 dp)

ii) $P(X > 3) = 1 - P(X \leq 3) = 1 - F(3) = 1 - (1 - e^{-\frac{3}{4}}) = e^{-0.75} = 0.472$ (3 dp)

iii) $P(1 < X < 3) = F(3) - F(1) = (1 - e^{-0.75}) - (1 - e^{-0.25})$ $= e^{-0.25} - e^{-0.75} = 0.221 + 0.472 = 0.31 \text{ (2 dp)}$

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Part (b): If three $x$ -values are chosen at random, find the probability that one will be less than 1 and the other two will be greater than 1.

Solution:

Let $Y$ = number less than 1 out of 3 values. Then $Y \sim B(3, 0.221)$

$P(Y = 1) = \binom{3}{1}(0.221)^1(1-0.221)^2 = 3 \times 0.221 \times 0.607 = 0.40 \text{ (2 dp)}$

Example 2: Mean, variance, and standard deviation

A random variable $X$ has an exponential distribution with $\lambda = 2$ .

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Part (a): Write down the mean and variance of $X$ .

Solution:

Using the standard results: $\text{Mean} = E(X) = \frac{1}{\lambda} = \frac{1}{2}$

$\text{Variance} = \text{Var}(X) = \frac{1}{\lambda^2} = \frac{1}{4}$

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Part (b): Find the probability that a randomly chosen value of $X$ will be within 1 standard deviation of the mean.

Solution:

Standard deviation: $\sqrt{\frac{1}{4}} = \frac{1}{2}$

We need: $P\left(\frac{1}{2} - \frac{1}{2} < X < \frac{1}{2} + \frac{1}{2}\right) = P(0 < X < 1)$

$P(0 < X < 1) = F(1) - F(0) = (1 - e^{-2 \times 1}) - 0 = 1 - e^{-2} = 0.865 \text{ (3 dp)}$

Example 3: Poisson process application

Random events occur at a rate of 4 per minute.

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Part (a): Write the probability density function, $f(t)$ , and the cumulative distribution function, $F(t)$ , of the random variable $T$ , the waiting time in minutes between events.

Solution:

The rate is $\mu = 4$ per minute, so $T$ follows an exponential distribution with parameter $\mu = 4$ .

$f(t) = 4e^{-4t}, \quad t \geq 0$

$F(t) = 1 - e^{-4t}, \quad t \geq 0$

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Part (b): Find the probability that, from the occurrence of one event, the waiting time until the next event will be greater than 15 seconds.

Solution:

Convert 15 seconds to minutes: $15 \text{ seconds} = 0.25 \text{ minutes}$

$P(T > 0.25) = 1 - F(0.25) = 1 - (1 - e^{-4 \times 0.25}) = e^{-1} = 0.37 \text{ (2 dp)}$

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Part (c): Calculate the mean and variance of the waiting time.

Solution:

Using the standard results with $\mu = 4$ :

$\text{Mean} = \frac{1}{\mu} = \frac{1}{4} \text{ minute}$

$\text{Variance} = \frac{1}{\mu^2} = \frac{1}{16} \text{ minute}^2$

Note: $\mu$ is the distribution parameter (rate), not the mean.

Problem-solving strategy

To solve problems using the exponential distribution:

Check the conditions: Ensure the problem involves waiting times between random events or another appropriate scenario for the exponential distribution.
Identify the parameter: Determine $\lambda$ (or $\mu$ ) from the given information. Remember:
- If given a rate, this is your parameter
- If given a mean waiting time, $\lambda = \frac{1}{\text{mean}}$
Choose the right formula:
- For probabilities: use $F(x) = 1 - e^{-\lambda x}$
- For mean: use $E(X) = \frac{1}{\lambda}$
- For variance: use $\text{Var}(X) = \frac{1}{\lambda^2}$
Use technology: Calculator or statistical tables can help find probabilities quickly.
Check units: Ensure all time units are consistent throughout your calculations.

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Exam tips:

The exponential distribution is "memoryless" – the probability of an event occurring in the next time interval does not depend on how long you've already waited
Always write $x > 0$ when stating the PDF
Remember that $P(X > x) = e^{-\lambda x}$ is often quicker to use than $1 - F(x)$
For Poisson process problems, the rate parameter of the exponential distribution equals the mean rate of the Poisson process

Remember!

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Key Points to Remember:

The probability density function is $f(x) = \lambda e^{-\lambda x}$ for $x > 0$ , where $\lambda$ is the rate parameter.
The cumulative distribution function is $F(x) = 1 - e^{-\lambda x}$ for $x > 0$ , making probability calculations straightforward without integration.
The mean and variance are $E(X) = \frac{1}{\lambda}$ and $\text{Var}(X) = \frac{1}{\lambda^2}$ , meaning the standard deviation equals the mean.
The exponential distribution models waiting times in a Poisson process: if events occur at rate $\mu$ per unit time, the waiting time between events follows an exponential distribution with parameter $\mu$ and mean $\frac{1}{\mu}$ .
The distribution has positive skew and is memoryless – past waiting time does not affect future probabilities.

Exponential Distribution (AQA A-Level Further Maths): Revision Notes