Difficulty: Intermediate | Prerequisites: Integration basics (you need to be comfortable with e^x and ln), the concept of a CDF, and familiarity with the Poisson distribution.
The exponential distribution is the first continuous distribution most students meet after the uniform. It models waiting times between events in a Poisson process, so it pairs naturally with the Poisson (which counts events in a fixed window). If the Poisson side feels unfamiliar, review the discrete distributions notes first. The exponential is also your introduction to the memoryless property in continuous time, a concept that appears again in queuing theory, reliability engineering, and survival analysis.
The exponential distribution models how long you wait between events that occur at a constant average rate. It has one parameter, lambda (the rate). Its mean and standard deviation are both 1/lambda. Its CDF has a clean closed form, so you can solve for exact probabilities and percentiles without tables. The memoryless property means that however long you have already waited, the remaining wait time has the same distribution as if you had just started.
Exponential distribution
A continuous distribution modelling the time between successive events in a Poisson process. Parameter: lambda (the rate, i.e. average number of events per unit time). Notation: X ~ Exp(lambda).
Think of it as: "How long until the next event?" where events arrive randomly at a steady average pace.
Rate parameter (lambda)
The average number of events per unit of time (or space). A higher lambda means events happen more frequently and waits are shorter.
In simple terms, this means: lambda = 0.2 breakdowns per day tells you the machine breaks down, on average, once every 5 days.
CDF (cumulative distribution function)
F(x) = P(X <= x). For the exponential: F(x) = 1 - e^(-lambda * x) for x >= 0. Gives the probability that the event has occurred by time x.
Think of it as: the running total of probability up to a given point.
Memoryless property
P(X > s + t | X > s) = P(X > t). Given that you have already waited s units without an event, the probability of waiting at least t more is the same as starting fresh.
In simple terms, this means: the exponential distribution does not "remember" how long you have been waiting. Past waiting time gives no information about the future.
Median
The value m where F(m) = 0.5, i.e. there is a 50% chance the event occurs before m and a 50% chance it occurs after. For the exponential: median = ln(2) / lambda.
Think of it as: the point where you are equally likely to have seen the event or not.
Setup
X = time between events (e.g. days between machine breakdowns)
Parameter: lambda (rate)
PDF: f(x) = lambda * e^(-lambda * x), for x >= 0
CDF: F(x) = 1 - e^(-lambda * x), for x >= 0
Survival function: P(X > x) = 1 - F(x) = e^(-lambda * x)
Expected value and standard deviation
E(X) = 1/lambda
Var(X) = 1/lambda^2
sigma_X = 1/lambda (the standard deviation equals the mean)
Worked example (Problem 4.2.3)
A machine breaks down with rate lambda = 0.2 breakdowns per day. X ~ Exp(0.2).
E(X) = 1/0.2 = 5 days between breakdowns on average
sigma_X = sqrt(1/0.2^2) = sqrt(25) = 5 days (equals the mean, which is always true for the exponential)
Median: solve F(x) = 0.5, so 1 - e^(-0.2x) = 0.5, giving x = ln(2)/0.2 = 3.466 days. The median is shorter than the mean because the exponential is right-skewed.
P(X >= 7) = 1 - F(7) = e^(-0.2 * 7) = e^(-1.4) = 0.2466. About a 25% chance of going at least 7 days without a breakdown.
The exponential CDF is one of the few you can invert by hand. This makes it a favourite for exam questions that ask "find the time by which there is a 50% (or any %) chance of the event occurring."
General method
Set F(x) = target probability: 1 - e^(-lambda * x) = p
Rearrange: e^(-lambda * x) = 1 - p
Take natural log: -lambda * x = ln(1 - p)
Solve: x = -ln(1 - p) / lambda
Example from Problem 4.2.3
Find the median (p = 0.5) with lambda = 0.2:
x = -ln(1 - 0.5) / 0.2 = -ln(0.5) / 0.2 = ln(2) / 0.2 = 0.6931 / 0.2 = 3.466 days
The exponential is the only continuous distribution with the memoryless property.
Formal statement: P(X > s + t | X > s) = P(X > t)
What it means in practice: Suppose a machine has been running for 6 days without breaking down. The probability it survives at least 2 more days is the same as the probability a brand-new machine survives 2 days. The 6 days already elapsed give no information.
Worked example (Problem 4.2.3, part e)
P(X >= 6 + 2 | X >= 6) = P(X >= 2) by the memoryless property.
P(X >= 2) = 1 - F(2) = e^(-0.2 * 2) = e^(-0.4) = 0.6703
So P(X >= 8 | X >= 6) = 0.67
This result surprises students who expect a machine that has already run for 6 days to be "due" for a breakdown. The exponential says otherwise: it has no memory of the past.
Property | Formula |
|---|---|
f(x) = lambda * e^(-lambda * x), x >= 0 | |
CDF | F(x) = 1 - e^(-lambda * x), x >= 0 |
Survival | P(X > x) = e^(-lambda * x) |
Mean | E(X) = 1/lambda |
Variance | Var(X) = 1/lambda^2 |
Std deviation | sigma = 1/lambda |
Median | ln(2)/lambda |
Percentile (p) | x = -ln(1 - p)/lambda |
Memoryless | P(X > s+t | X > s) = P(X > t) |
Reliability engineering: The time until a component fails is often modelled as exponential. This is why the memoryless property matters: it implies the component does not degrade with age (a strong assumption, but a useful starting model).
Queuing theory: The time between customer arrivals at a service counter. If arrivals follow a Poisson process, the inter-arrival times are exponential.
Radioactive decay: The time until the next decay event of a single unstable atom is exponential.
Students often confuse the rate lambda with the mean. The mean is 1/lambda, not lambda. If a machine breaks down 0.2 times per day on average, the mean time between breakdowns is 5 days, not 0.2 days.
The median is not the same as the mean for the exponential. Because the distribution is right-skewed, the median (ln(2)/lambda) is always less than the mean (1/lambda). Students who set E(X) = median on an exam lose marks.
The memoryless property does not mean "nothing has happened yet, so something is overdue." That is the gambler's fallacy. The exponential says the opposite: no matter how long you have waited, the future looks the same.
Students sometimes try to use the exponential PDF to compute P(X > a) directly. Use the survival function e^(-lambda * a) instead; integrating the PDF works but is slower and more error-prone under time pressure.
⚠️ "Find the median" is a near-certain exam question. Know the formula x = ln(2)/lambda and be ready to derive it from F(x) = 0.5.
⚠️ Conditional probability questions using the memoryless property appear frequently. Recognise the pattern: "given that the machine has already run for s days, find P(X > s + t)." Simplify immediately to P(X > t).
⚠️ Know that the standard deviation equals the mean for the exponential. Exams sometimes ask students to verify this as a quick check.
⚠️ Be ready to compute P(a <= X <= b) = F(b) - F(a) = e^(-lambdaa) - e^(-lambdab). This comes up when the question asks for the probability that the event occurs within a specific time window.
True or false: For the exponential distribution, the mean is always greater than the median. (True. The distribution is right-skewed, so the mean is pulled above the median.)
Fill in the blank: If lambda = 0.5, then E(X) = ______. (2.)
True or false: The exponential distribution can take negative values. (False. X >= 0 always.)
Fill in the blank: The memoryless property says P(X > s + t | X > s) = ______. (P(X > t).)
True or false: For an exponential random variable, the standard deviation equals the mean. (True. Both equal 1/lambda.)
Q: A light bulb has an exponential lifetime with a mean of 1000 hours. What is the rate parameter lambda?
A: lambda = 1/1000 = 0.001 failures per hour.
Q: Using the same light bulb, what is the probability it lasts more than 1200 hours?
A: P(X > 1200) = e^(-0.001 * 1200) = e^(-1.2) = 0.3012.
Q: A server receives requests at a rate of 3 per minute. What is the probability that the gap between two consecutive requests is less than 20 seconds (1/3 minute)?
A: X ~ Exp(3). P(X < 1/3) = 1 - e^(-3 * 1/3) = 1 - e^(-1) = 1 - 0.3679 = 0.6321.
Q: A machine has been running for 10 days without failure. If failures follow Exp(0.1), what is the probability it runs at least 3 more days?
A: By the memoryless property, P(X > 13 | X > 10) = P(X > 3) = e^(-0.1 * 3) = e^(-0.3) = 0.7408.
Q: Find the 90th percentile of an Exp(0.5) distribution.
A: x = -ln(1 - 0.9) / 0.5 = -ln(0.1) / 0.5 = 2.3026 / 0.5 = 4.605.
The exponential distribution is the continuous counterpart of the geometric distribution. Where the geometric counts discrete trials until the first success, the exponential measures continuous time until the first event. Both are memoryless. The Poisson distribution counts events in a fixed interval; the exponential measures the gaps between those events. If you sum n independent exponential random variables, you get a Gamma distribution, which generalises the exponential the same way the negative binomial generalises the geometric.
Exponential distribution, continuous probability, CDF, survival function, memoryless property, rate parameter, lambda, Poisson process, inter-arrival time, hazard rate, reliability, queuing theory, waiting time, Gamma distribution, percentile, median, intro probability and statistics, STAT homework