Difficulty: Intermediate | Prerequisites: Part 1 of these study notes (random variables, PDFs), integration by parts, improper integrals, L'Hopital's rule.
Once you know what a probability density function is, the natural next questions are: what shape does it take for common real-world situations, and how do you summarise a distribution with a single number? This section covers exponential distributions (the go-to model for waiting times and lifetimes), then introduces the expected value (mean), median, variance, and standard deviation. These summary statistics are the bridge between raw probability and the kind of claims you see in statistics: "the average lifetime is 10 years" or "about 68% of values fall within one standard deviation of the mean." You will need integration by parts and L'Hopital's rule.
Exponential distributions model how long you wait for something (a bus, a bulb to burn out, a phone call). Their defining feature is memorylessness: the probability of surviving another s hours does not depend on how long the object has already survived. The mean of an exponential distribution with parameter c is 1/c, and the median is (1/c) ln 2. Variance and standard deviation measure how spread out a distribution is around its mean.
Exponentially decreasing probability density function
A PDF of the form f(x) = ce^(-cx) for x >= 0 and f(x) = 0 for x < 0. Think of it as a curve that starts high and drops off rapidly, modelling the idea that short waits are common and long waits are rare.
Memoryless property
The probability that an object survives at least s + t hours, given it has survived t hours, equals the probability of surviving at least s hours from the start. In simple terms, the past does not affect the future probability. A radioactive particle that has lasted 100 years is no more likely to decay in the next hour than a brand-new one.
Expected value (mean)
The weighted average of all values of a continuous random variable, computed as the integral from negative infinity to positive infinity of x f(x) dx. Denoted mu or E(X). Think of it as the long-run average if you repeated the experiment endlessly.
Median
The value m that splits the probability in half: the integral from negative infinity to m of f(x) dx = 1/2. There is an equal chance of being above or below the median.
Variance
The expected value of (X - mu)^2, measuring how spread out the distribution is around the mean. Denoted Var(X). A higher variance means the values scatter more widely.
Standard deviation
The square root of the variance, denoted sigma. It is in the same units as the random variable itself, making it easier to interpret than variance.
The exponential PDF takes the form: f(x) = ce^(-cx) for x >= 0, and 0 for x < 0.
To verify it is a valid PDF: the integral from 0 to infinity of ce^(-cx) dx = 1 (use the limit definition of the improper integral).
The parameter c controls how quickly the curve decays. Larger c means faster decay (shorter expected wait).
Worked example (chip lifetimes): An electronics company models chip lifetime T in years with f(T) = 0.1e^(-0.1T) for T >= 0.
P(T > 2) = e^(-0.2) which is approximately 0.819. About 82% of chips last more than two years.
P(4 <= T <= 5) = e^(-0.4) - e^(-0.5) which is approximately 0.064. About 6% fail during the fifth year.
P(0 <= T <= 3) = 1 - e^(-0.3) which is approximately 0.259. About 259 of 1000 chips fail within three years.
Exponential distributions are memoryless: the probability that an object survives at least s + t hours, given it has already survived t hours, is the same as the unconditional probability of surviving s hours. The current age of a radioactive particle does not change the probability that it will survive another time period. This property is sometimes used as an approximation even when not strictly valid (e.g. predicting artificial hip replacement lifetimes).
The expected value of a continuous random variable X with PDF f is:
mu = E(X) = the integral from negative infinity to positive infinity of x f(x) dx.
This is a weighted average, where each value of X is weighted by f(X). It gives the long-run average.
For the exponential distribution: The mean is mu = 1/c. This is derived using integration by parts and L'Hopital's rule.
Once you know the mean mu of an exponential distribution, you can write the entire PDF: f(X) = (1/mu) e^(-X/mu) for X >= 0.
Worked example (non-exponential): For f(X) = (4/27)X^2(3 - X) on [0, 3], the expected value is E(X) = the integral from 0 to 3 of (4/27)X^3(3 - X) dX = 1.8.
The median m satisfies: the integral from negative infinity to m of f(X) dX = 1/2.
There is an equal likelihood of X falling below or above m.
For the exponential distribution: The median is m = (1/c) ln 2 = mu ln 2. Since ln 2 is approximately 0.693, the median is always less than the mean for an exponential distribution. The mean and median differ because the distribution is skewed to the right.
Variance measures dispersion around the mean:
Var(X) = the integral from negative infinity to positive infinity of (X - mu)^2 f(X) dX.
Standard deviation is the square root of the variance: sigma = sqrt(Var(X)).
Worked example: For the chip lifetime with f(T) = 0.1e^(-0.1T), the mean is mu = 10 and the variance is 100, giving sigma = 10. The probability that T lies within one standard deviation of the mean (i.e. 0 <= T <= 20) is 1 - e^(-2) which is approximately 0.865, so about 87% of chips fail within twenty years.
A useful identity: Var(X) = the integral of X^2 f(X) dX minus mu^2. This is often easier to compute than the definition directly.
f(x) = \begin{cases} 0 & x < 0 \\ ce^{-cx} & x \ge 0 \end{cases}Exponential PDF with parameter c.
\mu = \frac{1}{c}Mean of the exponential distribution.
f(X) = \frac{1}{\mu} e^{-X/\mu}, \quad X \ge 0Exponential PDF written in terms of its mean.
m = \mu \ln 2Median of the exponential distribution.
\mu = E(X) = \int_{-\infty}^{\infty} X f(X)\, dXExpected value (mean) of any continuous random variable.
\text{Var}(X) = \int_{-\infty}^{\infty} (X - \mu)^2 f(X)\, dXVariance.
\sigma_X = \sqrt{\text{Var}(X)}Standard deviation.
\text{Var}(X) = \int_{-\infty}^{\infty} X^2 f(X)\, dX - \mu^2Alternative variance formula (often simpler to compute).
Exponential distributions model lifetimes of light bulbs, radioactive particles, electronic components, tooth crowns, and artificial hip replacements. They also model waiting times: how long until a bus arrives, a bee pollinates a flower, or a phone call comes into a help desk. Expected value and standard deviation are the numbers that appear on product specifications ("mean time between failures: 10 years") and in medical studies ("mean systolic blood pressure: 119.7 mm Hg, standard deviation: 10.9 mm Hg").
Students often confuse the mean and the median of an exponential distribution. They are not the same. The median (mu ln 2) is always less than the mean (mu) because the distribution is right-skewed.
Students sometimes think the memoryless property means "the object will last forever." It does not. It means the remaining lifetime distribution is always the same, regardless of how old the object already is.
Students frequently forget to use integration by parts when computing the expected value of an exponential distribution. The integral of x e^(-cx) dx requires it.
Students assume variance and standard deviation are the same thing. They are not. Standard deviation is the square root of variance and is in the same units as X.
⚠️ Know how to write the exponential PDF from a given mean. If mu = 8, then f(T) = (1/8)e^(-T/8).
⚠️ Be able to compute P(a <= T <= b) for exponential distributions by evaluating definite integrals.
⚠️ Expect to derive the mean of the exponential distribution using integration by parts and L'Hopital's rule.
⚠️ The variance calculation for the exponential distribution (Example 10) is a common exam problem. Practice integrating (T - mu)^2 f(T) using repeated integration by parts.
⚠️ The alternative variance formula Var(X) = E(X^2) - mu^2 often saves time. Know it.
⚠️ The relationship between mean and median for the exponential (m = mu ln 2) is a likely short-answer item.
True or false: The mean of an exponential distribution with parameter c = 0.5 is 2.
True. mu = 1/c = 1/0.5 = 2.
Fill in the blank: Exponential distributions have the ______ property, meaning past survival does not affect future probability.
Memoryless.
True or false: For an exponential distribution, the median is always larger than the mean.
False. The median is mu ln 2, which is approximately 0.693 mu, so it is always less than the mean.
True or false: Standard deviation is measured in the same units as the random variable.
True.
Fill in the blank: Var(X) can also be computed as E(X^2) minus ______.
mu^2 (the square of the mean).
Q: A chip's lifetime is modelled by an exponential distribution with mean 8 years. Find the probability the chip fails within 5 years.
A: f(T) = (1/8)e^(-T/8). P(0 <= T <= 5) = the integral from 0 to 5 of (1/8)e^(-T/8) dT = [-e^(-T/8)] from 0 to 5 = 1 - e^(-5/8) = 1 - e^(-0.625) which is approximately 0.465, so about 47%.
Q: Find the mean of the random variable X with PDF f(X) = (4/27)X^2(3 - X) on [0, 3].
A: mu = E(X) = the integral from 0 to 3 of (4/27)X^3(3 - X) dX = (4/27)[(3/4)X^4 - (1/5)X^5] from 0 to 3 = (4/27)(243/4 - 243/5) = 1.8.
Q: Find the median of a random variable with exponential PDF f(X) = ce^(-cX) for X >= 0.
A: Set the integral from 0 to m of ce^(-cX) dX = 1/2. This gives 1 - e^(-cm) = 1/2, so e^(-cm) = 1/2, and m = (1/c) ln 2 = mu ln 2.
Q: An electronics company models chip lifetime with f(T) = 0.1e^(-0.1T). Find the standard deviation and the probability that T falls within one standard deviation of the mean.
A: The mean is mu = 1/0.1 = 10. The variance integral gives Var(T) = 100, so sigma = 10. P(0 <= T <= 20) = 1 - e^(-2) which is approximately 0.865.
Q: The digestion time of a fixed amount of food is exponentially distributed with mean 1 hour. What is the probability the food is digested in less than 30 minutes?
A: f(T) = e^(-T) for T >= 0. P(T < 0.5) = 1 - e^(-0.5) which is approximately 0.3935.
Exponential distributions connect to improper integrals (you evaluate limits as b approaches infinity throughout). The variance calculation relies on integration by parts, tying back to Chapter 8's core techniques. The mean and standard deviation introduced here are the same quantities used in normal distributions (covered in Part 3 of these notes), and the 68-95-99.7 rule for normal distributions is stated in terms of standard deviations from the mean.
Exponential distribution, exponentially decreasing PDF, memoryless property, expected value, mean, E(X), median, variance, Var(X), standard deviation, sigma, integration by parts, L'Hopital's rule, chip lifetime, waiting time, radioactive decay, mean time between failures, MTBF, Section 8.8, calculus probability, weighted average