Uniform Distribution: Continuous and Discrete for Exam P
Review both continuous and discrete uniform distributions and their Exam P applications.
Continuous Uniform Distribution
If X ~ Uniform(a, b), the PDF is f(x) = 1/(b - a) for a <= x <= b. The CDF is F(x) = (x - a)/(b - a). The mean is (a + b)/2 and the variance is (b - a)^2 / 12. The MGF is M(t) = (e^(tb) - e^(ta)) / (t(b - a)).
The uniform distribution models complete uncertainty within an interval. In insurance, it might model the time of an accident within a year (Uniform(0, 1)) or the location of damage along a structure of known length.
The uniform distribution is completely determined by two parameters: the lower bound a and the upper bound b. Every subinterval of the same length within [a, b] has the same probability. Specifically, P(c < X < d) = (d - c) / (b - a) for any a <= c < d <= b. This simple formula makes many probability calculations straightforward.
Higher moments of the Uniform(a, b) are E[X^n] = (b^(n+1) - a^(n+1)) / ((n+1)(b - a)). For the standard Uniform(0, 1), this simplifies to E[X^n] = 1/(n+1). So E[X] = 1/2, E[X^2] = 1/3, E[X^3] = 1/4, and so on. These values are useful for quickly computing variances and higher moments on Exam P.
Discrete Uniform Distribution
If X takes values 1, 2, ..., n with equal probability 1/n, then X follows a discrete uniform distribution. The mean is (n + 1)/2 and the variance is (n^2 - 1)/12. This distribution arises in Exam P when all outcomes are equally likely, such as rolling a fair die or randomly selecting a policy number.
For integer-valued uniform on {a, a+1, ..., b}, the mean is (a + b)/2 and the variance is ((b - a + 1)^2 - 1)/12.
The PMF of the discrete uniform on {1, 2, ..., n} is P(X = k) = 1/n for each k. The CDF is F(k) = k/n for integer values k in {1, ..., n}. Between integers, F is constant (a step function). To compute P(a <= X <= b) for the discrete uniform, simply count the integers in [a, b] and divide by n.
Example: A policy number is selected uniformly at random from {1, 2, ..., 100}. The probability that the number is between 30 and 50 inclusive is (50 - 30 + 1) / 100 = 21/100 = 0.21.
The Probability Integral Transform
If X is a continuous random variable with CDF F, then U = F(X) follows a Uniform(0, 1) distribution. This result is called the probability integral transform and has important applications in simulation and on Exam P.
The inverse also holds: if U ~ Uniform(0, 1) and F is a continuous, strictly increasing CDF, then X = F^(-1)(U) has CDF F. This inverse transform method is the basis for generating random variables from any continuous distribution using uniform random numbers.
On Exam P, this appears in problems asking you to find the distribution of F(X) or to determine what transformation of a uniform variable produces a given distribution. For instance, if U ~ Uniform(0, 1), then X = -ln(U)/lambda has an Exponential(lambda) distribution because F^(-1)(u) = -ln(1-u)/lambda and U and 1-U have the same distribution.
Order Statistics from Uniform Samples
Order statistics from Uniform(0, 1) have particularly elegant properties. If X_(1) <= X_(2) <= ... <= X_(n) are the order statistics from a sample of size n from Uniform(0, 1), then X_(k) has a Beta(k, n - k + 1) distribution with E[X_(k)] = k / (n + 1) and Var(X_(k)] = k(n - k + 1) / ((n + 1)^2(n + 2)).
A useful result: the spacings D_i = X_(i) - X_(i-1) (with X_(0) = 0 and X_(n+1) = 1) are related to exponential random variables. Specifically, the normalized spacings (n + 1 - i) * D_i are independent Exponential(1). This connection between uniform order statistics and exponentials occasionally appears in advanced Exam P problems.
Exam P Applications
Uniform distributions often appear in "mixed" problems. For example, given that a loss is uniformly distributed on (0, theta) and theta itself has some distribution, you may be asked to find the unconditional distribution or expected value of the loss. The double expectation formula E[X] = E[E[X|Theta]] = E[Theta/2] is typically the fastest approach. Problems involving order statistics from a uniform sample are also common on Exam P.
Another common setup: X and Y are jointly uniformly distributed on a region R. Then f(x, y) = 1 / (area of R) for (x, y) in R. Computing P((X, Y) in A) reduces to computing the area of A intersected with R divided by the area of R. This geometric probability approach appears frequently on Exam P.
Example: A point (X, Y) is chosen uniformly in the square [0, 2] x [0, 2]. Find P(X + Y < 3). The square has area 4. The region where x + y < 3 within the square is the entire square minus the triangle with vertices (1, 2), (2, 1), and (2, 2). That triangle has area 0.5 * 1 * 1 = 0.5. So P(X + Y < 3) = (4 - 0.5)/4 = 7/8.
Common Pitfalls
A frequent mistake is applying continuous uniform formulas to discrete settings, or vice versa. If X takes integer values from 1 to 6, its variance is (36 - 1)/12 = 35/12, not (6 - 1)^2/12 = 25/12. The discrete and continuous variance formulas are different.
Another error is forgetting that the uniform distribution has no memory in a special sense: it does not have the memoryless property. If X ~ Uniform(0, 10) and you know X > 3, then X | X > 3 ~ Uniform(3, 10), which is a different distribution from the original. The conditional distribution narrows the support.
Study Tips
For Exam P, memorize the mean, variance, and CDF for both continuous and discrete uniform distributions. Practice problems involving geometric probability (uniform distributions on regions in the plane), the probability integral transform, and conditional uniform distributions. The uniform distribution is often a building block in more complex problems, so fluency with its properties will serve you well across many question types. On exam day, uniform distribution problems typically require careful setup of integration limits or geometric reasoning, so take a moment to sketch the region before computing.