Continuous Random Variables: PDF, CDF, and Percentiles
Master continuous distributions, density functions, and percentile calculations for Exam P.
Probability Density Function
A continuous random variable X has a probability density function (PDF) f(x) satisfying f(x) >= 0 and the integral of f(x) over all real numbers equals 1. Unlike a PMF, f(x) is not a probability itself. Instead, P(a < X < b) = integral of f(x) from a to b. A consequence is that P(X = a) = 0 for any single point.
Common Exam P tasks include finding a normalizing constant. If f(x) = c * x^2 for 0 < x < 3, then integrating c * x^2 from 0 to 3 gives c * 9 = 1, so c = 1/9.
Because P(X = a) = 0 for continuous variables, it follows that P(a < X < b) = P(a <= X <= b) = P(a < X <= b) = P(a <= X < b). The inclusion or exclusion of endpoints does not matter. This is a key difference from discrete random variables, where endpoint inclusion changes the answer. On Exam P, you can freely add or remove equality signs in probability statements involving continuous variables.
The PDF can exceed 1 at some points; what matters is that the total area under the curve equals 1. For example, if X ~ Uniform(0, 0.5), then f(x) = 2 for 0 < x < 0.5. The value 2 is not a probability; it is a density. The probability over any interval is the area, which will always be at most 1.
CDF and Percentiles
The CDF is F(x) = P(X <= x) = integral of f(t) from negative infinity to x. F is non-decreasing with F(negative infinity) = 0 and F(positive infinity) = 1. To find the PDF from the CDF, differentiate: f(x) = F\'(x).
The p-th percentile (or quantile) x_p satisfies F(x_p) = p. For example, the median is the 50th percentile where F(x) = 0.5. Exam P problems often ask you to solve F(x) = 0.5 for x. If F(x) = x^3/27 on [0,3], then the median satisfies x^3/27 = 0.5, giving x = (13.5)^(1/3) = 2.381.
The interquartile range (IQR) is the difference between the 75th and 25th percentiles: IQR = x_{0.75} - x_{0.25}. This measure of spread is more robust to outliers than the standard deviation. While Exam P does not test IQR frequently, understanding percentiles thoroughly prepares you for related problems.
A useful property: F is continuous for a continuous random variable. Moreover, if f(x) > 0 on an interval, then F is strictly increasing on that interval, which guarantees that each percentile is unique. This is important when you need to invert the CDF to find quantiles.
Expected Value and Variance for Continuous Variables
E[X] = integral of x * f(x) dx and Var(X) = E[X^2] - (E[X])^2, analogous to the discrete case. Integration by parts is a common technique on Exam P. For survival-type distributions on [0, infinity), a useful shortcut is E[X] = integral from 0 to infinity of (1 - F(x)) dx, often called the survival function method for computing the mean.
The survival function method deserves special attention. If X >= 0, then E[X] = integral from 0 to infinity of S(x) dx, where S(x) = 1 - F(x) = P(X > x). This formula avoids computing the PDF entirely and is often simpler. For example, if S(x) = (1 - x/10)^3 for 0 < x < 10, then E[X] = integral from 0 to 10 of (1 - x/10)^3 dx. Using substitution u = 1 - x/10 gives E[X] = 10 * integral from 0 to 1 of u^3 du = 10 * (1/4) = 2.5.
For higher moments, E[X^n] = integral of x^n * f(x) dx. Alternatively, for non-negative random variables, E[X^n] = n * integral from 0 to infinity of x^(n-1) * S(x) dx. This generalized survival function formula is less commonly tested but occasionally saves time.
The Hazard Function
The hazard function (or failure rate) is h(x) = f(x) / S(x), where S(x) = 1 - F(x) is the survival function. It represents the instantaneous rate of failure at time x, given survival to time x. In actuarial science, this concept appears as the force of mortality.
Key relationships: S(x) = exp(-integral from 0 to x of h(t) dt) and f(x) = h(x) * S(x). The cumulative hazard function H(x) = integral from 0 to x of h(t) dt = -ln(S(x)). For the exponential distribution, the hazard function is constant: h(x) = lambda. This constant hazard characterizes the exponential distribution and connects to the memoryless property.
An increasing hazard function means the item is more likely to fail as it ages (like a car engine). A decreasing hazard function suggests the opposite (like software after debugging). Exam P may test your understanding of these interpretations.
Mixed Distributions
Some random variables are neither purely discrete nor purely continuous. A mixed distribution has both a discrete component (point masses) and a continuous component. For example, an insurance payment might be 0 with probability 0.7 (no claim) and follow a continuous distribution with probability 0.3 (a claim occurs).
For mixed distributions, the CDF has both jumps (at the point masses) and continuous segments. The expected value is computed by conditioning: E[X] = P(discrete part) * E[X | discrete] + P(continuous part) * E[X | continuous]. Exam P occasionally tests mixed distributions, often in the context of insurance payments with deductibles.
Common Pitfalls
A frequent mistake is forgetting to check the support of the PDF. The function f(x) = 0 outside the specified domain. When integrating to find probabilities or expectations, set the limits of integration to match the support, not to negative infinity and positive infinity blindly.
Another common error is confusing the PDF with the CDF. If a problem gives you F(x) and asks for a probability, use F directly: P(a < X < b) = F(b) - F(a). Do not differentiate F to get f and then integrate f; that is circular and wastes time.
Study Tips
Master integration techniques before Exam P, especially integration by parts, u-substitution, and partial fractions. Many continuous distribution problems reduce to evaluating integrals, and speed with integration translates directly to speed on the exam. Practice converting between PDF, CDF, and survival function representations. Build a reference sheet listing the PDF, CDF, mean, variance, and MGF for each named continuous distribution: Uniform, Exponential, Normal, Gamma, and Beta. On exam day, quickly identifying which distribution a problem involves can save several minutes per question.