Survival Functions and Hazard Rates for Exam P

The Survival Function

The survival function S(x) = P(X > x) = 1 - F(x) gives the probability that a random variable exceeds the value x. For lifetime data, S(t) represents the probability of surviving beyond time t. In insurance, the survival function describes the probability that a loss exceeds a given threshold, which is directly relevant to deductible and reinsurance calculations.

Properties of the survival function: S(0) = 1 (or more generally, S equals 1 at the left endpoint of the support), S(infinity) = 0, and S is non-increasing. The PDF can be recovered by differentiation: f(x) = -S'(x). This relationship means you can freely convert between the PDF and survival function representations.

A powerful identity for computing the mean of a non-negative random variable is E[X] = integral from 0 to infinity of S(x) dx. This avoids integration by parts and is often the fastest method on Exam P. The formula works because E[X] = integral from 0 to infinity of integral from 0 to x of f(t) dt dx, and switching the order of integration yields integral from 0 to infinity of (1 - F(t)) dt = integral from 0 to infinity of S(t) dt.

The survival function method extends to higher moments: E[X^n] = n * integral from 0 to infinity of x^(n-1) * S(x) dx for X >= 0. The most commonly used special case is E[X^2] = 2 * integral from 0 to infinity of x * S(x) dx, which gives the second moment needed for variance calculations.

Hazard Rate (Force of Mortality)

The hazard rate, also called the failure rate or force of mortality in actuarial notation, is defined as h(x) = f(x) / S(x). It represents the instantaneous rate of failure at time x, given survival to time x. In precise terms, h(x) * dx is approximately the probability of failing in the interval (x, x + dx) given survival to time x.

An equivalent definition is h(x) = -d/dx ln(S(x)). This differential form reveals the fundamental relationship between the hazard rate and the survival function. Integrating both sides gives -ln(S(x)) = integral from 0 to x of h(t) dt, which leads to the key formula S(x) = exp(-H(x)), where H(x) = integral from 0 to x of h(t) dt is the cumulative hazard function.

The complete chain of relationships connecting all four functions is:

• Given h(x): H(x) = integral of h from 0 to x, S(x) = exp(-H(x)), f(x) = h(x) * S(x)
• Given S(x): h(x) = -S'(x)/S(x), f(x) = -S'(x), F(x) = 1 - S(x)
• Given f(x): F(x) = integral of f, S(x) = 1 - F(x), h(x) = f(x)/S(x)

On Exam P, you should be able to start with any one of these functions and derive all the others.

Interpreting the Hazard Rate

The shape of the hazard function tells you about the aging or wear characteristics of the system being modeled:

• Constant hazard: h(x) = lambda. This characterizes the exponential distribution and implies the memoryless property. The system does not age; the chance of failure in the next instant is the same regardless of how long it has survived.
• Increasing hazard: h(x) increases with x. The system wears out over time, becoming more likely to fail as it ages. This models car engines, human mortality at older ages, and equipment subject to fatigue.
• Decreasing hazard: h(x) decreases with x. Early failures are more common than later ones. This models infant mortality in electronics, software bugs (which are found and fixed over time), and the "burn-in" period of components.
• Bathtub-shaped hazard: h(x) first decreases, then is roughly constant, then increases. This models the lifetime of many manufactured products: early defects, a stable useful life, and eventual wear-out.

Hazard Rates of Common Distributions

For the exponential distribution, h(x) = lambda (constant). This is the defining property of the exponential distribution among continuous distributions.

For the Weibull with S(x) = exp(-(x/theta)^tau), the hazard rate is h(x) = (tau/theta) * (x/theta)^(tau - 1). When tau > 1, the hazard is increasing. When tau < 1, it is decreasing. When tau = 1, it reduces to the constant hazard of the exponential.

For the Pareto with S(x) = (theta/(theta + x))^alpha, the hazard rate is h(x) = alpha / (theta + x), which is decreasing. This decreasing hazard is consistent with the heavy-tailed nature of the Pareto: losses that have already grown large have a lower instantaneous rate of "stopping."

For the Gamma(alpha, lambda) with alpha > 1, the hazard function increases from 0 to lambda (approaching the exponential rate). For alpha < 1, the hazard decreases from infinity to lambda. For alpha = 1 (exponential), h(x) = lambda is constant.

Worked Examples

Example 1: Given h(x) = 2x for x > 0, find S(x), f(x), and E[X]. The cumulative hazard is H(x) = integral from 0 to x of 2t dt = x^2. So S(x) = exp(-x^2) and f(x) = h(x) * S(x) = 2x * exp(-x^2). The mean is E[X] = integral from 0 to infinity of exp(-x^2) dx = sqrt(pi)/2 = 0.8862. This is a Rayleigh distribution.

Example 2: Given h(x) = 1/(10 - x) for 0 < x < 10, find S(x). H(x) = integral from 0 to x of 1/(10 - t) dt = -ln(10 - t) from 0 to x = -ln(10 - x) + ln(10) = ln(10/(10 - x)). So S(x) = exp(-ln(10/(10 - x))) = (10 - x)/10 = 1 - x/10. This is the survival function of a Uniform(0, 10) distribution. The increasing hazard h(x) = 1/(10 - x) reflects the fact that as x approaches the upper limit 10, the conditional probability of failure increases.

Example 3: A device has survival function S(t) = (1 + t)^(-2) for t > 0. Find the hazard rate and expected lifetime. f(t) = -S'(t) = 2(1 + t)^(-3). h(t) = f(t)/S(t) = 2/(1 + t), which is decreasing. E[T] = integral from 0 to infinity of (1 + t)^(-2) dt = 1. This is a Pareto(2, 1) distribution.

Common Pitfalls

A frequent error is confusing the hazard rate with the PDF. The hazard rate h(x) = f(x)/S(x) is not a probability density. It can exceed 1 (for example, h(x) = 2x exceeds 1 for x > 0.5), and it does not integrate to 1. The cumulative hazard H(x) = -ln(S(x)) ranges from 0 to infinity, not from 0 to 1.

Another mistake is forgetting the negative sign when recovering the PDF from the survival function: f(x) = -S'(x), not S'(x). Since S is non-increasing, S'(x) <= 0, so the negative sign ensures f(x) >= 0.

Students sometimes try to compute E[X] by integrating x * h(x), which is incorrect. The correct formula using the hazard rate would require going through the survival function: E[X] = integral of S(x) dx = integral of exp(-H(x)) dx.

Study Tips

For Exam P, memorize the relationships between S(x), f(x), F(x), h(x), and H(x). Practice converting between representations, starting from any one function. Know the hazard rates for the exponential (constant), Weibull (power function), and Pareto (decreasing). The survival function method for computing E[X] = integral of S(x) dx is especially important and should be your first choice whenever X >= 0. Work through problems that give you h(x) and ask for S(x), f(x), or E[X], as this is a common exam pattern. These problems are typically straightforward once you know the relationships, taking about two to three minutes each.