Recovering Hazard From Survival Probabilities

Why Recover Force from Survival?

A survival function describes the probability that a lifetime exceeds a given time. The force of mortality, in contrast, describes the instantaneous rate of failure among those who have survived to that time. Recovering the force from survival is useful when a model is supplied as a survival curve but a question asks for its local mortality rate. It also makes clear why two lifetime models can agree on a survival probability at one time yet have different forces there.

Let $T$ be a continuous lifetime measured from a specified origin, such as birth or policy issue. Write $S_0(t)=P(T>t)$. Assume that $S_0(t)>0$ and is differentiable at the time of interest. For the standard absolutely continuous lifetime models considered here, the density is $f_T(t)=-S_0'(t)$. At attained age $x$, the force of mortality is denoted by $\mu_x$.

Definition: The force of mortality at age $x$ is the instantaneous conditional rate of death at that age: $$ \mu_x=\lim_{h\downarrow 0}\frac{P(x<T\leq x+h\mid T>x)}{h}. $$ If $S_0$ is positive and differentiable, then $$ \mu_x=-\frac{S_0'(x)}{S_0(x)}=-\frac{d}{dx}\ln S_0(x)=\frac{f_T(x)}{S_0(x)}. $$ The force has units of inverse time, such as per year.

Deriving the Relationship

The conditional probability in the definition can be written using survival probabilities. Surviving beyond $x$ has probability $S_0(x)$. Surviving beyond $x+h$ has probability $S_0(x+h)$, so the probability of dying during $(x,x+h]$, conditional on being alive at $x$, is their difference divided by $S_0(x)$.

P(x<T\leq x+h\mid T>x) =\frac{S_0(x)-S_0(x+h)}{S_0(x)}.

Divide by $h$ and take the limit as $h$ decreases to zero. Differentiability gives the derivative of the survival function. Since $S_0(x)$ is positive, division by it is valid:

\begin{aligned} \mu_x &=\lim_{h\downarrow0}\frac{S_0(x)-S_0(x+h)}{hS_0(x)}\\ &=-\frac{S_0'(x)}{S_0(x)}. \end{aligned}

For an absolutely continuous lifetime, differentiating the probability $P(T>t)$ gives $f_T(t)=-S_0'(t)$. Substitution then gives $\mu_x=f_T(x)/S_0(x)$. This ratio is a conditional rate: the density of death at age $x$, relative to the probability of having survived to that age. It is not itself the probability of dying at the exact age $x$, which is zero for a continuous lifetime.

The logarithmic form, $\mu_x=-\frac{d}{dx}\ln S_0(x)$, is often the quickest way to differentiate a survival curve. It also shows that the force is nonnegative wherever the survival function is nonincreasing and differentiable. A formula obtained this way applies only where the survival function is positive and differentiable; a point of discontinuity or a point where the model has no ordinary derivative requires separate treatment.

Worked Example: Recovering Force and Density from a Survival Curve

A hypothetical lifetime, measured in years from policy issue, has survival function $S_0(t)=(1+t/6)^{-2}$ for $t\geq0$. Assume this is a continuous lifetime model. Find the survival probability, force of mortality, and density at $t=6$. Give the final rates to four significant figures.

Method. Differentiate the logarithm of the survival function to find the force, then use $f_T(t)=\mu_tS_0(t)$ to obtain the density. This avoids treating the survival probability itself as a mortality rate.

At $t=6$, the survival probability is

S_0(6)=(1+6/6)^{-2}=2^{-2}=\frac14=0.25.

Take logarithms and differentiate:

\ln S_0(t)=-2\ln(1+t/6), \qquad \mu_t=-\frac{d}{dt}\ln S_0(t)=\frac{2}{6+t}.

Therefore,

\mu_6=\frac{2}{12}=\frac16=0.1667\text{ per year}.

The density at age 6 is the force multiplied by the probability of reaching age 6:

f_T(6)=\mu_6S_0(6)=\frac16\cdot\frac14=\frac1{24} =0.04167\text{ per year}.

Check. Direct differentiation gives $S_0'(t)=-(1/3)(1+t/6)^{-3}$, so $-S_0'(6)=(1/3)(1/8)=1/24$, matching the density above. The common mistake is to report $S_0(6)=0.25$ as the force; it is a survival probability, not a rate.

Exponential Lifetime Model

For an exponential lifetime with parameter $\lambda>0$, measured in inverse years when time is measured in years, the survival function is $S_0(t)=e^{-\lambda t}$ for $t\geq0$. Applying the logarithmic derivative gives the force directly:

\mu_t=-\frac{d}{dt}(-\lambda t)=\lambda.

Thus the force is constant over time in this model. Its density is $f_T(t)=\lambda e^{-\lambda t}$. The parameter's units matter: if $\lambda$ is stated per year, $t$ must be measured in years. The exponential form and constant force apply under the exponential model assumption; they should not be applied automatically to an arbitrary survival function.

Worked Example: Exponential Force, Survival, and Density

A hypothetical lifetime follows an exponential model with $\lambda=0.04$ per year. Find the force at $t=5$, the probability of surviving beyond 5 years, and the density at 5 years. Report the probability and density to four decimal places and the force to four significant figures.

Method. Use the exponential survival function, then multiply its value by the force to obtain the density.

The force is constant, so

\mu_5=\lambda=0.04\text{ per year}.

The survival probability is

S_0(5)=e^{-0.04(5)}=e^{-0.20}=0.8187308\ldots\approx0.8187.

The density is the force times survival:

f_T(5)=0.04e^{-0.20}=0.0327492\ldots\approx0.0327\text{ per year}.

Check. Differentiating $S_0(t)=e^{-0.04t}$ gives $-S_0'(5)=0.04e^{-0.20}$, the same density. The force $0.04$ per year is not the probability of death during an entire year; it is an instantaneous rate.

Weibull Lifetime Model

A Weibull lifetime with shape parameter $k>0$ and scale parameter $\theta>0$ has survival function $S_0(t)=\exp[-(t/\theta)^k]$ for $t\geq0$. The scale $\theta$ has units of time, while $k$ is dimensionless. Differentiating the logarithm gives

\mu_t=-\frac{d}{dt}\left[-\left(\frac{t}{\theta}\right)^k\right] =\frac{k t^{k-1}}{\theta^k},\qquad t>0.

The shape parameter determines how the force changes with time. If $k=1$, the model reduces to an exponential lifetime with force $1/\theta$. If $k>1$, the force increases with time; if $0<k<1$, it decreases for $t>0$. At $t=0$, the force is zero for $k>1$, equals $1/\theta$ for $k=1$, and is unbounded in the limiting sense for $0<k<1$. The formula at positive $t$ should not be used to claim a finite force at zero in the last case.

The corresponding density follows by multiplying survival by force:

f_T(t)=\frac{k t^{k-1}}{\theta^k} \exp\left[-\left(\frac{t}{\theta}\right)^k\right].

Worked Example: Weibull Force and a Matched Survival Probability

Two hypothetical lifetime models are being compared, with time measured in years. Model W is Weibull with shape $k=2$ and scale $\theta=10$. Model E is exponential with $\lambda=0.05$ per year. At $t=5$, calculate each model's survival probability and force, and compare the results. Give probabilities to four decimal places and forces to four significant figures.

Method. Calculate the survival and force separately under each model. Matching survival values do not, by themselves, imply matching derivatives or forces.

For Model W, the survival probability at 5 years is

S_{0,W}(5)=\exp[-(5/10)^2]=e^{-0.25}=0.7788008\ldots\approx0.7788.

Its force is

\mu_{5,W}=\frac{2(5)^{2-1}}{10^2}=\frac{10}{100} =0.1000\text{ per year}.

For Model E,

S_{0,E}(5)=e^{-0.05(5)}=e^{-0.25} =0.7788008\ldots\approx0.7788, \qquad \mu_{5,E}=0.05\text{ per year}.

The two survival probabilities agree, but the Weibull force is twice the exponential force at that time: $0.1000/0.0500=2$. This occurs because the Weibull survival curve and the exponential survival curve have the same value at $t=5$ but different slopes there. The common mistake is to infer the force from one survival probability alone; recovering the force requires the local derivative of the survival curve.

Check. The exponents are both $-0.25$, confirming the equal survival probabilities. The Weibull force increases with time because its shape parameter is $2>1$, consistent with its being larger than the exponential force at this age.

Calculation Strategy and Exam Traps

When a problem supplies a survival function, first identify the time origin and units, then check that the function is positive and differentiable at the requested time. The logarithmic derivative is usually the cleanest route to the force. If the question also asks for density, use $f_T(t)=\mu_tS_0(t)$, or differentiate the survival function directly as a check.

Identify the model and time units.
Confirm whether time is measured from birth, issue, or another origin, and make the parameters' units consistent with that time scale.

Write the survival function clearly.
Use the given survival function, or state the exponential or Weibull form and parameters before differentiating.

Differentiate its logarithm.
Calculate $\mu_t=-\frac{d}{dt}\ln S_0(t)$, retaining inverse-time units.

Evaluate and verify.
Substitute the requested time. If density is needed, calculate $\mu_tS_0(t)$ and check it against $-S_0'(t)$.

A frequent exam trap is mixing up a probability and a rate. $S_0(t)$ is a probability of survival to time $t$; $\mu_t$ has units of inverse time and is a local conditional rate. Another trap is to use the Weibull scale as though it were a rate parameter: $\theta$ is measured in time, and the force formula contains $\theta^k$ in the denominator. Finally, the exponential model has a constant force because its survival function is exponential in time; that conclusion does not extend to the general Weibull model unless $k=1$.

Key takeaway: For a positive differentiable survival function, recover the force by taking the negative derivative of its logarithm. For an exponential lifetime the force is $\lambda$; for a Weibull lifetime it is $k t^{k-1}/\theta^k$. Always distinguish this rate from both survival probability and density.

Check Your Understanding

For each hypothetical model, derive or use the force formula before substituting values. The final answers include units where applicable.

A continuous lifetime has $S_0(t)=e^{-0.06t}$, with $t$ in years. Find the force at $t=5$, the survival probability at 5, and the density at 5. (Answer: $\mu_5=0.0600$ per year, $S_0(5)=0.7408$, and $f_T(5)=0.0444$ per year.)
A lifetime has $S_0(t)=(1+t/4)^{-3}$, with $t$ in years. Find $S_0(4)$, $\mu_4$, and $f_T(4)$. (Answer: $S_0(4)=0.1250$, $\mu_4=0.3750$ per year, and $f_T(4)=0.04688$ per year.)
A lifetime is Weibull with $k=2$ and $\theta=6$ years. At $t=3$, find survival, force, and density. (Answer: $S_0(3)=0.7788$, $\mu_3=0.1667$ per year, and $f_T(3)=0.1298$ per year.)
Compare a Weibull model with $k=2,\theta=8$ years and an exponential model with $\lambda=0.0625$ per year at $t=4$. Find each survival probability and force, and state whether the forces agree. (Answer: both survival probabilities are $e^{-0.25}=0.7788$; the Weibull force is $0.1250$ per year and the exponential force is $0.06250$ per year, so they do not agree.)
A lifetime has $S_0(t)=e^{-0.01t^2}$, where $t$ is in years. Find the force and density at $t=5$, and explain whether the force is increasing or decreasing for positive $t$. (Answer: $\mu_t=0.02t$, so $\mu_5=0.1000$ per year; $f_T(5)=0.07788$ per year; the force is increasing for $t>0$.)

What You'll Learn