Memorylessness under Constant Force

Why Memorylessness Matters

When a person or system has already survived for several years, a natural question is whether that history changes the distribution of its remaining lifetime. Under a constant force of mortality, the answer is no: conditional on survival so far, the remaining lifetime has the same exponential distribution as the original future lifetime. This special property makes conditional calculations simpler, but it depends on the constant-force assumption.

Let $T_x$ denote the future lifetime, in years, of a life aged $x$. Suppose the force of mortality remains constant at $\mu>0$ throughout the future. Then the survival probability for $t$ years is ${}_tp_x=e^{-\mu t}$. The key is to condition by dividing the probability of surviving the full period by the probability of surviving to the conditioning time.

Definition: A nonnegative future lifetime $T_x$ is memoryless if, for all $s,t\geq 0$ for which $P(T_x>s)>0$, $$ P(T_x>s+t\mid T_x>s)=P(T_x>t). $$ In words, after survival for $s$ years, the probability of surviving a further $t$ years is the same as it was at the start.

Deriving the Memoryless Property

Under the constant-force model, the survival function is ${}_tp_x=e^{-\mu t}$. By the definition of conditional probability, the probability of surviving a further $t$ years given survival to time $s$ is the probability of surviving to $s+t$, divided by the probability of surviving to $s$. Substitution gives:

P(T_x>s+t\mid T_x>s) =\frac{P(T_x>s+t)}{P(T_x>s)} =\frac{e^{-\mu(s+t)}}{e^{-\mu s}} =e^{-\mu t} =P(T_x>t).

The cancellation of $e^{-\mu s}$ is the reason the elapsed time disappears. It is valid because the same constant $\mu$ applies throughout both periods. Equivalently, if $R_s=T_x-s$ is the remaining lifetime conditional on $T_x>s$, then for $t\geq0$:

P(R_s>t)=e^{-\mu t}, \qquad P(R_s\leq t)=1-e^{-\mu t}.

Thus $R_s$ has the same exponential distribution, with rate $\mu$, as $T_x$. The lifetime itself has not become a new random variable independent of all past information; rather, its conditional remaining-lifetime distribution is unchanged by the fact of survival. If the force varies with time, the survival probabilities need not have the exponential form above, and this cancellation generally does not yield the memoryless property.

Theorem: If the future lifetime has constant force $\mu>0$, then its remaining lifetime after surviving $s$ years is exponential with rate $\mu$, for every $s\geq0$. In particular, $$ \frac{{}_{s+t}p_x}{{}_sp_x}=e^{-\mu t} $$ when the force is constant over the entire interval from time $0$ through $s+t$. The ratio requires ${}_sp_x>0$.

Conditional Probabilities for Future Intervals

The memoryless property immediately gives probabilities for several types of event. Conditional on surviving $s$ years, the probability of surviving at least another $t$ years is $e^{-\mu t}$, and the probability of dying within those next $t$ years is $1-e^{-\mu t}$. For $0\leq a<b$, the probability that the remaining lifetime falls between $a$ and $b$ years is:

P(a<R_s\leq b) =P(R_s>a)-P(R_s>b) =e^{-\mu a}-e^{-\mu b}.

The interval endpoints here measure time after survival to $s$. Equivalently, the event is that death occurs between times $s+a$ and $s+b$ from the original starting point, conditional on survival to $s$. This distinction prevents a common error: treating a remaining-time interval as though it began at time zero.

Identify the model and time units.
Confirm that the force is constant over the relevant periods, and express $\mu$, $s$, and the future duration in consistent units.

Translate the event into remaining time.
For example, “survive at least 3 more years” is $R_s>3$, while “die between 2 and 5 more years” is $2<R_s\leq5$.

Apply the exponential survival formula.
Use $P(R_s>t)=e^{-\mu t}$, or subtract two survival probabilities for an interval.

Check the conditioning and event.
Confirm that the time in each exponent is a future duration after the conditioning time, and that a death probability is between 0 and 1.

Worked Applications

Worked Example: Survival After a Period of Survival

A hypothetical life has a future lifetime with constant force $\mu=0.04$ per year. Given that the life has survived 10 years, calculate (a) the probability of surviving at least 5 more years and (b) the probability of dying during those next 5 years.

Method. Use the memoryless property. Conditional on surviving 10 years, the remaining lifetime still has an exponential distribution with rate $0.04$ per year.

(a) Survive another 5 years. The relevant duration is 5 years, not the total 15 years:

P(R_{10}>5)=e^{-0.04(5)}=e^{-0.20}=0.8187308.

To four decimal places, the probability is $0.8187$.

(b) Die within the next 5 years. This is the complement of surviving those 5 years:

P(R_{10}\leq5)=1-e^{-0.20}=1-0.8187308=0.1812692.

To four decimal places, the probability is $0.1813$. As a check, the two probabilities sum to $0.8187+0.1813=1.0000$. The common mistake avoided here is using the 15-year survival probability directly; the question is conditional on already having survived the first 10 years.

Worked Example: Death Within a Future Time Band

A hypothetical device has a lifetime $T$ with constant failure force $\mu=0.05$ per year. Given that it has operated for 8 years, find the probability that it fails more than 2 years but no more than 6 years from now.

Method. Let $R_8=T-8$ denote the remaining lifetime given $T>8$. The event is $2<R_8\leq6$. Under the exponential model, subtract the probability of surviving 6 more years from the probability of surviving 2 more years:

P(2<R_8\leq6)=e^{-0.05(2)}-e^{-0.05(6)}.

The two survival probabilities are:

e^{-0.10}=0.9048374, \qquad e^{-0.30}=0.7408182.

Therefore,

P(2<R_8\leq6)=0.9048374-0.7408182=0.1640192.

To four decimal places, the answer is $0.1640$. The result is between 0 and 1, as required. A common error is to calculate the unconditional probability of failure between years 10 and 14 from the original start and forget the condition $T>8$. The remaining-time interval is 2 to 6 years; memorylessness lets us calculate it directly from that interval.

Worked Example: Finding the Force From a Survival Probability

A hypothetical future lifetime has constant force $\mu$. Its probability of surviving 10 years is $0.5$. Given survival to year 7, calculate the probability of dying more than 5 but no more than 8 years after year 7.

Step 1: Find the force. The 10-year survival probability satisfies $e^{-10\mu}=0.5$. Taking natural logarithms gives:

-10\mu=\ln(0.5), \qquad \mu=-\frac{\ln(0.5)}{10} =\frac{\ln 2}{10} =0.0693147\text{ per year}.

Step 2: Translate the event. The specified event is $5<R_7\leq8$. It concerns the remaining lifetime after year 7, not death between years 5 and 8 from the original start.

Step 3: Calculate the interval probability.

P(5<R_7\leq8) =e^{-5\mu}-e^{-8\mu} =2^{-1/2}-2^{-4/5}.

Numerically, $2^{-1/2}=0.7071068$ and $2^{-4/5}=0.5743492$. Thus:

P(5<R_7\leq8)=0.7071068-0.5743492=0.1327576.

To four decimal places, the answer is $0.1328$. The check is that the probability is positive because surviving 5 years is more likely than surviving 8 years. The common mistake avoided is treating the given 10-year survival probability as though it directly gave the requested conditional interval probability; first it must determine $\mu$.

Common Mistakes and Model Limits

Using elapsed time instead of remaining time. After conditioning on survival to $s$, a future duration $t$ uses $e^{-\mu t}$, not $e^{-\mu(s+t)}$.
Forgetting to condition. The unconditional probability of survival to $s+t$ is not the same as the probability of surviving another $t$ years given survival to $s$. The conditional calculation divides by ${}_sp_x$.
Reversing the subtraction for an interval. For $a<b$, $P(a<R_s\leq b)=e^{-\mu a}-e^{-\mu b}$, which is nonnegative because $e^{-\mu a}\geq e^{-\mu b}$.
Applying memorylessness without a constant force. If the force changes over time, the ratio ${}_{s+t}p_x/{}_sp_x$ need not equal ${}_tp_x$. Use the appropriate survival probabilities for the model rather than assuming the elapsed time cancels.

The practical value of memorylessness is that it replaces a conditional calculation involving two survival probabilities with the same exponential survival formula applied to the remaining duration. It is a property of the exponential model, not a general rule that past survival never matters. In applications, verify the force assumption and read each time interval relative to the point at which survival is given.

Check Your Understanding

For each problem, assume the stated constant force applies throughout the relevant period. Give probabilities to four decimal places.

A hypothetical lifetime has constant force $\mu=0.03$ per year. Given survival to year 6, what is the probability of surviving at least 4 more years? (Answer: $0.8869$)
A hypothetical component has constant failure force $\mu=0.08$ per year. Given that it has survived 5 years, what is the probability it fails within the next 2 years? (Answer: $0.1479$)
A future lifetime has constant force $\mu=0.04$ per year. Given survival for 10 years, find the probability the remaining lifetime is more than 2 but no more than 6 years. (Answer: $0.1365$)
A hypothetical lifetime has constant force and a 9-year survival probability of $0.7$. Given survival to year 9, find the probability of surviving 3 more years. (Answer: $0.8879$)
A hypothetical device has constant failure force $\mu=0.10$ per year and has operated for 4 years. Find the conditional probability it fails more than 3 but no more than 6 years from now. (Answer: $0.1920$)

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