Exponential Survival Probabilities for Lifetimes

What an Exponential Survival Probability Means

In a lifetime model, the survival probability answers a direct question: what is the probability that the lifetime lasts beyond a stated time? The time might be the age an individual reaches, the number of years a device remains in service, or the duration after an insurance policy is issued. Careful interpretation matters because “survive to age 70” and “survive another 10 years” refer to different starting points unless the individual is currently age 60.

Throughout this tutorial, assume a hypothetical lifetime follows an exponential distribution with rate $\lambda>0$. The rate must be expressed in the reciprocal of the time unit used for the lifetime: a rate per year must be paired with durations in years. No interest rate or discounting is used in these probability calculations.

Definition: Let $T$ be a nonnegative lifetime measured from a stated starting point, and suppose $T$ has an exponential distribution with rate $\lambda$. Its survival function is $$ S_T(t)=P(T>t)=e^{-\lambda t}, \qquad t\geq 0. $$ The corresponding probability of ending by time $t$ is $$ P(T\leq t)=1-e^{-\lambda t}. $$ When $T$ is a lifetime measured from birth, its survival function may be written $S_0(t)$. For a future lifetime $T_x$ of a person already age $x$, the probability of surviving a further $t$ time units is written ${}_tp_x=P(T_x>t)$.

The distinction between the two notations is about the starting point. $S_0(t)$ describes survival from the original time origin through duration $t$. In contrast, ${}_tp_x$ describes survival for a further $t$ units, conditional on having reached age $x$. Under an exponential model for $T_x$ with rate $\lambda$, ${}_tp_x=e^{-\lambda t}$.

Why the Exponential Formula Is a Survival Probability

For an exponential lifetime, the cumulative distribution function gives the probability that the lifetime has ended by time $t$. Survival is the complementary event: the lifetime is greater than $t$. Subtracting the cumulative probability from one gives the survival function.

F_T(t)=P(T\leq t)=1-e^{-\lambda t} \quad\Longrightarrow\quad S_T(t)=P(T>t)=1-F_T(t)=e^{-\lambda t}.

This also explains the relationship between reaching an age and a future-lifetime probability. If $T_0$ is age at death, then reaching age $b$ from age zero has probability $S_0(b)$. If a person has already reached age $a$, the probability of reaching a later age $b$ is conditional:

P(T_0>b\mid T_0>a) =\frac{P(T_0>b)}{P(T_0>a)} =\frac{S_0(b)}{S_0(a)}, \qquad b>a.

The numerator event $T_0>b$ is contained in the denominator event $T_0>a$, which is why the conditional-probability formula reduces to this ratio. If $S_0(t)=e^{-\lambda t}$, then

\frac{S_0(b)}{S_0(a)} =\frac{e^{-\lambda b}}{e^{-\lambda a}} =e^{-\lambda(b-a)}.

Thus the relevant duration for a person already age $a$ who is to reach age $b$ is $b-a$, not $b$. This is a central translation step in actuarial lifetime problems.

Worked Survival Calculations

Worked Example: Survival of a Hypothetical Medical Device

A hypothetical medical device has a lifetime $T$, measured in years from installation. Assume $T$ is exponential with rate $\lambda=0.04$ per year. What is the probability that the device operates for more than 15 years, and what is the probability it fails by 15 years?

Model and method. The exponential survival formula applies because the stated lifetime is modeled as exponential. The time and rate units match: years and per year. There is no interest rate or mortality assumption, because this is a device lifetime rather than a human lifetime.

The probability of operating beyond 15 years is

P(T>15)=e^{-0.04(15)} =e^{-0.60} \approx 0.5488.

The probability of failure by 15 years is the complement:

P(T\leq 15)=1-P(T>15) =1-0.5488 =0.4512.

Check. The two probabilities sum to $0.5488+0.4512=1.0000$, as required. The calculation is rounded to four decimal places.

Common mistake avoided. A rate of $0.04$ per year is not a 4% probability of surviving each year that can simply be multiplied by 15. The exponential model uses $e^{-\lambda t}$, giving $e^{-0.04(15)}$.

Worked Example: Survival from One Age to a Later Age

For a hypothetical life insurance calculation, suppose the future lifetime $T_{58}$ of a person currently age 58 is exponential with rate $\lambda=0.05$ per year. Find the probability the person survives to age 63, the probability the person survives to age 68, and the probability of surviving from age 63 to age 68 conditional on reaching age 63.

Model and method. The assumed mortality model is exponential for future lifetime at age 58, with rate $0.05$ per year. No interest rate is used. Age 63 is 5 years after age 58; age 68 is 10 years after age 58.

First, survival to age 63 means surviving a further 5 years:

P(T_{58}>5)=e^{-0.05(5)} =e^{-0.25} \approx 0.7788.

Survival to age 68 means surviving a further 10 years from age 58:

P(T_{58}>10)=e^{-0.05(10)} =e^{-0.50} \approx 0.6065.

For the conditional probability, use the ratio of the probability of reaching age 68 to the probability of reaching age 63:

P(\text{reach 68}\mid \text{reach 63}) =\frac{0.6065}{0.7788} \approx 0.7788.

Using the unrounded exponential expressions gives the same result directly:

\frac{e^{-0.50}}{e^{-0.25}} =e^{-0.25} \approx 0.7788.

Check. The probability of surviving the additional five years after age 63 is less than one, and it matches the probability for a five-year interval under the stated exponential model. All probabilities are rounded to four decimal places.

Common mistake avoided. Using $e^{-0.05(68)}$ would treat the person as starting at age zero under a model with that origin. This problem specifies future lifetime at age 58, so the durations are 5 and 10 years, respectively.

Worked Example: Using a Median to Find Survival at a Target Age

A hypothetical insurer models the future lifetime $T_{65}$ of a person age 65 as exponential. The assumed median future lifetime is 15 years. Find the model rate and the probability of surviving to age 90. Also find the probability of surviving from age 80 to age 90, conditional on reaching age 80.

Model and method. The lifetime distribution is exponential, measured in years from age 65. No interest rate is used. By definition, the median $m$ satisfies $P(T_{65}\leq m)=0.5$, so the survival probability at the median is $0.5$.

Set the exponential survival probability at 15 years equal to $0.5$, then solve for $\lambda$:

e^{-15\lambda}=0.5 \quad\Longrightarrow\quad -15\lambda=\ln(0.5) \quad\Longrightarrow\quad \lambda=\frac{\ln 2}{15} \approx 0.04621 \text{ per year}.

Age 90 is 25 years after age 65. Therefore,

P(T_{65}>25) =e^{-(\ln 2/15)(25)} =2^{-25/15} \approx 0.3150.

Age 80 is 15 years after age 65, so the conditional probability of then surviving to age 90 is the probability of surviving the next 10 years:

P(\text{reach 90}\mid \text{reach 80}) =\frac{P(T_{65}>25)}{P(T_{65}>15)} =\frac{2^{-25/15}}{0.5} =2^{-10/15} \approx 0.6300.

Check. The probability of reaching age 90 from age 65 is smaller than the probability of reaching age 80 and then surviving the next 10 years conditional on reaching age 80. Numerically, $0.3150=0.5(0.6300)$, up to rounding. The rate is shown to five significant figures and probabilities to four decimal places.

Common mistake avoided. The 15-year median is a median future lifetime from age 65, not an age of 15. The target age 90 corresponds to a duration of $90-65=25$ years.

Interpreting the Result and Avoiding Age Errors

A survival probability is always attached to a specified origin and duration. Before substituting into the exponential formula, identify what $T$ measures and when its clock starts. For an individual currently age $x$, a target age $y>x$ requires future duration $y-x$. For a device installed today, “operates beyond 12 years” uses duration 12 directly.

When an age-origin survival function $S_0$ is provided, reaching a later age after already attaining an earlier age requires a conditional probability. Under a general survival model, use the ratio $S_0(y)/S_0(x)$. Under an exponential model, this ratio simplifies to $e^{-\lambda(y-x)}$. Do not substitute the target age alone into a future-lifetime formula unless the model’s time origin is actually age zero.

The model’s assumptions also limit what can be concluded. The formula $e^{-\lambda t}$ applies to a lifetime modeled as exponential with positive rate $\lambda$, for $t\geq 0$, using consistent time units. It does not establish that a particular real population or product has an exponential lifetime distribution. The probabilities here are model-based results for hypothetical examples.

Key takeaway: For an exponential lifetime, survival beyond duration $t$ is $e^{-\lambda t}$. First identify the starting point, then calculate the elapsed duration to the target. If survival is conditional on already reaching an age, use the conditional probability or survival-function ratio.

Check Your Understanding

For each hypothetical problem, identify the relevant duration before calculating. Probabilities are rounded to four decimal places where needed.

A hypothetical pump has an exponential lifetime measured from installation, with rate $0.03$ per month. Find the probability it lasts more than 24 months. (Answer: $e^{-0.72}\approx 0.4868$)
A hypothetical future lifetime at age 45 is exponential with rate $0.04$ per year. Find the probability of surviving to age 52. (Answer: $e^{-0.28}\approx 0.7558$)
A hypothetical age-at-death model has $S_0(t)=e^{-0.015t}$. Given that a person has reached age 60, find the probability of reaching age 75. (Answer: $e^{-0.225}\approx 0.7985$)
A hypothetical future lifetime is exponential with median 10 years. Find the probability that it exceeds 15 years. (Answer: $2^{-1.5}\approx 0.3536$)
A hypothetical component lifetime from installation is exponential with rate $0.08$ per year. Given that the component has lasted 4 years, find the probability it lasts at least 6 more years. (Answer: $e^{-0.48}\approx 0.6188$)

What You'll Learn