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Risk Modelling and Survival Analysis · Concepts of survival models

Future Lifetime Random Variable T_x and Curtate K_x Explained

Updated 11 October 2026 · Fact-checked

T_x is the random variable for the time a life now aged x will live from today. Its distribution function is F_x(t) = P(T_x ≤ t), and its survival function is S_x(t) = 1 − F_x(t) = P(T_x > t). K_x = ⌊T_x⌋ is the whole number of years completed. Solve questions by writing the event in terms of T_x or K_x first.

Understand Future Lifetime Random Variable

Think of a person aged exactly x today. We do not know when they will die. So we treat the time until death as a random variable and call it T_x, the complete future lifetime. It is measured in years and is continuous, taking any value t ≥ 0.

T_x has a distribution function F_x(t) = P(T_x ≤ t). This is the probability that the life dies within t years. The survival function is S_x(t) = P(T_x > t) = 1 − F_x(t). This is the probability the life is still alive t years from now. In life table notation, S_x(t) = tp_x and F_x(t) = tq_x. At t = 0, S_x(0) = 1. S_x(t) never increases as t grows.

The probability density function is f_x(t) = d/dt F_x(t) = −d/dt S_x(t). It gives the density of death at time t. In actuarial notation it equals tp_x · μ(x+t), where μ is the force of mortality. Probabilities over an interval come from integrating f_x(t), or more simply from differences of S_x.

The curtate future lifetime K_x is the number of complete years lived from age x. It is the integer part of T_x: K_x = ⌊T_x⌋. It is discrete, taking values 0, 1, 2, … So K_x = k means T_x lies in the interval [k, k+1). Hence P(K_x = k) = kp_x · q_(x+k) = kp_x − (k+1)p_x. Note that K_x is always less than or equal to T_x, and the gap between them is the fraction of the final year lived.

A useful link: T_x is the age at death minus x, given survival to x. So S_x(t) = S_0(x+t) ÷ S_0(x), where S_0 is the survival function from birth. This is a conditional probability. It is how you move between a newborn's survival function and that of someone aged x.

Key rules to remember

Distribution function of T_x
F_x(t) = P(T_x ≤ t) = tq_x
Probability of death within t years for a life aged x.
Survival function of T_x
S_x(t) = P(T_x > t) = tp_x = 1 − F_x(t)
S_x(0) = 1 and S_x is non-increasing.
Density of T_x
f_x(t) = F_x'(t) = −S_x'(t) = tp_x · μ(x+t)
Valid where the derivative exists, for t ≥ 0.
Interval probability
P(s < T_x ≤ t) = F_x(t) − F_x(s) = sp_x − tp_x
For 0 ≤ s < t. Equivalently the integral of f_x from s to t.
Deferred death probability
s|tq_x = P(s < T_x ≤ s+t) = sp_x − (s+t)p_x = sp_x · tq_(x+s)
Survive s years, then die within the next t.
Link to survival from birth
tp_x = S_0(x+t) ÷ S_0(x)
Conditional on being alive at age x.
Curtate future lifetime
K_x = ⌊T_x⌋; P(K_x = k) = kp_x · q_(x+k) = kp_x − (k+1)p_x
k = 0, 1, 2, …
Survival of K_x
P(K_x ≥ k) = P(T_x ≥ k) = kp_x
Also P(K_x > k) = (k+1)p_x.

How to solve Future Lifetime Random Variable questions

Use this method for any question that defines a lifetime distribution and asks for a probability, density or distribution function.

  1. 1Write down the random variable asked about: T_x (continuous) or K_x (integer).
  2. 2Translate the event into T_x or K_x. For K_x, convert to T_x: K_x = k means k ≤ T_x < k+1; K_x ≥ k means T_x ≥ k.
  3. 3Identify what you are given: F_x(t), S_x(t), f_x(t), or S_0 from birth.
  4. 4Convert to the function you need using S_x(t) = 1 − F_x(t) and f_x(t) = −S_x'(t). Differentiate to go from S or F to f; integrate to go back.
  5. 5If the age changes, use S_x(t) = S_0(x+t) ÷ S_0(x), or tp_x = (x+t)p_0 ÷ xp_0.
  6. 6For an interval or deferred event, use differences of survival probabilities, such as sp_x − tp_x.
  7. 7Substitute numbers and check: the answer is between 0 and 1, S_x(0) = 1, and the density is non-negative.
  8. 8State the answer with the notation and the assumption used.

Quickest way: Survival-function-first shortcut

When to use it: When the question gives a formula for S_0(x) or S_x(t) and asks for a probability, expected value input or density.

  1. Get S_x(t) first. Every other function follows from it.
  2. For any probability, write it as a difference of two S values, or a ratio of S values for conditional events.
  3. Only differentiate when the question asks for the density or force of mortality.
  4. For K_x, write P(K_x = k) = S_x(k) − S_x(k+1) directly.
  5. Sanity check S_x(0) = 1 before moving on.

Common mistakes in Future Lifetime Random Variable

  • Treating P(K_x = k) as kp_x.

    kp_x looks like the probability of living k years, so it feels like the same thing.

    Fix: kp_x is P(K_x ≥ k). For exactly k, subtract: P(K_x = k) = kp_x − (k+1)p_x.

  • Confusing P(K_x > k) with P(K_x ≥ k).

    Discrete inequalities shift by one, and the strict sign is easy to miss.

    Fix: Convert to T_x. K_x ≥ k means T_x ≥ k, so probability kp_x. K_x > k means K_x ≥ k+1, so probability (k+1)p_x.

  • Forgetting to divide by S_0(x) when starting at age x.

    Students plug x+t into the birth survival function and stop.

    Fix: T_x is conditional on survival to x. Always use S_0(x+t) ÷ S_0(x).

  • Writing f_x(t) = dS_x(t)/dt without the minus sign.

    Differentiating is automatic and the sign is forgotten.

    Fix: S is decreasing, so its derivative is negative. Use f_x(t) = −S_x'(t), and check the density is non-negative.

  • Saying T_x is the age at death.

    Notation mixes x and x+T_x.

    Fix: T_x is time lived from age x. Age at death is x + T_x.

  • Using integrals over the wrong range for P(s < T_x ≤ t).

    Students integrate from 0 by habit.

    Fix: Integrate f_x from s to t, or just compute F_x(t) − F_x(s).

Worked examples

Example 1

For a newborn, the survival function is S_0(x) = 1 − x/100 for 0 ≤ x ≤ 100. For a life aged 20, find (a) the survival function S_20(t), (b) the density f_20(t), and (c) P(T_20 > 30).

Show the solution
  1. Use S_20(t) = S_0(20+t) ÷ S_0(20) for 0 ≤ t ≤ 80.
  2. S_0(20) = 1 − 20/100 = 0.8.
  3. S_0(20+t) = 1 − (20+t)/100 = (80 − t)/100.
  4. So S_20(t) = ((80 − t)/100) ÷ 0.8 = (80 − t)/80 = 1 − t/80.
  5. Density: f_20(t) = −d/dt (1 − t/80) = 1/80 for 0 ≤ t ≤ 80.
  6. (c) P(T_20 > 30) = S_20(30) = 1 − 30/80 = 50/80 = 0.625.

Answer: S_20(t) = 1 − t/80, f_20(t) = 1/80 on [0, 80], and P(T_20 > 30) = 0.625.

Example 2

For a life aged x, tp_x = e^(−0.02t) for t ≥ 0. Find (a) P(K_x = 2), (b) P(K_x ≥ 3), and (c) P(1 < T_x ≤ 4).

Show the solution
  1. (a) P(K_x = 2) = 2p_x − 3p_x.
  2. 2p_x = e^(−0.04) = 0.960789.
  3. 3p_x = e^(−0.06) = 0.941765.
  4. Difference = 0.960789 − 0.941765 = 0.019024.
  5. (b) P(K_x ≥ 3) = P(T_x ≥ 3) = 3p_x = e^(−0.06) = 0.941765.
  6. (c) P(1 < T_x ≤ 4) = 1p_x − 4p_x = e^(−0.02) − e^(−0.08).
  7. e^(−0.02) = 0.980199 and e^(−0.08) = 0.923116.
  8. Difference = 0.057083.

Answer: (a) 0.0190, (b) 0.9418, (c) 0.0571 (to four decimal places).

Exam tips

  • Always convert K_x events to T_x events before writing any probability. It removes most inequality errors.
  • Show the link S_x(t) = S_0(x+t) ÷ S_0(x) explicitly. Examiners give method marks for the conditional step.
  • Keep full precision in exponentials until the final line, then round to the accuracy asked.
  • In multiple-choice questions, test candidate answers with S_x(0) = 1 and non-negativity of f_x to eliminate wrong options quickly.
  • Use the notation tp_x, tq_x and μ(x+t) in written answers, and define any symbol you introduce.

Practice questions from Concepts of survival models

Future Lifetime Random Variable: frequently asked questions

What is the difference between T_x and K_x?

T_x is the exact time lived from age x and is continuous. K_x is the whole number of years completed, K_x = ⌊T_x⌋, and is discrete. K_x is always less than or equal to T_x.

How do I get the survival function from the distribution function of T_x?

Use S_x(t) = 1 − F_x(t). This holds because S_x(t) = P(T_x > t) is the complement of P(T_x ≤ t). To get the density, differentiate F_x(t), or take the negative derivative of S_x(t).

Is tp_x the same as S_x(t)?

Yes. tp_x is the standard actuarial notation for P(T_x > t), which is the survival function of T_x evaluated at t. Likewise tq_x = F_x(t).

How is P(K_x = k) calculated?

It equals the probability of surviving k years and then dying in year k+1: kp_x · q_(x+k). Equivalently it is kp_x − (k+1)p_x.