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Actuarial Mathematics for Modelling · Gross random future loss

Percentile Premium and Probability of Loss in Actuarial Mathematics

Updated 11 October 2026 · Fact-checked

The probability of loss is Pr(L > 0), where L is the future loss random variable. The percentile premium is the smallest premium for which Pr(L > 0) is at most a chosen level, such as 5%. Write L in terms of time of death, find when L > 0, then use the survival distribution.

Understand Percentile Premium and Probability of Loss

The future loss random variable L is the present value of outgo minus the present value of income, for one policy, at the policy start. Outgo is benefits and expenses. Income is premiums. Because the time of death is random, L is random. A loss happens when L > 0.

The equivalence principle sets E[L] = 0. That gives a premium that breaks even on average. But a single policy can still lose money. The insurer may want more certainty than an average gives.

The percentile premium principle does this. You pick a small probability α, such as 5%. You then find the smallest premium P so that Pr(L > 0) ≤ α. In words, the insurer makes a profit (or at least no loss) with probability at least 1 − α. This equals saying 0 is at or above the (1 − α) percentile of L.

The key step is always the same. Write L as a function of the future lifetime (T for continuous, K for curtate). For a whole life or term assurance, the present value of the benefit falls as death gets later. So L is a decreasing function of T for a single premium, and a loss happens only if death is early: T < t₀ for some cut-off t₀. Then Pr(L > 0) = Pr(T < t₀), which you read from the survival model.

For discrete (K-based) problems, L takes only a few values. Premium moves the loss values up and down, so the probability of loss is a step function of P. Check each case carefully. For large portfolios, a normal approximation using E[L] and Var(L) is used instead.

Key rules to remember

Future loss
L = PV of outgo − PV of income
Define it for one policy at the start date. Include expenses if the question is a gross premium one.
Probability of loss
Pr(loss) = Pr(L > 0)
A loss means L is strictly positive. L = 0 is not a loss.
Percentile premium principle
smallest P such that Pr(L > 0) ≤ α, equivalently Pr(L ≤ 0) ≥ 1 − α
α is given in the question, for example 0.05 for a 5% chance of loss.
Cut-off lifetime (L decreasing in T)
Pr(L > 0) = Pr(T < t₀), where L(t₀) = 0
Solve L = 0 for t₀. Then use Pr(T < t₀) = 1 − ₜ₀pₓ.
Constant force of mortality
Pr(T < t) = 1 − e^(−μt)
Use only when μ is constant.
Normal approximation for L
Pr(L > 0) ≈ 1 − Φ((0 − E[L]) ÷ √Var(L))
Use only when told to, or for a large portfolio where the sum of losses is close to normal.

How to solve Percentile Premium and Probability of Loss questions

Use this method for any question on the probability of loss or a percentile premium.

  1. 1Write the loss L for one policy: PV of benefits and expenses minus PV of premiums, using the stated interest rate and premium pattern.
  2. 2Express L as a function of the time of death (T or K). Treat the premium P as unknown if you need to find it.
  3. 3Find which death times give L > 0. For a single premium and decreasing L, solve L = 0 to get the cut-off t₀ and note loss occurs for T < t₀.
  4. 4Convert that event to a probability using the survival model: Pr(T < t₀) = 1 − ₜ₀pₓ, or add up the probabilities of the relevant K values.
  5. 5For a probability of loss, substitute the given premium and compute Pr(L > 0).
  6. 6For a percentile premium, set Pr(L > 0) = α (or find the step in P where the probability drops to α or below). Solve for t₀ first, then for P.
  7. 7For discrete cases, list each possible loss value against its probability and pick the smallest P that meets the limit. Check the boundary case.
  8. 8State the answer with units, and say what the probability means in words.

Quickest way: Cut-off time method

When to use it: Single premium or level premium contracts where the loss falls as death gets later, with a given survival model. This is the usual exam case.

  1. Find t₀ from the percentile: Pr(T < t₀) = α, so t₀ = the α-quantile of T.
  2. For constant μ, t₀ = −ln(1 − α) ÷ μ.
  3. Put L = 0 at t = t₀ and solve for P directly. For a single premium with sum assured S, P = S × v^t₀.
  4. Check one value of T below t₀ to confirm L > 0 there.

Common mistakes in Percentile Premium and Probability of Loss

  • Using the equivalence premium (E[L] = 0) when the question asks for a percentile premium.

    Both are premium principles and students default to the one they know best.

    Fix: Read the wording. If it mentions a probability of loss or a percentage, set Pr(L > 0) ≤ α and ignore E[L].

  • Reversing the inequality, using T > t₀ for a loss.

    Students forget that an earlier death means the benefit is discounted less, so the loss is larger.

    Fix: Test a very small T. If L is positive there, loss happens for T below the cut-off.

  • Treating L = 0 as a loss.

    The boundary looks like a loss, but only L > 0 is one.

    Fix: Loss means L > 0 strictly. At the boundary premium the loss at the cut-off is exactly zero.

  • Using the wrong tail: Pr(L > 0) = α but then using 1 − α for t₀.

    Mixing up the percentile of L with the percentile of T.

    Fix: Work out which tail of T gives a loss first. If loss is for small T, then Pr(T < t₀) = α directly.

  • Discounting from the death time to the wrong date, or using K instead of K + 1 for death at the end of the year.

    Rushing the loss function for discrete benefits.

    Fix: For a sum paid at the end of the year of death, the discount power is K + 1. Write L for each K value before using it.

  • Using a normal approximation for one policy with a very skewed loss.

    The normal formula is easy, so students apply it everywhere.

    Fix: Use the exact distribution of L for a single policy unless the question says to approximate.

Worked examples

Example 1

A two-year term assurance pays ₹1,00,000 at the end of the year of death. It is bought with a single premium P at issue. Interest is 5% per year. The probability of death in year 1 is 0.03, and the probability of death in year 2 given survival to the start of year 2 is 0.04. Ignore expenses. (a) Find the probability of loss if P = ₹92,000. (b) Find the smallest premium for which the probability of loss is at most 5%.

Show the solution
  1. Loss: L = 1,00,000 v^(K+1) − P if K = 0 or 1, and L = −P if the policyholder survives two years.
  2. Discounted benefits: 1,00,000 ÷ 1.05 = ₹95,238.10 for death in year 1. 1,00,000 ÷ 1.05² = 1,00,000 ÷ 1.1025 = ₹90,702.95 for death in year 2.
  3. Probabilities: Pr(death in year 1) = 0.03. Pr(death in year 2) = 0.97 × 0.04 = 0.0388. Survival has probability 1 − 0.03 − 0.0388 = 0.9312.
  4. (a) With P = 92,000: year 1 death gives L = 95,238.10 − 92,000 = 3,238.10 > 0. Year 2 death gives L = 90,702.95 − 92,000 = −1,297.05 < 0. Survival gives L = −92,000. So loss happens only for death in year 1.
  5. Pr(loss) = 0.03.
  6. (b) If P < 90,702.95, loss happens for death in either year: probability 0.03 + 0.0388 = 0.0688, which is above 0.05.
  7. If P is at least 90,702.95 but below 95,238.10, loss happens only for death in year 1: probability 0.03, which is at most 0.05. At P = 90,702.95 the year 2 loss is exactly 0, which is not a loss.
  8. So the smallest premium is ₹90,702.95.

Answer: (a) The probability of loss is 0.03 (3%). (b) The percentile premium is ₹90,702.95, about ₹90,703.

Example 2

A whole life assurance pays ₹10,00,000 immediately on death. The single premium is paid at issue. The force of mortality is constant at μ = 0.02 and the force of interest is δ = 0.05. Ignore expenses. (a) Find the probability of loss if the single premium is ₹8,50,000. (b) Find the smallest single premium so that the probability of loss is at most 5%.

Show the solution
  1. Loss: L = 10,00,000 e^(−0.05T) − P. L decreases as T increases, so a loss happens when T is small.
  2. (a) Solve L = 0 with P = 8,50,000: e^(−0.05 t₀) = 0.85. So t₀ = −ln(0.85) ÷ 0.05 = 0.162519 ÷ 0.05 = 3.2504 years.
  3. Pr(loss) = Pr(T < 3.2504) = 1 − e^(−0.02 × 3.2504) = 1 − e^(−0.065008) = 1 − 0.93707 = 0.0629.
  4. (b) We need Pr(T < t₀) = 0.05. So t₀ = −ln(0.95) ÷ 0.02 = 0.051293 ÷ 0.02 = 2.5647 years.
  5. Premium: P = 10,00,000 × e^(−0.05 × 2.5647) = 10,00,000 × e^(−0.128233).
  6. e^(−0.128233) = 0.87965. So P = ₹8,79,650 (to the nearest ₹50).
  7. Check: a premium above ₹8,50,000 should give a lower loss probability than 6.29%. It does, since ₹8,79,650 gives exactly 5%.

Answer: (a) Probability of loss ≈ 0.063 (6.3%). (b) The percentile premium is about ₹8,79,650.

Exam tips

  • Write L in full before doing anything else. Most marks in written questions are for setting L up correctly.
  • Say in one line which direction L moves with T. It shows the examiner you know why the loss event is T < t₀.
  • For discrete problems, tabulate K, probability and L. It makes the step in the probability of loss easy to see.
  • Keep five or six decimal places in the exponent and quantile steps. Rounding t₀ early can shift the premium noticeably.
  • In MCQs, test the options: put the premium into L and check whether Pr(L > 0) is at most the stated level.

Practice questions from Gross random future loss

Percentile Premium and Probability of Loss: frequently asked questions

What is the percentile premium principle?

It sets the premium so that the probability of a loss on the policy is no more than a chosen level α. You find the smallest P with Pr(L > 0) ≤ α. It focuses on the chance of loss, not the average loss.

How do I find the premium so the probability of loss is less than 5%?

Write L as a function of the time of death and find the cut-off time t₀ where L = 0. Set Pr(T < t₀) = 0.05 (if loss occurs for early death) and solve for t₀ from the survival model. Then substitute t₀ back into L = 0 and solve for P.

Is the percentile premium higher than the equivalence premium?

Usually yes, when the percentile level is strict, such as a 5% chance of loss. The equivalence premium only makes the expected loss zero. It can leave a much larger chance of loss. This is not guaranteed for every contract, so compare the numbers.

Is a loss when L > 0 or L ≥ 0?

A loss is L > 0. If L = 0 the insurer neither gains nor loses. This matters at the boundary in discrete problems.