Actuarial Mathematics for Modelling · Gross random future loss
Mean and Variance of Gross Future Loss
Updated 11 October 2026 · Fact-checked
Gross future loss is the present value of benefits and expenses minus the present value of premiums. Its mean is EPV of outgo minus EPV of income, and setting it to zero gives the equivalence premium. For whole life, write the loss as a constant times v^(K+1) plus a constant. The variance is that constant squared times Var(v^(K+1)).
Understand Mean and Variance of Future Loss
A policy is uncertain because you do not know when the policyholder will die. The gross future loss L is a random variable that captures this. It is the present value at issue of everything the insurer pays out (benefits and expenses) minus the present value of everything it receives (premiums). If death comes early, the loss is large. If death comes late, the loss can be negative, which is a profit.
Because L depends on the curtate future lifetime K (or the future lifetime T), you find its mean and variance the same way as for any function of a random variable. The mean is the expected present value (EPV) of outgo minus the EPV of income. You already know how to compute EPVs from assurance and annuity values such as A_x and ä_x.
The equivalence principle says you choose the premium so that E[L] = 0. That means EPV of premiums equals EPV of benefits plus expenses. The premium found this way is the equivalence principle premium. The mean tells you the loss is zero on average. It says nothing about how risky the policy is. That is the job of the variance.
For a whole life policy with premiums paid in advance, the premium annuity can be written using the discount rate: ä_{K+1|} = (1 − v^(K+1)) ÷ d. Substitute this into L and the whole loss becomes a constant times v^(K+1) plus another constant. Adding a constant does not change variance. Multiplying by a constant c multiplies variance by c². So the variance depends only on Var(v^(K+1)) = ²A_x − (A_x)², scaled by the square of the coefficient.
The same idea works for any contract where L is a linear function of one present value factor. For other contracts, such as term assurance or endowment, the loss has more than one piece. You then work out the variance piece by piece, as covered in the related topics.
Key rules to remember
- Gross future loss
- L = PV of benefits + PV of expenses − PV of gross premiums
- All present values are at policy issue and are random because they depend on K (or T).
- Mean of the loss
- E[L] = EPV of benefits + EPV of expenses − EPV of premiums
- This is a signed number. A negative value means the premium is more than enough on average.
- Equivalence principle
- E[L] = 0, so EPV of premiums = EPV of benefits + EPV of expenses
- Solve this equation for the premium P. Use the policy's own expense and premium assumptions.
- Whole life loss, sum assured S paid at end of year of death
- L = S v^(K+1) + I + (f − P) ä_{K+1|}
- P is the annual gross premium in advance. I is an initial expense at time 0. f is an expense at the start of every policy year including the first.
- Whole life loss in simplified form
- L = (S + (P − f)/d) v^(K+1) + I − (P − f)/d
- Uses ä_{K+1|} = (1 − v^(K+1)) ÷ d. The term I − (P − f)/d is a constant.
- Mean for whole life
- E[L] = S A_x + I − (P − f) ä_x
- Use ä_x = (1 − A_x) ÷ d when only A_x is given.
- Variance for whole life
- Var(L) = (S + (P − f)/d)² × [²A_x − (A_x)²]
- The initial expense I and other constants do not affect the variance. ²A_x is A_x calculated at force of interest 2δ, that is at rate of interest 2i + i².
- Net premium special case (no expenses, equivalence premium)
- Var(L) = (S ÷ (1 − A_x))² × [²A_x − (A_x)²]
- Applies only when I = 0, f = 0 and P = S A_x ÷ ä_x, so that S + P/d = S ÷ (1 − A_x).
How to solve Mean and Variance of Future Loss questions
Use this method for any question that asks for the mean, variance or equivalence premium of a future loss.
- 1Write down the contract: benefit amount, when it is paid, premium term, and premium timing (in advance or arrears).
- 2List the expenses and when they occur. Separate the initial expense at time 0 from the regular expense at each premium date.
- 3Write L in words, then in symbols: PV of benefits + PV of expenses − PV of premiums, using K+1 or T as the random time.
- 4For the mean, replace each random present value by its expected value, for example v^(K+1) by A_x and ä_{K+1|} by ä_x. If the question asks for the premium, set the mean to zero and solve.
- 5For the variance, rewrite L as (constant) × v^(K+1) + (another constant). For a whole life policy use ä_{K+1|} = (1 − v^(K+1)) ÷ d.
- 6Compute Var(L) = (coefficient of v^(K+1))² × [²A_x − (A_x)²]. Take ²A_x at the doubled force of interest, that is the rate 2i + i², unless the question gives it.
- 7Check the result: the variance must be positive, the units are rupees squared, and the standard deviation should be a sensible size compared with S. State the standard deviation if asked.
- 8Write the notation, the working and the final answer clearly, with any assumptions stated.
Quickest way: Coefficient method for whole life variance
When to use it: Use it for a whole life (or similar single-factor) loss with premiums in advance, when A_x, ²A_x and the interest rate are given.
- Find d = i ÷ (1 + i).
- Compute the coefficient c = S + (P − f) ÷ d. Here P − f is the premium net of the regular expense.
- Compute Var(L) = c² × (²A_x − A_x²). Do not bother with the constant part of L.
- If P is the equivalence premium with no expenses, skip finding P and use c = S ÷ (1 − A_x) directly.
- For the mean, use E[L] = S A_x + I − (P − f) × (1 − A_x) ÷ d.
Common mistakes in Mean and Variance of Future Loss
Using (A_x)² in place of ²A_x, or forgetting to subtract (A_x)².
The notations look alike and students rush to the formula.
Fix: Remember that ²A_x is the EPV of v^(2(K+1)), a number you are given or compute at the rate 2i + i². Always write ²A_x − (A_x)² as the variance of v^(K+1).
Forgetting to square the coefficient of v^(K+1).
Students remember that variance is linked to the multiplier but treat it like the mean.
Fix: Var(c X + k) = c² Var(X). Write c² explicitly before you multiply.
Putting the initial expense or other constants into the coefficient.
Students include every term in the loss when they build the coefficient.
Fix: Only terms that multiply v^(K+1) belong in the coefficient. The initial expense at time 0 is a constant and drops out of the variance, though it stays in the mean.
Using ä_x in the variance in place of the rewritten form, or mixing up the sign of P − f.
Students try to apply a variance rule directly to ä_{K+1|}, which is itself random and linked to v^(K+1).
Fix: Always substitute ä_{K+1|} = (1 − v^(K+1)) ÷ d first. The premium net of expense, P − f, adds to S in the coefficient for a loss written as outgo minus income.
Treating a non-zero mean as an error when the premium is not the equivalence premium.
Students expect E[L] to always be zero.
Fix: E[L] = 0 only when P is the equivalence premium. If P is given, E[L] can be positive or negative.
Using the wrong interest rate for ²A_x, for example doubling i to 2i.
Students confuse doubling the force of interest with doubling the rate.
Fix: Doubling δ gives v² = e^(−2δ), so the rate is 2i + i² (because (1 + i)² = 1 + 2i + i²). Use that when ²A_x must be recomputed.
Worked examples
Example 1
A whole life policy is issued to a life aged x with sum assured ₹1,00,000 payable at the end of the year of death. Premiums are payable annually in advance for life. There are no expenses. Assume i = 5%, A_x = 0.30 and ²A_x = 0.11. The premium is the equivalence principle premium. Find (a) the premium, (b) E[L], and (c) Var(L) and the standard deviation of L.
Show the solution
- d = 0.05 ÷ 1.05, so 1/d = 21. Then ä_x = (1 − A_x) ÷ d = 0.70 × 21 = 14.7.
- (a) Equivalence principle: P ä_x = S A_x. So P = 1,00,000 × 0.30 ÷ 14.7 = 30,000 ÷ 14.7 = ₹2,040.82 (to the nearest paisa).
- (b) By construction E[L] = 0. Check: S A_x − P ä_x = 30,000 − 2,040.82 × 14.7 = 30,000 − 30,000 = 0.
- (c) L = S v^(K+1) − P ä_{K+1|} = (S + P/d) v^(K+1) − P/d.
- P/d = 2,040.82 × 21 = 42,857.14. So the coefficient is 1,00,000 + 42,857.14 = 1,42,857.14. This equals S ÷ (1 − A_x) = 1,00,000 ÷ 0.7, which confirms it.
- Var(v^(K+1)) = ²A_x − (A_x)² = 0.11 − 0.09 = 0.02.
- Var(L) = (1,42,857.14)² × 0.02 = 2.0408163 × 10^10 × 0.02 ≈ 4.0816 × 10^8.
- Standard deviation = 1,42,857.14 × √0.02 = 1,42,857.14 × 0.141421 ≈ ₹20,203.
Answer: P ≈ ₹2,040.82; E[L] = 0; Var(L) ≈ 4.08 × 10^8 (rupees squared); standard deviation ≈ ₹20,203.
Example 2
A whole life policy with sum assured ₹5,00,000 payable at the end of the year of death is issued to a life aged x. Gross premiums are ₹6,000 a year payable in advance for life. Expenses are ₹4,000 at issue plus ₹200 at the start of every policy year including the first. Assume i = 4%, A_x = 0.25 and ²A_x = 0.08. Find (a) E[L], (b) Var(L) and its standard deviation, and (c) the annual premium that makes E[L] = 0.
Show the solution
- d = 0.04 ÷ 1.04, so 1/d = 26. Then ä_x = (1 − 0.25) × 26 = 19.5.
- Write the loss: L = S v^(K+1) + I + (f − P) ä_{K+1|}, with S = 5,00,000, I = 4,000, f = 200, P = 6,000. So P − f = 5,800.
- (a) E[L] = S A_x + I − (P − f) ä_x = 5,00,000 × 0.25 + 4,000 − 5,800 × 19.5 = 1,25,000 + 4,000 − 1,13,100 = ₹15,900.
- (b) Rewrite L = (S + (P − f)/d) v^(K+1) + I − (P − f)/d.
- (P − f)/d = 5,800 × 26 = 1,50,800. The coefficient is 5,00,000 + 1,50,800 = 6,50,800.
- Var(v^(K+1)) = 0.08 − 0.25² = 0.08 − 0.0625 = 0.0175.
- Var(L) = (6,50,800)² × 0.0175 = 4.2354064 × 10^11 × 0.0175 ≈ 7.412 × 10^9.
- Standard deviation = √(7.412 × 10^9) ≈ ₹86,093.
- (c) Set E[L] = 0: 1,25,000 + 4,000 = (P − 200) × 19.5. So P − 200 = 1,29,000 ÷ 19.5 = 6,615.38, and P = ₹6,815.38.
Answer: (a) E[L] = ₹15,900 (positive, so ₹6,000 is below the equivalence premium on average). (b) Var(L) ≈ 7.41 × 10^9 and the standard deviation ≈ ₹86,093. (c) The equivalence premium is about ₹6,815.38 a year.
Exam tips
- Write the loss in words first. Marks are often given for a correct loss definition, even if the arithmetic goes wrong later.
- Say clearly which assumptions you use: timing of the benefit (end of year of death), premiums in advance, and expenses at each date.
- Check whether the question gives ²A_x. If not, you must recompute A_x at the rate 2i + i², using the given mortality basis.
- In MCQs, test each option for the right structure: the variance should include a squared coefficient and the factor ²A_x − (A_x)². Options that miss either are wrong.
- In the computer-based paper, build A_x, ²A_x and ä_x from the life table in separate cells or variables, then calculate the coefficient and variance in a final step so you can check each part.
Practice questions from Gross random future loss
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Mean and Variance of Future Loss in other exams
The same ground in other exams, if you are preparing for more than one or want another angle on it.
Mean and Variance of Future Loss: frequently asked questions
Why does a zero expected loss give the equivalence principle premium?
The equivalence principle defines the premium as the one that makes EPV of premiums equal to EPV of benefits plus expenses. That is the same as E[L] = 0 because E[L] is EPV of outgo minus EPV of income. So solving E[L] = 0 for P gives that premium.
Why does the initial expense not appear in the variance?
The initial expense is paid at time 0 whatever happens, so it is a constant in L. Adding a constant to a random variable does not change its variance. It does change the mean.
How do I get ²A_x if it is not given?
²A_x is the EPV of a whole life assurance of 1 calculated at the rate of interest 2i + i², using the same mortality as A_x. In practice you recompute the assurance sum with v replaced by v². In exams it is usually given or the table supports it.
Does the variance formula work for term or endowment policies?
Not in the simple form shown here. The loss then has different forms depending on whether death occurs inside or after the term. You work out the variance by splitting the cases, or use a known relationship between the assurances and annuities involved.