Thiele's differential equation, explained term by term

CM1 · Actuarial Mathematics · 8 min read

1The equation

For a policy with sum assured S payable immediately on death, premiums paid continuously at rate P a year, force of interest δ_t and force of mortality μ_x+t:

Three terms on the right. Read as sentences they are obvious, which is the point of this page.

2Term by term

P — premium income. Money is arriving at P a year, so the reserve grows at P a year from that source alone. Nothing subtle.

δ_t · _tV — interest. The reserve is invested, so it earns the force of interest on itself. The bigger the reserve, the bigger this term, which is why reserves accelerate as they build.

− μ_x+t · (S − _tV) — the cost of cover, and the only term worth thinking about. Over a short interval the probability of death is μx+t dt. If the policyholder dies, the insurer pays S — but it was already holding _tV against this policy, and that is released. So the extra money the death costs is S − tV, not S.

That quantity has a name: the death strain at risk. It is the amount the insurer is genuinely exposed to on this policy at this moment, and it is the single most useful idea in the equation.

3What the equation tells you about the shape of a reserve

Look at the last term again. Early in a policy's life the reserve is small, so the death strain at risk is nearly the whole sum assured and the cost-of-cover term is at its largest. The reserve therefore grows slowly at first. As the reserve builds, the death strain at risk falls, the cost of cover falls with it, and more of the premium and more of the interest is left over — so the reserve grows faster and faster.

That is a result you can state without solving anything, and it is the kind of answer an exam question asking you to "comment" is looking for. It also explains the endpoint of an endowment assurance: as _tV approaches S, the death strain at risk approaches zero, the policy is no longer providing any real cover, and the reserve is simply a savings pot about to be paid out.

4Where it comes from

You do not need to reproduce the derivation to use the equation, but one pass through it makes the equation impossible to misremember. Consider the reserve over a short interval from t to t + dt. Starting with _tV, the insurer:

• receives P dt in premiums;
• earns δt · tV dt in interest;
• with probability μx+t dt, pays S and releases the reserve — a net cost of (S − tV) μx+t dt.

So t+dtV = tV + [P + δt tV − μx+t(S − tV)] dt, and dividing by dt and letting it go to zero gives Thiele. Every term in the equation is one line of that list.

5Solving it in an exam

Closed-form solutions exist only for convenient assumptions. What a question will actually ask for is a numerical solution by small steps, which is the same recursion you have just derived with dt replaced by a finite h:

This is the Euler method. Work out the bracket at the start of the step, multiply by the step length, add. Then do it again. The question will tell you the step size and the boundary condition, and the two boundary conditions you will meet are 0V = 0 for a policy priced by the equivalence principle and nV = S at the maturity of an endowment assurance.

The step size matters, and a question may ask you to show that. Each step assumes the derivative is constant across it, which it is not, so a bigger step is a cruder answer. Halving the step roughly halves the error in this method, and that relationship is worth knowing because it tells you whether a discrepancy between two runs is the method or a mistake.

A worked example

A whole life policy with sum assured ₹10,00,000 payable immediately on death, premiums paid continuously at ₹12,000 a year, a force of interest of δ = 0.05, and a force of mortality of μ = 0.01 taken as constant over the first year. The reserve at issue is zero.

Find the reserve after one year using Euler steps of h = 0.5.

1. 1From t = 0 to t = 0.5

   Evaluate the bracket at t = 0, where ₀V = 0:

   P = 12,000
   δ · ₀V = 0.05 × 0 = 0
   μ · (S − ₀V) = 0.01 × (10,00,000 − 0) = 10,000
   bracket = 12,000 + 0 − 10,000 = 2,000
   _0.5V ≈ 0 + 0.5 × 2,000 = ₹1,000

   Of ₹12,000 of premium a year, ₹10,000 is being consumed by the cost of cover. Only ₹2,000 a year is building the reserve, which is why the reserve on a policy like this starts out so small.

2. 2From t = 0.5 to t = 1

   Now _tV is no longer zero, so both the interest term and the death strain at risk have changed:

   P = 12,000
   δ · _0.5V = 0.05 × 1,000 = 50
   μ · (S − _0.5V) = 0.01 × (10,00,000 − 1,000) = 0.01 × 9,99,000 = 9,990
   bracket = 12,000 + 50 − 9,990 = 2,060
   ₁V ≈ 1,000 + 0.5 × 2,060 = ₹2,030

3. 3Why the second step is bigger than the first

   The bracket grew from 2,000 to 2,060 — ₹50 of it interest earned on the reserve, and ₹10 of it the cost of cover falling because the death strain at risk dropped from ₹10,00,000 to ₹9,99,000. Both effects point the same way and both get stronger as the reserve builds. That is the acceleration the equation predicts, visible after half a year on a round-numbers example.

4. 4What a coarser step would have given

   One step of h = 1, evaluating the bracket once at t = 0:

   ₁V ≈ 0 + 1 × 2,000 = ₹2,000

   ₹2,000 against ₹2,030 — a 1.5% difference from the step size alone, and it is an understatement in both cases, because the single step assumes the derivative stays at its lowest value for the whole year. Smaller steps would keep raising the answer towards the true one. If an exam gives you a step size, that is why: the answer depends on it, and the examiner needs to know which answer to expect.

5. 5The sanity checks

   The reserve started at zero, as the equivalence principle requires. It is positive and growing. It is a small fraction of the sum assured after one year, which is right for a whole life policy that has barely begun. And the cost-of-cover term is almost the whole premium, which is the honest description of what this policy is in its first year: nearly all protection, almost no saving.

Where this fits in CM1

This page explains one idea from CM1 Actuarial Mathematics. The full CM1 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.

Other CM1 topics explained

• Simple and compound interest, and the principle that rules out one of them
• Present value, the discount factor, and why d is not i
• Nominal rates payable p times a year, and the loan that costs more than it says
• Force of interest explained, and how it relates to i, v and d
• When the interest rate changes: steps, and a force that varies
• Continuous payment streams, and why they are worth more than they look
• Annuity functions in CM1: immediate, due, deferred and temporary
• Perpetuities and increasing annuities: payments that never stop, or never stay level
• Money rates, real rates and inflation: the exact link, not the subtraction
• Real yields from an inflation index, and index-linked payments
• Loan schedules: the prospective and retrospective methods
• Life table functions explained: l(x), d(x), q(x) and p(x)
• Net premiums and the equivalence principle, worked from first principles
• Reserves in CM1: prospective, retrospective, and why they agree

Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.