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Economic Modelling · Ruin theory

Adjustment Coefficient and Lundberg's Inequality for Ruin Probability

Updated 11 October 2026 · Fact-checked

The adjustment coefficient R is the positive root of the Lundberg equation λ M_X(r) = λ + cr in the compound Poisson model, where c is the premium rate. Lundberg's inequality then gives ψ(u) ≤ e^(−Ru). Solve for R, then plug in the initial surplus u.

Understand Adjustment Coefficient and Lundberg's Inequality

In the classical compound Poisson surplus model, the surplus at time t is U(t) = u + ct − S(t). Here u is the initial surplus, c is the premium income per unit time, and S(t) is total claims up to time t. Claims arrive as a Poisson process with rate λ. Individual claims X are independent and identically distributed.

Ruin means the surplus falls below zero at some time. The ruin probability ψ(u) is the chance this ever happens, starting from surplus u. Exact formulas exist only for a few claim distributions, so we need a simple bound that works for any claim distribution with a moment generating function.

The adjustment coefficient R measures how fast the ruin probability falls as u grows. It is the positive solution of the Lundberg equation: λ M_X(r) = λ + cr. Think of it as the point where the exponential moment of the surplus change equals one: E[e^(−r(c·T − X))] = 1 for one inter-claim period T. A larger R means a safer business.

Lundberg's inequality says ψ(u) ≤ e^(−Ru) for all u ≥ 0. So ruin probability shrinks at least exponentially with initial surplus. The proof idea: e^(−rU(t)) is a martingale when r = R. Stopping it at the ruin time and using the fact that surplus at ruin is negative gives the bound. You should know this outline, but the exam mostly tests the calculation.

A positive root exists only if the premium loading is positive, that is c > λ E[X]. With zero or negative loading, ruin is certain and R does not exist. Also M_X(r) must exist at the root, which needs a light-tailed claim distribution.

Key rules to remember

Surplus process
U(t) = u + ct − S(t)
S(t) is the compound Poisson total claims with Poisson rate λ.
Premium with loading
c = (1 + θ) λ E[X]
θ is the relative security loading. Need θ > 0 for R to exist.
Lundberg equation
λ M_X(r) = λ + cr, equivalently λ(M_X(r) − 1) = cr
R is the unique positive root. r = 0 is always a trivial root, so ignore it.
Lundberg's inequality
ψ(u) ≤ e^(−Ru)
Valid for u ≥ 0 in the classical compound Poisson model.
Exponential claims, mean 1/β
R = β − λ/c = βθ/(1 + θ)
Here the claim density is βe^(−βx) and c = (1 + θ)λ/β.
Exact ruin for exponential claims
ψ(u) = (1/(1 + θ)) e^(−Ru)
The bound e^(−Ru) is correct in the exponent but overstates the constant.
Exponential MGF
M_X(r) = β/(β − r) for r < β
Root must satisfy r < β.

How to solve Adjustment Coefficient and Lundberg's Inequality questions

Use this method for any question asking for R or a bound on ruin probability.

  1. 1Write down λ, the premium rate c, and the claim distribution. If the loading θ is given, compute c = (1 + θ) λ E[X].
  2. 2Check that c > λ E[X]. If not, R does not exist.
  3. 3Write the claim MGF M_X(r) and note the range of r where it exists.
  4. 4Substitute into λ M_X(r) = λ + cr.
  5. 5Simplify. Cancel the trivial root r = 0 and solve for the positive root in the valid range of r.
  6. 6Write ψ(u) ≤ e^(−Ru) and substitute the value of u.
  7. 7State your assumptions and give a short interpretation, for example that a larger R means faster decay of ruin probability.

Quickest way: Shortcut for exponential and gamma-type claims

When to use it: Use when claims are exponential, or when the MGF gives a quadratic after clearing denominators.

  1. For exponential claims with parameter β, use R = β − λ/c directly. Check it with the loading form βθ/(1 + θ).
  2. For other MGFs, multiply through by the denominators of M_X(r) to get a polynomial in r.
  3. Divide out the factor r, since r = 0 is a root, then solve the quadratic.
  4. Reject any root that is not positive or that lies outside the MGF range.
  5. If the equation cannot be solved by hand, use Newton-Raphson with a starting point. In CS2 Paper A, show two or three iterations. In the computer-based Paper B (R), you can code the iteration.

Common mistakes in Adjustment Coefficient and Lundberg's Inequality

  • Accepting r = 0 as the answer.

    r = 0 always satisfies the Lundberg equation, so it appears as a root.

    Fix: Divide by r or state clearly that you need the positive root.

  • Using the wrong premium rate c.

    Students forget that c is per unit time and must include λ, or they use E[X] with loading only.

    Fix: Compute c = (1 + θ) λ E[X] and check the units against λ.

  • Giving a root larger than the MGF limit.

    For exponential claims M_X(r) exists only for r < β, but the algebra may produce other roots.

    Fix: Check every root against the range where M_X(r) exists.

  • Saying ψ(u) = e^(−Ru).

    The bound looks like a formula, so it is quoted as equality.

    Fix: Write ψ(u) ≤ e^(−Ru). Equality fails in general; for exponential claims the exact value is smaller by the factor 1/(1 + θ).

  • Cancelling λ incorrectly when simplifying the Lundberg equation.

    Students write M_X(r) = 1 + (1 + θ) r, or leave λ on one side, when c already contains λ. Since c = (1 + θ) λ E[X], λ cancels completely and E[X] must stay.

    Fix: Substitute c into λ(M_X(r) − 1) = cr, then divide both sides by λ. This gives M_X(r) = 1 + (1 + θ) E[X] r. Use this form only when c = (1 + θ) λ E[X].

Worked examples

Example 1

Claims arrive as a Poisson process with rate λ = 2 per year. Claim sizes are exponential with mean ₹50,000. Premiums are received at rate c = ₹1,25,000 per year. Find the adjustment coefficient and an upper bound for ruin probability when the initial surplus is ₹2,00,000.

Show the solution
  1. Work in units of ₹1,000. Then E[X] = 50 so β = 1/50 = 0.02, and c = 125, λ = 2.
  2. Check loading: λ E[X] = 100 < 125, so θ = 0.25 and R exists.
  3. Lundberg equation: 2 × 0.02/(0.02 − r) = 2 + 125r.
  4. Multiply through by (0.02 − r): 0.04 = (2 + 125r)(0.02 − r) = 0.04 − 2r + 2.5r − 125r².
  5. This gives 0 = 0.5r − 125r², so r(0.5 − 125r) = 0.
  6. The positive root is r = 0.5/125 = 0.004. Check with R = β − λ/c = 0.02 − 2/125 = 0.02 − 0.016 = 0.004. It lies below β = 0.02.
  7. Here u = 200 in units of ₹1,000, so Ru = 0.004 × 200 = 0.8.
  8. Bound: ψ(u) ≤ e^(−0.8) = 0.4493.

Answer: R = 0.004 per ₹1,000 (that is, 0.000004 per rupee). ψ(₹2,00,000) ≤ e^(−0.8) ≈ 0.449.

Example 2

In a compound Poisson model the claim amount X takes the value 1 with probability 0.5 and the value 3 with probability 0.5. The Poisson rate is λ = 1 and the premium rate is c = 3. Show that the adjustment coefficient R satisfies the equation 0.5e^r + 0.5e^(3r) = 1 + 3r and find R numerically to two decimal places if it lies between 0.2 and 0.4.

Show the solution
  1. E[X] = 0.5 × 1 + 0.5 × 3 = 2. Then λ E[X] = 2 < 3, so the loading is positive and R exists.
  2. M_X(r) = 0.5e^r + 0.5e^(3r).
  3. Lundberg equation: 0.5e^r + 0.5e^(3r) = 1 + 3r.
  4. Define f(r) = 0.5e^r + 0.5e^(3r) − 1 − 3r.
  5. f(0.2) = 0.5(1.22140) + 0.5(1.82212) − 1 − 0.6 = 0.61070 + 0.91106 − 1.6 = −0.07824.
  6. f(0.4) = 0.5(1.49182) + 0.5(3.32012) − 1 − 1.2 = 0.74591 + 1.66006 − 2.2 = 0.20597.
  7. Try r = 0.3: 0.5(1.34986) + 0.5(2.45960) − 1 − 0.9 = 0.67493 + 1.22980 − 1.9 = 0.00473.
  8. Try r = 0.29: 0.5(1.33643) + 0.5(2.38691) − 1 − 0.87 = 0.66822 + 1.19346 − 1.87 = −0.00832.
  9. The root lies between 0.29 and 0.30. Linear interpolation: 0.29 + 0.01 × 0.00832/(0.00832 + 0.00473) = 0.29 + 0.0064 = 0.2964.
  10. So R ≈ 0.30 to two decimal places.

Answer: The Lundberg equation is 0.5e^r + 0.5e^(3r) = 1 + 3r. Its positive root is R ≈ 0.30, so ψ(u) ≤ e^(−0.30u) approximately.

Exam tips

  • Always state the model assumptions: compound Poisson claims, constant premium rate, independent claims and u ≥ 0.
  • Show the substitution into the Lundberg equation before you simplify. Method marks sit there.
  • Check that the loading is positive and that your root lies inside the MGF range.
  • For discrete or mixed claims, expect a numerical root. Show two or three trial values and interpolate.
  • When asked to compare two risks or the effect of reinsurance, compare the values of R. The larger R gives the smaller bound.

Practice questions from Ruin theory

Adjustment Coefficient and Lundberg's Inequality: frequently asked questions

What does the adjustment coefficient tell you?

It tells you how quickly the upper bound on ruin probability falls as initial surplus rises. A larger R means ruin is less likely for the same surplus. It depends on the claim distribution, the claim rate and the premium rate.

Why is r = 0 ignored in the Lundberg equation?

At r = 0 both sides of the equation equal λ, so it is always a root and carries no information. The adjustment coefficient is defined as the unique positive root. It exists only when premiums exceed expected claims.

Is Lundberg's inequality exact?

No. It is an upper bound: ψ(u) ≤ e^(−Ru). For exponential claims the exact value is e^(−Ru)/(1 + θ), which is smaller. The bound gets the rate of decay right but not the constant.

Does the adjustment coefficient exist for heavy-tailed claims?

Not in general. If the MGF of the claim size does not exist for any positive r, such as for Pareto claims, there is no adjustment coefficient and Lundberg's inequality does not apply.