Economic Modelling · Simple models for credit risk

Markov Rating Transition Models for Credit Risk

Updated 11 October 2026 · Fact-checked

A Markov rating transition model treats a bond's credit rating as a Markov chain. A transition matrix P gives one-period probabilities of moving between ratings, with default as an absorbing state. The n-period probabilities are the entries of Pⁿ. Multiply P by itself n times and read the default column.

Understand Markov Rating Transition Models

A credit rating, such as AAA, A, BBB or Default, summarises how likely a borrower is to pay. Ratings change over time. A rating transition model describes these changes with probabilities.

The model assumes the rating follows a Markov chain. The states are the rating classes plus a default state. The Markov property says the next rating depends only on the current rating, not on the earlier path. A bond that was downgraded last year has the same future probabilities as one that has always held that rating.

The probabilities are stored in a transition matrix P. Entry p(i,j) is the probability of moving from rating i to rating j in one period. Each row is a probability distribution, so each row sums to 1. Default is usually an absorbing state: once there, the chain stays there, so its row is (0, 0, ..., 1).

For several periods, use the Chapman-Kolmogorov equations. The n-step matrix is Pⁿ, assuming the chain is time-homogeneous, meaning P is the same every period. The entry in row i, column Default of Pⁿ is the probability of defaulting within n periods starting from rating i. Because default is absorbing, this is also the probability of being in default at time n.

The model is simple and easy to estimate from historical rating data. But its assumptions are strong. Real ratings show momentum, depend on the economic cycle, and vary by industry. You should be able to state these limits in the exam.

Key rules to remember

One-step transition probability: p(i,j) = P(X(t+1) = j | X(t) = i); Entries of matrix P. Every row sums to 1 and every entry is ≥ 0.
Markov property: P(X(t+1) = j | X(t) = i, X(t−1), ..., X(0)) = P(X(t+1) = j | X(t) = i); Future depends on the present rating only.
Chapman-Kolmogorov: p(i,j; n) = Σk p(i,k; m) × p(k,j; n−m); In matrix form Pⁿ = Pᵐ × Pⁿ⁻ᵐ. Valid for a time-homogeneous chain.
Multi-period matrix: P(n) = Pⁿ; Default column of Pⁿ gives cumulative default probability by time n, if default is absorbing.
Distribution at time n: π(n) = π(0) × Pⁿ; π(0) is a row vector of starting probabilities. Multiply row vector on the left.
Absorbing default state: p(D,D) = 1; Default row is (0, ..., 0, 1). Defaults then accumulate and never reverse.

How to solve Markov Rating Transition Models questions

Use this method for any question on rating transition matrices.

1List the states and check the matrix. Each row must sum to 1. Note whether default is absorbing and whether the chain is time-homogeneous.
2Identify what is asked: a one-period probability, a probability over n periods, or a distribution starting from a mix of ratings.
3For n periods, decide whether to compute Pⁿ in full or only the rows or paths you need.
4Multiply matrices carefully. Entry (i,j) of a product is row i of the first times column j of the second.
5Read the answer from the right entry. Default within n periods is row i, Default column of Pⁿ.
6For a starting mix, multiply the row vector π(0) by Pⁿ.
7For a default in a specific year only, subtract: P(default in year n) = P(default by n) − P(default by n−1).
8Check the result: rows still sum to 1, the default probability is non-decreasing in n, and the answer is between 0 and 1. State assumptions explicitly.

Quickest way: Track only the vector you need

When to use it: When the question asks about one starting rating over two or three periods. Avoid full matrix powers.

Write the starting rating as a row vector with 1 in that state.
Multiply this vector by P to get the distribution after one period. Only one row is used.
Multiply the result by P again for period two. Repeat as needed.
Read the default entry. This is quicker than computing the whole of Pⁿ.
Check that the vector entries still sum to 1.

Common mistakes in Markov Rating Transition Models

Raising each entry of P to the power n instead of using matrix multiplication.
Squaring numbers feels natural, and calculators can do it entrywise.
Fix: Pⁿ means P × P × ... × P. Use row-by-column multiplication.
Multiplying a starting vector on the wrong side, as P × π.
Column vector habits from other topics.
Fix: With row-vector probabilities and row-sum-1 matrices, write π(0) × Pⁿ.
Treating default as a state that can be left.
Forgetting that default is absorbing in the simple model.
Fix: Set the default row to (0, ..., 0, 1) and note that default probability accumulates.
Reading the default probability in a single year as the default column of Pⁿ.
Mixing up cumulative and marginal default probability.
Fix: The default column of Pⁿ is cumulative by time n. Subtract the previous cumulative value for year n only.
Forgetting to state the time-homogeneity and Markov assumptions.
Students jump straight to calculation.
Fix: Write the assumptions in one line before computing. Written papers give marks for them.
Giving weak limitations such as 'the model is simple'.
Not linking limits to credit behaviour.
Fix: Give specific points: rating momentum breaks the Markov property, transition rates vary with the economic cycle, matrices are not constant, ratings lag market information, and rare transitions are estimated from little data.

Worked examples

Example 1

A bond is rated A or B or in default (D). The annual transition matrix is: from A: A 0.90, B 0.08, D 0.02; from B: A 0.10, B 0.80, D 0.10; from D: D 1. Find the probability that a bond now rated A has defaulted by the end of two years. Assume time-homogeneous Markov.

Show the solution

Assumptions: the rating follows a time-homogeneous Markov chain and D is absorbing.
Start in A, so the vector is (1, 0, 0).
After year 1: (0.90, 0.08, 0.02).
After year 2, probability in D = 0.90 × 0.02 + 0.08 × 0.10 + 0.02 × 1.
Compute: 0.018 + 0.008 + 0.02 = 0.046.

Answer: 0.046

Example 2

Using the same matrix, a portfolio is 60% rated A and 40% rated B. Find the proportion in default after two years, and the probability that a bond now rated B defaults in year 2 only.

Show the solution

Year 1 vector from A: (0.90, 0.08, 0.02). From B: (0.10, 0.80, 0.10).
Probability in D after two years from A is 0.046 (from example above).
From B: year 2 default = 0.10 × 0.02 + 0.80 × 0.10 + 0.10 × 1 = 0.002 + 0.08 + 0.10 = 0.182.
Portfolio proportion in default by year 2 = 0.60 × 0.046 + 0.40 × 0.182 = 0.0276 + 0.0728 = 0.1004.
For B, default in year 2 only = cumulative by 2 minus cumulative by 1 = 0.182 − 0.10 = 0.082.

Answer: Portfolio default proportion after two years = 0.1004. For a B bond, default in year 2 only = 0.082.

Exam tips

Write the assumptions (Markov, time-homogeneous, absorbing default) first. They earn marks and cost seconds.
In MCQs, use the starting-vector method on one row. Do not compute the full matrix power.
Check row sums after each multiplication to catch arithmetic slips.
Be ready to criticise the model: rating momentum, economic cycle dependence, sparse data for rare transitions, and rating agency lag.
In computer-based questions, show the matrix power in code, such as repeated matrix multiplication, and state the entry you read.

Practice questions from Simple models for credit risk

Markov Rating Transition Models in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

Markov Rating Transition Models: frequently asked questions

How do I calculate multi-period default probability from a transition matrix?

Raise the matrix to the power n using matrix multiplication. The entry in your starting-rating row and the Default column is the probability of default by time n. This holds when default is absorbing and the chain is time-homogeneous.

Why is default treated as an absorbing state?

In the simple model, a defaulted borrower stays in default. This makes default probabilities accumulate over time and keeps the model easy to handle.

What are the main limitations of a Markov model for credit ratings?

Real rating changes often show momentum, so the past matters. Transition probabilities change with the economic cycle and differ by industry. Ratings can lag market information, and rare moves are estimated from little data.

What is the difference between cumulative and annual default probability?

Cumulative default probability by year n is the default entry of Pⁿ. The probability of defaulting in year n only is that value minus the cumulative value for year n−1.