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CAS Exam 9 · Credit Risk Modeling and Management · Tutorial 723 of 995

Multi-Period Migration Through Matrix Powers

Derive multi-period rating probabilities from transition matrices, work through matrix-power calculations, and identify the assumptions that make those calculations valid.

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What You'll Learn

  • Interpret a transition matrix using a consistent row-vector convention.
  • Calculate two- and three-period migration probabilities by matrix multiplication.
  • Derive why matrix powers sum over all possible intermediate rating paths.
  • Propagate a portfolio’s starting rating distribution across multiple periods.
  • Distinguish homogeneous matrix powers from products of period-specific matrices.
  • Assess the Markov and time-homogeneity assumptions behind multi-period projections.

Why Multi-Period Migration Requires More Than One-Step Probabilities

A one-period rating transition matrix describes the probability of moving from each current rating to each possible rating one period later. For a multi-year horizon, the question is how probabilities accumulate across every possible sequence of intermediate ratings. Matrix multiplication provides a systematic way to add those paths without listing them individually.

The key calculation is a matrix power when the same transition probabilities apply in every period. That calculation is useful only when its assumptions are understood: in particular, the rating process must be modeled as Markov and the one-period transition matrix must remain constant over the horizon. These methods are aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.

Definition: Let the rating states be \(1,\ldots,K\). A one-period transition matrix \(T\) has entries \(T_{ij}=P(X_{t+1}=j\mid X_t=i)\), where \(X_t\) is the rating state at time \(t\). Thus each row corresponds to the current state, each column to the next state, and \(T_{ij}\geq 0\) with \(\sum_{j=1}^{K}T_{ij}=1\) for every row \(i\). This is a row-stochastic matrix.

Use row vectors for rating distributions. If \(\boldsymbol{\pi}_0\) gives the probabilities of being in each state at the starting date, then after one period the distribution is \(\boldsymbol{\pi}_1=\boldsymbol{\pi}_0T\). In a homogeneous model, the same matrix applies in each period, so after \(n\) periods \(\boldsymbol{\pi}_n=\boldsymbol{\pi}_0T^n\). The entry \((T^n)_{ij}\) is the probability of being in state \(j\) after \(n\) periods, conditional on starting in state \(i\).

Why Matrix Powers Add All Intermediate Paths

The two-period result follows from conditioning on the state after the first period. To be in state \(j\) after two periods, the process must first reach some intermediate state \(k\), then move from \(k\) to \(j\). The events associated with distinct intermediate states are mutually exclusive, so their probabilities add.

$$ P(X_{t+2}=j\mid X_t=i) =\sum_{k=1}^{K}P(X_{t+1}=k\mid X_t=i)P(X_{t+2}=j\mid X_{t+1}=k) =\sum_{k=1}^{K}T_{ik}T_{kj} =(T^2)_{ij}. $$

The second equality uses the Markov property: conditional on the current state \(k\), the next-state distribution does not depend on earlier states. The final equality is precisely the definition of matrix multiplication. Repeating the argument gives \(T^n\) for \(n\) periods. Equivalently, each entry of \(T^n\) is the sum, over all sequences of \(n-1\) intermediate states, of the product of the one-period probabilities along each path.

Theorem: If the rating process is Markov and time-homogeneous with one-period transition matrix \(T\), then the \(n\)-period transition matrix is \(T^n\), for any positive integer \(n\). Every row of \(T^n\) remains a probability distribution: its entries are nonnegative and sum to 1.

The row-sum result is also a useful calculation check. Since each row of \(T\) sums to 1, \(T\mathbf{1}=\mathbf{1}\), where \(\mathbf{1}\) is a column vector of ones. Therefore \(T^n\mathbf{1}=T^{n-1}(T\mathbf{1})=T^{n-1}\mathbf{1}=\mathbf{1}\). Small discrepancies can result from rounding displayed entries; calculations should use unrounded values until the final reporting step.

Worked Example: Two-Year Default Probability and a Three-Year Rating Distribution

A hypothetical rating model has states A, B, and default D. Default is absorbing. The same annual transition matrix applies in both years:

$$ T= \begin{pmatrix} 0.7000&0.2000&0.1000\\ 0.1000&0.6000&0.3000\\ 0&0&1.0000 \end{pmatrix}. $$

Find the probability that an obligor starting in B is in default after two years. Then find the three-year rating distribution for an obligor starting in A. Use the homogeneous Markov model. Method: square the transition matrix, then multiply the resulting row for A by \(T\) once more.

Two-year calculation. The probability of default after two years from B is the sum over the possible intermediate states:

$$ (T^2)_{B,D} =(0.1000)(0.1000)+(0.6000)(0.3000)+(0.3000)(1.0000) =0.0100+0.1800+0.3000 =0.4900. $$

The three terms correspond to B→A→D, B→B→D, and B→D→D. In particular, the final term is included because default is absorbing.

Matrix square. Computing the other entries by the same row-by-column method gives:

$$ T^2= \begin{pmatrix} 0.7000(0.7000)+0.2000(0.1000) & 0.7000(0.2000)+0.2000(0.6000) & 0.7000(0.1000)+0.2000(0.3000)+0.1000(1.0000)\\ 0.1000(0.7000)+0.6000(0.1000) & 0.1000(0.2000)+0.6000(0.6000) & 0.1000(0.1000)+0.6000(0.3000)+0.3000(1.0000)\\ 0&0&1.0000 \end{pmatrix} = \begin{pmatrix} 0.5100&0.2600&0.2300\\ 0.1300&0.3800&0.4900\\ 0&0&1.0000 \end{pmatrix}. $$

Each row sums to 1. For the three-year result from A, multiply the A row of \(T^2\), \((0.5100,0.2600,0.2300)\), by \(T\):

$$ \begin{aligned} P(X_3=A\mid X_0=A)&=0.5100(0.7000)+0.2600(0.1000)=0.3570+0.0260=0.3830,\\ P(X_3=B\mid X_0=A)&=0.5100(0.2000)+0.2600(0.6000)=0.1020+0.1560=0.2580,\\ P(X_3=D\mid X_0=A)&=0.5100(0.1000)+0.2600(0.3000)+0.2300(1.0000)\\ &=0.0510+0.0780+0.2300=0.3590. \end{aligned} $$

The three-year distribution is \((0.3830,0.2580,0.3590)\), and \(0.3830+0.2580+0.3590=1.0000\), as required. The method avoids omitting paths or treating a default during an earlier year as if the obligor could return to a performing rating.

Propagating a Starting Portfolio Distribution

The matrix \(T^n\) gives a result conditional on a specific starting rating. A portfolio distribution is handled by weighting the rows by the starting probabilities. If \(\boldsymbol{\pi}_0\) is a row vector, then \(\boldsymbol{\pi}_n=\boldsymbol{\pi}_0T^n\). The entries of \(\boldsymbol{\pi}_n\) are the portfolio’s modeled proportions or probabilities in each state; they sum to 1 when the initial vector does.

Worked Example: Updating a Portfolio Mix Over Three Periods

A hypothetical portfolio has 40% of its obligors in rating A and 60% in rating B. The two-state annual transition matrix is the same in every period:

$$ \boldsymbol{\pi}_0=(0.4000,0.6000), \qquad T= \begin{pmatrix} 0.9000&0.1000\\ 0.2000&0.8000 \end{pmatrix}. $$

Find the distribution after two and three periods. Method: calculate \(T^2\) and \(T^3\), then apply each matrix to the initial row vector.

First, square \(T\):

$$ T^2= \begin{pmatrix} 0.9000(0.9000)+0.1000(0.2000)&0.9000(0.1000)+0.1000(0.8000)\\ 0.2000(0.9000)+0.8000(0.2000)&0.2000(0.1000)+0.8000(0.8000) \end{pmatrix} = \begin{pmatrix} 0.8300&0.1700\\ 0.3400&0.6600 \end{pmatrix}. $$

Applying this matrix to the starting distribution gives:

$$ \begin{aligned} \boldsymbol{\pi}_2 &=(0.4000,0.6000) \begin{pmatrix}0.8300&0.1700\\0.3400&0.6600\end{pmatrix}\\ &=(0.4000(0.8300)+0.6000(0.3400),\; 0.4000(0.1700)+0.6000(0.6600))\\ &=(0.3320+0.2040,\;0.0680+0.3960)\\ &=(0.5360,0.4640). \end{aligned} $$

Next, multiply \(T^2T\):

$$ T^3= \begin{pmatrix} 0.8300(0.9000)+0.1700(0.2000)&0.8300(0.1000)+0.1700(0.8000)\\ 0.3400(0.9000)+0.6600(0.2000)&0.3400(0.1000)+0.6600(0.8000) \end{pmatrix} = \begin{pmatrix} 0.7810&0.2190\\ 0.4380&0.5620 \end{pmatrix}. $$

Therefore:

$$ \begin{aligned} \boldsymbol{\pi}_3 &=(0.4000(0.7810)+0.6000(0.4380),\; 0.4000(0.2190)+0.6000(0.5620))\\ &=(0.3124+0.2628,\;0.0876+0.3372)\\ &=(0.5752,0.4248). \end{aligned} $$

Both distributions sum to 1. The method avoids applying a one-period matrix only once when the requested horizon is longer; the result is specifically three applications of the same annual model.

When a Matrix Power Is—and Is Not—Appropriate

The matrix-power calculation rests on two distinct assumptions. The Markov assumption says that, conditional on the current rating, the next rating is independent of earlier ratings. The time-homogeneity assumption says that the transition probabilities do not change from one period to another. A model may satisfy one assumption but not the other.

  • If the process is Markov and homogeneous, use \(T^n\).
  • If it is Markov but transition probabilities vary by period, use the ordered product of the period-specific matrices.
  • If current rating alone does not contain enough information to determine the next-state distribution, a rating-only Markov matrix may be inadequate without expanding the state description or changing the model.

For period-specific matrices \(T_1,T_2,\ldots,T_n\), with \(T_1\) applying first, the multi-period transition matrix is \(T_1T_2\cdots T_n\). The order matters because matrix multiplication is generally not commutative. Replacing these matrices by a power of one chosen matrix imposes time homogeneity rather than merely simplifying notation.

Worked Example: Why the Order of Period-Specific Matrices Matters

A hypothetical two-rating process has annual transition matrices

$$ T_1= \begin{pmatrix}0.8000&0.2000\\0.3000&0.7000\end{pmatrix}, \qquad T_2= \begin{pmatrix}0.9000&0.1000\\0.4000&0.6000\end{pmatrix}. $$

The first matrix applies in year 1 and the second in year 2. Find the two-year distribution for an obligor starting in rating A. Method: multiply in chronological order, \(T_1T_2\), and select the A row.

$$ T_1T_2= \begin{pmatrix} 0.8000(0.9000)+0.2000(0.4000)&0.8000(0.1000)+0.2000(0.6000)\\ 0.3000(0.9000)+0.7000(0.4000)&0.3000(0.1000)+0.7000(0.6000) \end{pmatrix} = \begin{pmatrix}0.8000&0.2000\\0.5500&0.4500\end{pmatrix}. $$

Starting in A, the two-year distribution is \((0.8000,0.2000)\). The row sums are 1. If the order were incorrectly reversed, the A row of \(T_2T_1\) would be \((0.9000(0.8000)+0.1000(0.3000),\,0.9000(0.2000)+0.1000(0.7000))=(0.7500,0.2500)\), a different answer. The method avoids assuming a single constant matrix when the model specifies different annual transition probabilities.

Common Mistakes and Model Limitations

A frequent exam trap is to calculate a multi-period probability by multiplying just one transition probability along a single path. That gives the probability of that path, not the probability of reaching the destination by all possible paths. For a two-period transition from \(i\) to \(j\), the required value is \(\sum_k T_{ik}T_{kj}\), not merely \(T_{ij}^2\), unless the model happens to make all other paths impossible.

Another common error is to confuse the row and column conventions. Under the convention used here, distributions are row vectors and update by multiplication on the right: \(\boldsymbol{\pi}_1=\boldsymbol{\pi}_0T\). A column-vector convention is also possible, but its matrix orientation must be defined consistently; mixing the two conventions can produce incorrect results even when the arithmetic is accurate.

Finally, a valid matrix power is not evidence that the model is realistic. Rating transitions may vary with the economic environment, and the current rating may not capture all relevant information about an obligor. The calculation describes the implications of the stated Markov and homogeneity assumptions; it does not establish that those assumptions hold. Estimation of transition probabilities and their uncertainty are separate modeling questions.

Key takeaway: Under a time-homogeneous Markov model, \(T^n\) gives the \(n\)-period transition probabilities because matrix multiplication sums the probabilities of all intermediate-state paths. Check row sums, use the stated vector convention, and use ordered products rather than matrix powers when the transition matrix changes by period.

Check Your Understanding

For each problem, use the row-stochastic convention and retain four decimal places in the final probabilities.

  1. A hypothetical two-state model has \(T=\begin{pmatrix}0.8500&0.1500\\0.2500&0.7500\end{pmatrix}\). Find the two-period distribution for an obligor starting in state 1. (Answer: \((0.7600,0.2400)\))
  2. A hypothetical model has performing state C and absorbing default state D, with \(T=\begin{pmatrix}0.9000&0.1000\\0&1.0000\end{pmatrix}\). Under the homogeneous Markov assumption, find the probability of default within three periods for an obligor starting in C. (Answer: \(0.2710\))
  3. A portfolio starts with distribution \((0.3000,0.7000)\) and has homogeneous transition matrix \(T=\begin{pmatrix}0.9000&0.1000\\0.2000&0.8000\end{pmatrix}\). Find its distribution after two periods. (Answer: \((0.4870,0.5130)\))
  4. For a two-state process, the first and second period matrices are \(T_1=\begin{pmatrix}0.8000&0.2000\\0.3000&0.7000\end{pmatrix}\) and \(T_2=\begin{pmatrix}0.9000&0.1000\\0.4000&0.6000\end{pmatrix}\). Find the two-period distribution for a start in state 2, using the correct order. (Answer: \((0.5500,0.4500)\))
  5. A homogeneous model has \(T=\begin{pmatrix}0.8000&0.2000\\0.1000&0.9000\end{pmatrix}\). Find the two-period probability of being in state 2 given a start in state 1, showing that paths through both possible intermediate states are included. (Answer: \(0.3400\))
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