Estimating a Credit Transition Matrix with a Weighted Base and Stress Blend
Article Quant Q&A · Author: Ashley C.
Summary
The document describes fitting a credit-rating transition matrix as a convex blend of a base matrix and a stress matrix. A single scalar weight determines the blend, and the proposed objective is the sum of squared cell-by-cell differences from an observed transition matrix. The example implements this objective in R and minimizes it with the `optim` function using BFGS.
The supplied example reports a fitted weight near 42% and a total squared error near 2.56% under that implementation. These values differ slightly from the questioner's Excel Solver result, illustrating that numerical optimization settings or precision can affect reported estimates. The document gives a practical method for this particular one-parameter least-squares fit, but does not discuss uncertainty in the estimated weight, weighting matrix cells by observation counts, or constraints that may be needed if the input matrices are not valid transition matrices. Its application is credit-risk modeling rather than trading strategy.
Key ideas
- A blended transition matrix can be formed by weighting a base matrix and a stress matrix with one scalar parameter.
- The scalar can be estimated by minimizing the sum of squared differences between fitted and observed matrix entries.
- R's `optim` function provides a general-purpose way to minimize this objective.
- The reported R estimate differs slightly from the Excel Solver result, so optimization and precision choices matter.
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Full text
# Fitting transition matrices in R by solving for coefficient
# Fitting transition matrices in R by solving for coefficient
I'm using various matrices to impute a fitted transition matrix for credit ratings by solving for a variable [S]. Essentially the idea is to determine a base matrix and stress matrix to compare to a single point of time [t1] to create a fitted transition matrix [t1 fitted]. With the data below, [t1 fitted] and [t1 variance] is derived from the following formula:
```
[t1 fitted] = (1-S)*[Base Matrix] + S*[Stress Matrix]
[t1 variance] = ([t1] - [t1 fitted])^2
```
I would like to know how or a direction I can take to solve for this [S] variable where the sum of [t1 variance] is the lowest value. I'd appreciate any direction or solution (functions to use or sample code) as I am not too familiar with R functions/packages and simply recreating the approach as a challenge of the results.
Thanks for your time.
Solution from the sample data below where I derived it using Excel's Solver function by using the GRG Nonlinear method solving for the min value of [S]. (Excel document can be provided if necessary)
```
[S]=41.8962055525792%
sum[t1 variance]=2.56220853052315%
```
```
****************************** Base Matrix (t1-t4) ******************************
From/To: AAA AA A BBB BB B CCC Default
AAA 91.90% 7.39% 0.72% 0.00% 0.00% 0.00% 0.00% 0.00%
AA 1.13% 91.26% 7.09% 0.31% 0.21% 0.00% 0.00% 0.00%
A 0.10% 2.56% 91.19% 5.33% 0.62% 0.21% 0.00% 0.00%
BBB 0.00% 0.21% 5.36% 87.94% 5.46% 0.83% 0.10% 0.10%
BB 0.00% 0.11% 0.43% 5.00% 85.12% 7.33% 0.43% 1.59%
B 0.00% 0.11% 0.11% 0.54% 5.97% 82.19% 2.17% 8.90%
CCC 0.00% 0.44% 0.44% 0.87% 2.51% 5.90% 67.80% 22.05%
Default 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 100.00%
******************************* Stress Matrix (t2) ******************************
From/To: AAA AA A BBB BB B CCC Default
AAA 90.79% 8.29% 0.72% 0.10% 0.10% 0.00% 0.00% 0.00%
AA 0.10% 91.22% 7.85% 0.62% 0.10% 0.10% 0.00% 0.00%
A 0.92% 2.36% 90.04% 5.44% 0.72% 0.31% 0.10% 0.10%
BBB 0.00% 0.32% 5.94% 86.95% 5.30% 1.17% 0.12% 0.21%
BB 0.00% 0.11% 0.66% 7.69% 80.55% 8.79% 0.99% 1.21%
B 0.00% 0.11% 0.23% 0.45% 6.47% 82.75% 4.09% 5.90%
CCC 0.23% 0.00% 0.23% 1.25% 2.28% 12.86% 60.64% 22.53%
Default 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 100.00%
*********************************** t1 Matrix ***********************************
From/To: AAA AA A BBB BB B CCC Default
AAA 94.16% 5.84% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
AA 4.56% 90.89% 4.34% 0.11% 0.00% 0.11% 0.00% 0.00%
A 0.00% 10.10% 87.04% 2.67% 0.06% 0.00% 0.13% 0.00%
BBB 0.00% 0.19% 7.42% 88.44% 3.67% 0.28% 0.00% 0.00%
BB 0.00% 0.00% 0.19% 8.25% 83.49% 7.32% 0.75% 0.00%
B 0.10% 0.00% 0.00% 0.20% 6.54% 83.71% 9.46% 0.00%
CCC 0.00% 0.00% 0.00% 0.00% 0.00% 8.13% 68.62% 23.25%
Default 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 100.00%
******************************** t1 Fitted Matrix *******************************
From/To: AAA AA A BBB BB B CCC Default
AAA 91.43% 7.76% 0.72% 0.04% 0.04% 0.00% 0.00% 0.00%
AA 0.70% 91.25% 7.41% 0.44% 0.16% 0.04% 0.00% 0.00%
A 0.45% 2.48% 90.71% 5.38% 0.66% 0.25% 0.04% 0.04%
BBB 0.00% 0.25% 5.60% 87.52% 5.40% 0.97% 0.11% 0.15%
BB 0.00% 0.11% 0.52% 6.12% 83.21% 7.94% 0.66% 1.43%
B 0.00% 0.11% 0.16% 0.51% 6.18% 82.43% 2.97% 7.65%
CCC 0.10% 0.25% 0.35% 1.03% 2.41% 8.81% 64.80% 22.25%
Default 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 100.00%
******************************* t1 Variance Matrix ******************************
From/To: AAA AA A BBB BB B CCC Default
AAA 0.07% 0.04% 0.01% 0.00% 0.00% 0.00% 0.00% 0.00%
AA 0.15% 0.00% 0.09% 0.00% 0.00% 0.00% 0.00% 0.00%
A 0.00% 0.58% 0.13% 0.07% 0.00% 0.00% 0.00% 0.00%
BBB 0.00% 0.00% 0.03% 0.01% 0.03% 0.00% 0.00% 0.00%
BB 0.00% 0.00% 0.00% 0.05% 0.00% 0.00% 0.00% 0.02%
B 0.00% 0.00% 0.00% 0.00% 0.00% 0.02% 0.42% 0.58%
CCC 0.00% 0.00% 0.00% 0.01% 0.06% 0.00% 0.15% 0.01%
Default 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00% 0.00%
```
## Answer by John Buonagurio (score 2)
https://quant.stackexchange.com/a/17105
You can do this using the `optim` function in R. One possible solution is as follows:
```
base <- c(0.9190, 0.0739, 0.0072, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000,
0.0113, 0.9126, 0.0709, 0.0031, 0.0021, 0.0000, 0.0000, 0.0000,
0.0010, 0.0256, 0.9119, 0.0533, 0.0062, 0.0021, 0.0000, 0.0000,
0.0000, 0.0021, 0.0536, 0.8794, 0.0546, 0.0083, 0.0010, 0.0010,
0.0000, 0.0011, 0.0043, 0.0500, 0.8512, 0.0733, 0.0043, 0.0159,
0.0000, 0.0011, 0.0011, 0.0054, 0.0597, 0.8219, 0.0217, 0.0890,
0.0000, 0.0044, 0.0044, 0.0087, 0.0251, 0.0590, 0.6780, 0.2205,
0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 1.0000)
stress <- c(0.9079, 0.0829, 0.0072, 0.0010, 0.0010, 0.0000, 0.0000, 0.0000,
0.0010, 0.9122, 0.0785, 0.0062, 0.0010, 0.0010, 0.0000, 0.0000,
0.0092, 0.0236, 0.9004, 0.0544, 0.0072, 0.0031, 0.0010, 0.0010,
0.0000, 0.0032, 0.0594, 0.8695, 0.0530, 0.0117, 0.0012, 0.0021,
0.0000, 0.0011, 0.0066, 0.0769, 0.8055, 0.0879, 0.0099, 0.0121,
0.0000, 0.0011, 0.0023, 0.0045, 0.0647, 0.8275, 0.0409, 0.0590,
0.0023, 0.0000, 0.0023, 0.0125, 0.0228, 0.1286, 0.6064, 0.2253,
0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 1.0000)
t1 <- c(0.9416, 0.0584, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000,
0.0456, 0.9089, 0.0434, 0.0011, 0.0000, 0.0011, 0.0000, 0.0000,
0.0000, 0.1010, 0.8704, 0.0267, 0.0006, 0.0000, 0.0013, 0.0000,
0.0000, 0.0019, 0.0742, 0.8844, 0.0367, 0.0028, 0.0000, 0.0000,
0.0000, 0.0000, 0.0019, 0.0825, 0.8349, 0.0732, 0.0075, 0.0000,
0.0010, 0.0000, 0.0000, 0.0020, 0.0654, 0.8371, 0.0946, 0.0000,
0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0813, 0.6862, 0.2325,
0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 0.0000, 1.0000)
base <- matrix(base, nrow=8, ncol=8, byrow=TRUE)
stress <- matrix(stress, nrow=8, ncol=8, byrow=TRUE)
t1 <- matrix(t1, nrow=8, ncol=8, byrow=TRUE)
sum.variance <- function(s, base, stress, t1) {
t1.fitted <- (1-s)*base + s*stress
t1.variance <- (t1 - t1.fitted)^2
sum(t1.variance)
}
res <- optim(par=0.5, fn=sum.variance, method="BFGS", base=base, stress=stress, t1=t1)
sprintf("[S]=%f%%", res$par*100)
sprintf("sum[t1 variance]=%f%%", res$value*100)
```
```
[1] "[S]=41.872702%"
[1] "sum[t1 variance]=2.561079%"
```Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.