Show how empirical data can be used to estimate the Gauss+ model parameters.

FRM® Part II · Market Risk Measurement and Management · The Vasicek and Gauss+ Models · FRM2.MR.16.d

The idea

Gauss+ parameters can be estimated from historical rate data, using a time series of short, medium and long rates. For each factor, regress the change over one period on the gap between its target and its level, or on its own lagged level with an intercept. In the gap form the slope equals k × dt, where dt is the length of a period in years, so the annual speed of reversion is Speed = slope ÷ dt. Using the fitted intercept α and the slope β on the lagged level, the speed of reversion in annual terms is k = −β ÷ dt. Once the speed is known, the long-run level that the factor reverts to follows from both coefficients as Level = −α ÷ β. The standard deviation of the regression residuals estimates the annual volatility times the square root of dt, so dividing by that root converts it to an annual figure: Volatility = residual standard deviation ÷ √dt. The most common error is forgetting to convert per-period estimates to annual terms. Slow factors, especially the long one, are estimated imprecisely, which is the exam caveat.

Formulas

Speed of mean reversion from a gap regression

k = slope ÷ dt

slope: the coefficient from regressing the factor's change on its gap to target
dt: the data period in years

The slope is per period; dividing by dt annualises it. A positive slope signals mean reversion.

Long-run level from a regression on the lagged level

θ = −α ÷ β

α: the regression intercept
β: the slope on the lagged level, negative under mean reversion

Used with change = α + β × lagged level; in the same form the speed is k = −β ÷ dt.

Annual volatility from regression residuals

σ = s ÷ √dt

s: the standard deviation of the per-period residuals
dt: the data period in years

Residuals are per period; divide by √dt, not dt, to annualise.

A practice question

From the question bank, with the reason each wrong option is wrong.

How can Gauss+ speed-of-reversion parameters be estimated from empirical data?

Regress the change in each factor on the gap between its target and its level; the slope equals speed times period. - correct
Set each speed equal to the standard deviation of the factor, on the logic that volatile factors revert faster.Why not: Confuses speed of reversion with volatility; speeds come from the regression slope, not the standard deviation.
Take the average level of each factor over the sample; this gives the speed of reversion directly.Why not: The sample average relates to the long-run level, not to the speed of reversion.
Regress the level of the short rate on the level of the long rate, and read the correlation as the speed.Why not: Correlation between levels does not measure mean-reversion speed; the regression must relate changes to the gap.

From the discrete form, change = k x gap x dt + noise. Regressing the change on the gap gives a slope of k x dt, and k = slope / dt.

A harder one

Weekly changes in the short rate are regressed on the weekly gap (m - r), in the same units, giving a slope of 0.0154. Using 52 weeks per year, the estimated annual speed k_r is closest to:

0.19Why not: Annualises with 12 periods per year when the data are weekly.
0.80 - correct
0.11Why not: Multiplies by the square root of 52, which is the scaling for volatility, not for the speed.
0.0003Why not: Divides by 52 instead of multiplying, which makes the speed smaller instead of larger.

The slope equals k x dt with dt = 1/52, so k = 0.0154 x 52 = 0.80 per year.

The hardest tier

An analyst fits the Gauss+ long factor on ten years of data and finds that the slope on its gap is positive but statistically insignificant. What is the most appropriate conclusion?

The long factor has no mean reversion, so the model should be changed to one without a long-run level.Why not: Treats an imprecise estimate as proof of no mean reversion; it only shows the data cannot tell.
Ten years cannot pin down the speed and long-run level of such a slow factor, so judgement or longer data is needed. - correct
The long factor is so fast that the data cannot see it, so its speed should be set high.Why not: Reverses the logic; a very slow factor, not a fast one, produces a weak, insignificant slope.
The result proves the regression is misspecified, so the gaps should be replaced by the levels of the three rates themselves.Why not: A weak slope does not show misspecification; it reflects limited information about a very persistent factor.

A very slow factor moves little toward its target during the sample, so the regression has little information about k and theta. Estimates can be unstable. The analyst should use longer data, impose priors or use judgement.

In the app: 6 practice questions on this outcome, spaced review that brings it back before you forget it.

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