Nonlinear responses
Session 07 · Worked solutions
1 State weights and timing
Because e^{1.5}\simeq4.4817 and e^{-1.5}\simeq0.2231,
F(-1)\simeq0.8176,\quad F(0)=0.5,\quad F(1)\simeq0.1824.
Low growth receives the high recession weight, and opposite standardized states have complementary weights. This is an immediate direction check for the sign in the exponential. The implementation’s stable formula also returns finite weights at states of magnitude 1000.
With three lags, fitted equations are dated t=4,5,6,7,8. Their state observations are s_3,s_4,s_5,s_6,s_7, respectively. Using s_4 for the equation at t=4 would use the contemporaneous state; using s_2 would introduce an extra unintended lag.
Full-sample standardization uses observations later than some historical forecast origins to estimate the mean and standard deviation. It is a well-defined retrospective normalization, but not the same information set available at each origin. Moreover, the raw state itself must have a documented timing construction. Lagging a centered moving average once does not necessarily remove all future observations from that average.
2 Whitening a likelihood
The standardized residuals are 2/2=1 and 3/3=1. The negative Gaussian log likelihood is therefore
\mathcal L=\log(2\pi)+\log2+\log3+\tfrac12(1+1) \simeq4.6296.
Dropping the constant subtracts \log(2\pi)\simeq1.8379, giving about 2.7918. The minimizing parameter values do not change when the same fixed-sample constant is omitted, but reported objective levels cannot be compared until the convention is reconciled. This is precisely the adjustment used in the classroom likelihood audit.
For a common T\times k design X and constant n\times n covariance \Sigma, the GLS normal equations for the coefficient matrix D are X'(Y-XD)\Sigma^{-1}=0. Multiplying on the right by \Sigma gives X'Y-X'XD=0, hence D=(X'X)^{-1}X'Y. The production implementation uses QR to solve the equivalent system rather than forming that inverse. Equal regime covariances make weights irrelevant to the covariance step; they do not remove state interactions from a mean design that already contains them.
3 A state that changes after impact
Initially s_{t-1}=-1<0, so both paths use coefficient 0.8 at impact. The unshocked value is y_t^0=-0.8; the shocked value is y_t^1=-0.8+1=0.2. Their impact difference is one.
The histories then imply different next states. The unshocked path remains in the high-persistence regime, giving y_{t+1}^0=0.8(-0.8)=-0.64. The shocked path has crossed zero and uses coefficient 0.2, giving y_{t+1}^1=0.2(0.2)=0.04. The horizon-one difference is
0.04-(-0.64)=0.68.
Freezing the initial coefficient at 0.8 on both paths would instead give the difference 0.8\times1=0.8. The disagreement is not a simulation error. One calculation lets the shock change the future regime; the other deliberately prevents that channel. In a generalized response with random future innovations, the comparison would also integrate over future innovation paths under a stated conditional distribution.
4 Shock size and asymmetry
The response function at this horizon is r(\delta)=-0.5\delta-0.2\delta^2. Thus
| Shock | Linear part | Quadratic part | Total |
|---|---|---|---|
| +1 | -0.5 | -0.2 | -0.7 |
| -1 | +0.5 | -0.2 | +0.3 |
| +2 | -1.0 | -0.8 | -1.8 |
| -2 | +1.0 | -0.8 | +0.2 |
Doubling the positive shock changes the response from -0.7 to -1.8, not -1.4. Both the positive and negative shocks have a negative quadratic component. A large negative shock can therefore have a smaller positive effect than a small negative shock in this example.
Adding and subtracting unit-shock responses recovers
C_ha=\frac{r(1)-r(-1)}{2}=\frac{-0.7-0.3}{2}=-0.5, \qquad C_hb=\frac{r(1)+r(-1)}{2}=\frac{-0.7+0.3}{2}=-0.2.
Each response compares the chosen shock with zero, holding the history and other innovations fixed. It is not a deviation from the unconditional mean of the squared innovation. For a unit-variance shock, E(\eta^2)=1, whereas the zero-shock baseline has \eta^2=0.