Structural dynamic factor models
Session 10 · Worked solutions
These solutions calculate the covariance lost under rank reduction and the units of a normalized policy response. A scalar filtering example follows a missing measurement through the prediction step; a final calculation applies the factor-count penalty to three candidate models.
1 What rank reduction discards
The retained impact matrix is
R=\begin{pmatrix}3&0\\0&2\\0&0\end{pmatrix},\qquad RR'=\operatorname{diag}(9,4,0).
The discarded covariance is \operatorname{diag}(0,0,1). The retained trace share is (9+4)/(9+4+1)=13/14\simeq0.928571. It is incorrect to write RR'=\Sigma_F because the third eigenvalue is positive.
With K_q=(e_1,e_2) and M_q=\operatorname{diag}(3,2), the retained standardized shocks for (3,2,5)' are w=M_q^{-1}K_q'\varepsilon=(1,1)'. Reconstruction gives Rw=(3,2,0)', leaving discarded innovation (0,0,5)'. A low-rank approximation can explain a large fraction of unconditional covariance while missing a large component of a particular realized observation.
2 Policy-shock units
The reporting multiplier is 0.5/0.2=2.5. At horizon four, the GDP response becomes 2.5(-0.003)=-0.0075 log points, or approximately -0.75 percent. The exact proportional change would be 100(e^{-0.0075}-1)\simeq-0.7472 percent.
The original shock series still has unit variance. If the impulse response is rescaled, its innovation-unit interpretation changes and the multiplier must be recorded. The other shock columns describe different disturbances, so applying the policy multiplier to them does not standardize them to an economically comparable intervention. The replication retains unit-variance shocks and stores policyScale separately.
3 A missing observation
At date one, the innovation is one and F_1=1+0.25=1.25. Thus K_1=1/1.25=0.8, the posterior mean is 0.8, and its variance is (1-0.8)1=0.2. The Gaussian log-likelihood contribution is
\ell_1=-\tfrac12[\log(2\pi)+\log(1.25)+1/1.25] \simeq-1.430510.
At date two, prediction gives mean 0.8(0.8)=0.64 and variance 0.8^2(0.2)+0.36=0.488. No measurement is available, so these are also the posterior moments and the measurement log-likelihood contribution is zero.
At date three, the predicted mean is 0.8(0.64)=0.512 and variance is 0.8^2(0.488)+0.36=0.67232. Now
K_3=\frac{0.67232}{0.67232+0.25}\simeq0.728944,
so the zero observation yields posterior mean 0.512(1-K_3)\simeq0.138780 and variance 0.67232(1-K_3)\simeq0.182236. Deleting the missing second row would apply only one transition between the first and third measurements and produce a different, incorrectly timed answer.
4 The factor penalty
The penalty per factor is
\frac{100+200}{100\cdot200}\log(100) =0.015\log(100)\simeq0.069078.
Consequently,
\begin{aligned} IC_2(1)&=\log(0.50)+0.069078\simeq-0.624070,\\ IC_2(2)&=\log(0.46)+2(0.069078)\simeq-0.638373,\\ IC_2(3)&=\log(0.43)+3(0.069078)\simeq-0.636737. \end{aligned}
The minimum is at two factors. The third factor improves reconstruction, but not enough to compensate for its additional penalty. Unpenalized reconstruction error would select three among these candidates and, more generally, would keep falling as the retained PCA rank increases. This is precisely why reconstruction fit alone cannot determine a parsimonious factor dimension.