25. RBC: Neoclassical Growth Model with Labor Productivity Shock

In this section, we are going to discuss the steady state and transitional dynamics of a Neoclassical Growth Model that has labor productivity \(Z_t(s^t)\) in it. In this set-up, the labor productivity series could be:

• completely deterministic: \(Z_t(s^t)=(1+g)^t s_t\) with \(\ln s_{t+1}=\rho\ln s_t\), \(s_0\) given;
• or stochastic with deterministic trend: \(Z_t(s^t)=(1+g)^t s_t\) with \(\ln s_{t+1}=\rho\ln s_t+\upsilon_{t+1}\), \(s_0\) given and \(\upsilon_{t+1}\) stochastic;
• or stochastic with stochastic trend: \(Z_t(s^t)=Z_{t-1}(s^{t-1})s_t\).

So, in the latter two cases, it is no longer deterministic how the economy would evolve starting from any point off the steady state, because there might be new shocks happening along the way of convergence. Basically the idea is that we model the RBC as the stochastic convergence pattern of a calibrated model. Since the model itself embeds shocks to labor productivity, we are theoretically assuming a specific pattern or structure (generating process) of the labor productivity shock series.

25.1 The model

25.1.1 Set-up

• Discrete periods \(t=0,1,2,\ldots\)
• State variable \(s_t\): the information revealed in period \(t\).
• History \(s^t=(s_0,s_1,\ldots,s_t)\): the accumulated information from period 0 up to period \(t\). \(s^{t+1}=\{s^t,s_{t+1}\}\) is a successor history of \(s^t\). We can denote \(s^{t+1}>s^t\).
• The period \(t\) consumption \(C_t(s^t)\), labor supply \(H_t(s^t)\), productivity \(Z_t(s^t)\) and period \(t+1\) capital \(K_{t+1}(s^t)\) are all functions of history \(s^t\). (This is a more general assumption than first-order Markov process; we can also, as we will do later, assume first-order Markov process for simplicity. But at the beginning, let's keep it general.)
• Probability distribution over history in ex ante perspective in period 0: \(\pi_t(s^t)\in[0,1]\) for \(\forall t\) and \(\forall s^t\). Bayes' rule: the conditional probability of \(s^{t+1}|s^t\) is \(\dfrac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\).
• Measurability condition: we say the system is measurable w.r.t. history if only one decision can be made based on a given history. In other words, \(C_t(s^t),H_t(s^t),K_{t+1}(s^t)\) are all well-defined functions.
• Utility function: \(u(C_t(s^t),H_t(s^t))=\ln C_t(s^t)-v(H_t(s^t))\) where \(v(\cdot)\) is increasing and convex.
• Production function: for the purpose of this model, we actually only need the production function to be CRS. But for simplicity, we will assume Cobb-Douglas production function

$$ F\big(K_t(s^{t-1}),Z_t(s^t)\cdot H_t(s^t)\big)=\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)\cdot H_t(s^t)\big]^{1-\alpha} $$

where \(Z_t(s^t)\) is the productivity of labor. Notice that \(Z_t(s^t)\), unlike endogenous functions \(K_{t+1}(s^t),H_t(s^t)\) and \(C_t(s^t)\), is an exogenously determined functional form out of control, which could possibly the maximum state in the history or the sum of all past states etc.

25.1.2 Social planner's problem

The social planner's problem is to maximize the expected value (in perspective of period 0) of the additively time separable discounted utility of the infinitely living representative household, i.e.

$$ \max_{\{C_t(s^t),H_t(s^t),K_{t+1}(s^t)\}_{t=0}^{\infty}}\sum_{t=0}^{\infty}\sum_{s^t}\beta^t\pi_t(s^t)\big[\ln C_t(s^t)-v(H_t(s^t))\big] \tag{25.1} $$

$$ \text{s.t.}\quad K_{t+1}(s^t)=\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)H_t(s^t)\big]^{1-\alpha}+(1-\delta)K_t(s^{t-1})-C_t(s^t)\ \text{for }\forall t,\forall s^t \tag{25.2} $$

\(K_0=K_0(s^{-1})\) given.

25.1.3 The Lagrangian

$$ \begin{aligned} \mathcal{L}=&\sum_{t=0}^{\infty}\sum_{s^t}\beta^t\pi_t(s^t)\big[\ln C_t(s^t)-v(H_t(s^t))\big]\\ &+\sum_{t=0}^{\infty}\sum_{s^t}\beta^t\pi_t(s^t)\lambda_t(s^t)\Big\{\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)H_t(s^t)\big]^{1-\alpha}+(1-\delta)K_t(s^{t-1})-C_t(s^t)-K_{t+1}(s^t)\Big\} \end{aligned} \tag{25.3} $$

where \(\lambda_t(s^t)\) is the Lagrangian multiplier for the budget constraint in period \(t\) with history \(s^t\).

25.1.4 First order conditions

The f.o.c. of (25.3) are:

$$ [C_t(s^t)]\quad \frac{1}{C_t(s^t)}=\lambda_t(s^t) \tag{25.4} $$

$$ [H_t(s^t)]\quad v'(H_t(s^t))=(1-\alpha)\lambda_t(s^t)\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)\big]^{1-\alpha}\big[H_t(s^t)\big]^{-\alpha} \tag{25.5} $$

$$ [K_{t+1}(s^t)]\quad \lambda_t(s^t)=\beta\sum_{s^{t+1}>s^t}\lambda_{t+1}(s^{t+1})\frac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\Big\{\alpha\big[K_{t+1}(s^t)\big]^{\alpha-1}\big[Z_{t+1}(s^{t+1})H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)\Big\} \tag{25.6} $$

$$ [\lambda_t(s^t)]\quad K_{t+1}(s^t)=\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)H_t(s^t)\big]^{1-\alpha}+(1-\delta)K_t(s^{t-1})-C_t(s^t)\ \text{for }\forall t,\forall s^t $$

Using both (25.4) and (25.5), we can get the stacked f.o.c., which is also the intra-temporal indifference condition:

$$ v'(H_t(s^t))C_t(s^t)=(1-\alpha)\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)\big]^{1-\alpha}\big[H_t(s^t)\big]^{-\alpha} \tag{25.7} $$

where the LHS is \(MRS_t\) and the RHS is \(MPL_t\).

25.1.5 Euler equation and transversality condition

The Euler equation (EE) can be obtained by using (25.4) and (25.6):

$$ \frac{1}{C_t(s^t)}=\beta\sum_{s^{t+1}>s^t}\frac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\cdot\frac{\alpha\big[K_{t+1}(s^t)\big]^{\alpha-1}\big[Z_{t+1}(s^{t+1})H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)}{C_{t+1}(s^{t+1})} \tag{25.8} $$

Or equivalently, we can write the EE in expectation form, i.e.

$$ \frac{1}{C_t(s^t)}=\beta\mathbb{E}_{s^{t+1}}\left[\frac{\alpha\big[K_{t+1}(s^t)\big]^{\alpha-1}\big[Z_{t+1}(s^{t+1})H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)}{C_{t+1}(s^{t+1})}\,\Big|\,s^t\right] \tag{25.9} $$

And the transversality condition (TC) is

$$ \lim_{T\to\infty}\beta^T\lambda_T(s^{T-1})K_T(s^{T-1})=0\ \Rightarrow\ \lim_{T\to\infty}\beta^T\frac{1}{C_T(s^{T-1})}K_T(s^{T-1})=0 $$

Note that the transversality condition is not necessary for optimality of the solution, but it ensures that the capital stock won't explode along time.

25.1.6 Summary of system of equations

• Intra-temporal indifference condition (stacked f.o.c.):

$$ v'(H_t(s^t))C_t(s^t)=(1-\alpha)\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)\big]^{1-\alpha}\big[H_t(s^t)\big]^{-\alpha} \tag{25.10} $$

• Inter-temporal indifference condition (EE):

$$ \frac{1}{C_t(s^t)}=\beta\sum_{s^{t+1}>s^t}\frac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\cdot\frac{\alpha\big[K_{t+1}(s^t)\big]^{\alpha-1}\big[Z_{t+1}(s^{t+1})H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)}{C_{t+1}(s^{t+1})} \tag{25.11} $$

• Resource constraint:

$$ K_{t+1}(s^t)=\big[K_t(s^{t-1})\big]^{\alpha}\big[Z_t(s^t)H_t(s^t)\big]^{1-\alpha}+(1-\delta)K_t(s^{t-1})-C_t(s^t)\ \text{for }\forall t,\forall s^t \tag{25.12} $$

• Transversality condition:

$$ \lim_{T\to\infty}\beta^T\frac{1}{C_T(s^{T-1})}K_T(s^{T-1})=0 \tag{25.13} $$

Note that (25.10) and (25.11) are for optimality; (25.12) is for feasibility; and (25.13) rules out "crazy" result (i.e. accumulate capital without consuming anything).

25.2 More assumptions on productivity trend

To compute saddle path using shooting algorithm or do the log-linearization in the neighborhood of balanced growth path as we did in section 24.6, we need to impose some assumption on the exogenous function \(Z_t(s^t)\). Two commonly used assumptions are:

• Deterministic trend: \(Z_t(s^t)=(1+g)^t s_t\). The unchanging growth rate (i.e. the deterministic trend) for productivity is \(1+g\); shock \(s_t\) in period \(t\) can only affect period \(t\), and the trend is not affected; from the perspective in period 0, the productivity growth is deterministic except some local fluctuations.
• Stochastic trend: \(Z_t(s^t)=Z_{t-1}(s^{t-1})s_t\). The productivity in period \(t\) is the accumulated effects of all past realized stochastic shocks; from the perspective in period 0, the productivity growth is completely stochastic depending on the sequence of shocks to come.

Notice that if we use the deterministic trend assumption plus an extremely persistent \(s_t\) sequence (i.e. auto-correlation approximately 1 or almost unit root), it is almost the same as using the stochastic trend assumption with i.i.d. \(s_t\) sequence. Saying they are the same means that they give us the same \(Z_t(s^t)\) sequence and the economy would behave the same under those two assumptions. So, we can stick to the deterministic trend assumption \(Z_t(s^t)=(1+g)^t s_t\) hereafter.

25.3 Detrend the variables and rewrite the system of equations

Same as before in the Neoclassical Growth Model, we can detrend the variables by \(c_t(s^t)\equiv\dfrac{C_t(s^t)}{(1+g)^t}\) and \(k_t(s^{t-1})\equiv\dfrac{K_t(s^{t-1})}{(1+g)^t}\). And we can rewrite the system of equations (25.10), (25.11), and (25.12) as

$$ v'(H_t(s^t))c_t(s^t)=(1-\alpha)\big[k_t(s^{t-1})\big]^{\alpha}s_t^{1-\alpha}\big[H_t(s^t)\big]^{-\alpha} \tag{25.14} $$

$$ \frac{1+g}{c_t(s^t)}=\beta\sum_{s^{t+1}>s^t}\frac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\cdot\frac{\alpha\big[k_{t+1}(s^t)\big]^{\alpha-1}\big[s_{t+1}H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)}{c_{t+1}(s^{t+1})} \tag{25.15} $$

$$ (1+g)k_{t+1}(s^t)=\big[k_t(s^{t-1})\big]^{\alpha}\big[s_t H_t(s^t)\big]^{1-\alpha}+(1-\delta)k_t(s^{t-1})-c_t(s^t)\ \text{for }\forall t,\forall s^t \tag{25.16} $$

Since \((1+g)^t\) is deterministic, the solution to the system of equations (25.14), (25.15), and (25.16) is the optimal solution (balanced growth path) for the economy.

25.4 Two cases of the shock process \(\{s_t\}_{t=0}^{\infty}\)

We can obtain the balanced growth path as the intersection point of two locus in the phase diagram and compute the saddle path using shooting algorithm following exactly the same procedure as in section 24.6. Here we will not present that redundant procedure. Instead, we will proceed and discuss two cases of the shock process \(\{s_t\}_{t=0}^{\infty}\) for log-linearization.

25.4.1 Deterministic case

Suppose that \(\ln s_{t+1}=\rho\ln s_t\), \(s_0\) given. Then the system of equations (25.14), (25.15), and (25.16) are all deterministic, and this problem becomes trivial, which is the same as the Neoclassical Growth Model with the exception that here we have included a deterministic sequence of productivity that grows at a different rate than \(g\). Using exactly the same logic as in subsection 24.6.4, we can do the log-linearization to the system of equations (25.14), (25.15), and (25.16). Assume \(v(H)=\dfrac{\gamma\varepsilon}{1+\varepsilon}H^{\frac{1+\varepsilon}{\varepsilon}}\) (\(\varepsilon>0\), \(v'(H)=\gamma H^{1/\varepsilon}\), \(\varepsilon\) is the Frisch elasticity of labor supply). We want to log-linearize the following system of equations

$$ c_t(s^t)\gamma\big[H_t(s^t)\big]^{\frac{1}{\varepsilon}}=(1-\alpha)\big[k_t(s^{t-1})\big]^{\alpha}s_t^{1-\alpha}\big[H_t(s^t)\big]^{-\alpha} \tag{25.17} $$

$$ \frac{1+g}{c_t(s^t)}=\beta\sum_{s^{t+1}>s^t}\frac{\pi_{t+1}(s^{t+1})}{\pi_t(s^t)}\cdot\frac{\alpha\big[k_{t+1}(s^t)\big]^{\alpha-1}\big[s_{t+1}H_{t+1}(s^{t+1})\big]^{1-\alpha}+(1-\delta)}{c_{t+1}(s^{t+1})} \tag{25.18} $$

$$ (1+g)k_{t+1}(s^t)=\big[k_t(s^{t-1})\big]^{\alpha}\big[s_t H_t(s^t)\big]^{1-\alpha}+(1-\delta)k_t(s^{t-1})-c_t(s^t) \tag{25.19} $$

After first-order Taylor approximation:

• Define \(\phi_t=(\hat k_t,\hat c_t,\hat s_t)'\) where \(\hat k_t\equiv\ln k_t-\ln k\), \(\hat c_t\equiv\ln c_t-\ln c\), and \(\hat s_t\equiv\ln s_t-\ln s\) and \(k,c,s\) are steady state values. Then (25.17), (25.18), and (25.19) gives us

$$ \phi_{t+1}=\mathbf{A}\phi_t \tag{25.20} $$

where \(\mathbf{A}\) is a \(3\times3\) matrix and the steady state solution satisfies \(\phi_t=\mathbf{0}\).

• Denote the three eigenvalues of \(\mathbf{A}\) by \(\lambda_1,\lambda_2,\lambda_3\) and the corresponding eigenvectors by \(\mathbf{v}_1,\mathbf{v}_2,\mathbf{v}_3\), so \(\mathbf{A}=\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\mathbf{v}_3\end{bmatrix}\text{diag}(\lambda_1,\lambda_2,\lambda_3)\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\mathbf{v}_3\end{bmatrix}^{-1}\). Given \(k_0,s_0\) (and thus \(\phi_0\)), denote \(\begin{bmatrix}\mu_1&\mu_2&\mu_3\end{bmatrix}\equiv\begin{bmatrix}\mathbf{v}_1&\mathbf{v}_2&\mathbf{v}_3\end{bmatrix}^{-1}\phi_0\), then

$$ \phi_t=\begin{pmatrix}\hat k_t\\ \hat c_t\\ \hat s_t\end{pmatrix}=\mu_1\lambda_1^t\mathbf{v}_1+\mu_2\lambda_2^t\mathbf{v}_2+\mu_3\lambda_3^t\mathbf{v}_3 $$

• Plug in the data \(g=0.005\), \(\beta=0.989\), \(\alpha=0.4\), \(\delta=0.014\), \(\varepsilon=1\), \(\rho=0.95\), \(\gamma=0.00152\Rightarrow H=23\), and we get

$$ \begin{pmatrix}\hat k_{t+1}\\ \hat c_{t+1}\\ \hat s_{t+1}\end{pmatrix}=\begin{bmatrix}1.024 & 0.088 & 0.064\\ -0.013 & 0.989 & 0.023\\ 0 & 0 & 0.95\end{bmatrix}\begin{pmatrix}\hat k_t\\ \hat c_t\\ \hat s_t\end{pmatrix} $$

The three eigenvalues are \(\lambda_1=0.968\), \(\lambda_2=1.044\) and \(\lambda_3=0.95\). It has to be the case that \(\mu_2=0\) if we want to get to the steady state, which is achieved by having appropriate \(\hat c_0\) for given \(\hat k_0\) and \(\hat s_0\) (this also pins down \(\mu_1\) and \(\mu_3\)). Otherwise the solution will explode as the eigenvalue \(\lambda_2\) associated with \(\mu_2\) is greater than one. We also must have that \(\mu_1\mathbf{v}_1\ne0\) and \(\mu_3\mathbf{v}_3\ne0\). Since we already know \(\mathbf{v}_1\) and \(\mathbf{v}_3\), we have that

$$ \hat c_t=0.632\hat k_t+0.186\hat s_t \tag{25.21} $$

Since we know \(\mathbf{A}\) and (25.21), we have that

$$ \hat k_{t+1}=0.968\hat k_t+0.098\hat s_t \tag{25.22} $$

Also, plug (25.21) into the first-order Taylor approximation of (25.17), we get

$$ \hat H_t=-0.160\hat k_t+0.273\hat s_t \tag{25.23} $$

clearly, the labor supply is higher to offset the lower-than-steady-state capital, and labor supply is higher when the positive productivity shock takes place. We can also get

$$ \hat x_t=-0.653\hat k_t+0.2474\hat s_t \tag{25.24} $$

which comes from the law of motion of capital \(x_t=k_{t+1}-(1-\delta)k_t\). Finally, don't forget

$$ \hat s_{t+1}=0.95\hat s_t \tag{25.25} $$

Figure 16 (Transitional Dynamics of the Log-linearized System, paraphrased): suppose we start at \(\hat s_1<0\). Horizontal axis is \(t\). Driven by the productivity \(\hat s_t\), the variables' paths: \(\hat H_t\) (and similarly \(\hat x_t\)) jumps up first and then declines back; \(\hat s_t\) and \(\hat k_t\) rise monotonically from negative values back to \(0\); \(\hat c_t\) falls first and then rises back from some negative value. All variables converge smoothly to the steady state \(0\).

25.4.2 Stochastic case

Here the stochastic doesn't mean changing the trend assumption for \(Z_t(s^t)\). We still assume deterministic trend in \(Z_t(s^t)\), i.e. \(Z_t(s^t)=(1+g)^t s_t\). Instead, the stochastic here means adding stochastic component to the state variable sequence, i.e.

$$ \ln s_{t+1}=\rho\ln s_t+\upsilon_{t+1},\quad s_0\text{ given and }\upsilon_{t+1}\text{ stochastic} $$

Then, (25.21), (25.22), (25.23) and (25.24) don't change. The only thing that changes in the log-linearization calibration is (25.25), which becomes

$$ \hat s_{t+1}=0.95\hat s_t+\upsilon_{t+1} $$

This feature of the log-linearization makes the model very easy to incorporate randomness in the shocks without changing too much from the simple deterministic case. However, the unchanging (25.21), (25.22), (25.23) and (25.24) also is a drawback of the log-linearization because as long as the mean of \(\upsilon_{t+1}\) is zero, (25.21), (25.22), (25.23) and (25.24) won't change, and thus the variance of \(\upsilon_{t+1}\) plays no role in the system at all, which is not that reasonable. But if we include higher order Taylor approximation (not linearization any more), the higher order moment of \(\upsilon_{t+1}\) could enter the model.

25.5 Hodrick-Prescott (HP) Filter

Since we used detrended variables in the model, we should also use detrended data. HP Filter gives us the flexibility of detrending the data to a different level. Here we will be using a slightly different notation as on page 196.

25.5.1 Definition of HP Filter trend

Given \(y_t\) for \(t=1,\ldots,T\) (e.g. quarterly real-GDP or quarterly log real-GDP), the Hodrick-Prescott trend \(\tau_t\) for \(t=1,\ldots,T\) is defined per the minimization of penalty function (the first term is deviation from trend, the second is change in trend):

$$ \min_{\tau_t}\ \sum_{t=1}^{T}(y_t-\tau_t)^2+\lambda\sum_{t=2}^{T-1}\big((\tau_{t+1}-\tau_t)-(\tau_t-\tau_{t-1})\big)^2 $$

where \(\lambda\) is a smoothing parameter that puts weights on the penalty of two parts. The cyclical component (detrended variable) is defined as \(y_t^c=y_t-\tau_t\).

25.5.2 Smoothing parameter \(\lambda\) provides flexibility in detrending

Consider two extreme cases for \(\lambda\):

$$ \lambda=\infty\ \Rightarrow\ \tau_t=\hat y t\ (\text{i.e. no change in trend}) $$

$$ \lambda=0\ \Rightarrow\ \tau_t=y_t\ (\text{trivial trend}) $$

So, if \(\lambda\) is very large, the researcher really doesn't want the trend to change, so he ends up assuming linear trend and using deviation to explain for all the difference between \(y_t\) and the trend. On the contrary, if \(\lambda\) is very small, then the research doesn't mind changing the trend all the time, so the deviation becomes zero and trend is always changing to match the real data \(y_t\).

25.5.3 Solving for the HP Filter trend

\(\tau_t\) appears in four terms. So the f.o.c. for \(\tau_t\), \(t=3,\ldots,T-2\) is

$$ \begin{aligned} &2\lambda\big((\tau_t-\tau_{t-1})-(\tau_{t-1}-\tau_{t-2})\big)-4\lambda\big((\tau_{t+1}-\tau_t)-(\tau_t-\tau_{t-1})\big)\\ &-2(y_t-\tau_t)+2\lambda\big((\tau_{t+2}-\tau_{t+1})-(\tau_{t+1}-\tau_t)\big)=0 \end{aligned} $$

which gives

$$ \begin{aligned} &\lambda\big(2\tau_t-4\tau_{t-1}+2\tau_{t-2}-4\tau_{t+1}+8\tau_t-4\tau_{t-1}+2\tau_{t+2}-4\tau_{t+1}+2\tau_t\big)=2(y_t-\tau_t)\\ \Rightarrow\ &y_t=\tau_t+\lambda\big(\tau_{t-2}-4\tau_{t-1}+6\tau_t-4\tau_{t+1}+\tau_{t+2}\big)\\ \Rightarrow\ &(6\lambda+1)\tau_t=y_t-\lambda\big(\tau_{t-2}-4\tau_{t-1}-4\tau_{t+1}+\tau_{t+2}\big)\\ \Rightarrow\ &\tau_t=\frac{y_t-\lambda\big(\tau_{t-2}-4\tau_{t-1}-4\tau_{t+1}+\tau_{t+2}\big)}{6\lambda+1} \end{aligned} \tag{25.26} $$

the f.o.c. for \(\tau_t\), \(t=1,2,T-1,T\) are slightly different and not important when \(T\) is large.

Or equivalently, we can stack all \(y_t\)'s into a \(T\times1\) vector \(\mathbf{y}^t\) and all \(\tau_t\)'s into a \(T\times1\) vector \(\boldsymbol\tau^t\), and based on (25.26) and f.o.c. for \(t=1,2,T-1,T\), we can write the following matrix equation

$$ \mathbf{y}^t=\mathbf{M}\boldsymbol\tau^t $$

where \(\mathbf{M}\) is a \(T\times T\) invertible matrix. Then, we can solve for \(\boldsymbol\tau^t\) by

$$ \boldsymbol\tau^t=\mathbf{M}^{-1}\mathbf{y}^t $$

and \(\mathbf{y}^t-\boldsymbol\tau^t\) gives us the detrended series.