Why are Bayesian trend-cycle decompositions of US real GDP so different?

Abstract

This paper provides an underlying reason for why recent Bayesian trend-cycle decompositions of US real GDP differ despite using identical unobserved components models. We stress that a pitfall in estimating unobserved components models accounts for the divergence in the empirical conclusions. Our results also show that the decline in the long-run growth rate of real GDP has been slow and gradual rather than abrupt during the post-World War II period.

Appendix: Posterior simulation for latent state variables

To recover the Markovian property of the unobserved state variables, we cast the UC model in Eqs. (1), (2), and (3) as a state-space representation as follows:

$$\begin{aligned} y_t= & {} H'\beta _{t}, \end{aligned}$$

(A.1)

$$\begin{aligned} \beta _t= & {} \delta _{t} + F\beta _{t-1} + G\epsilon _t, \end{aligned}$$

(A.2)

where

$$\begin{aligned} H= & {} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}, \ \beta _t = \begin{bmatrix} x_t \\ z_t \\ z_{t-1} \end{bmatrix}, \ \delta _{t} = \begin{bmatrix} \mu _{s_t} \\ 0 \\ 0 \end{bmatrix} , \ F = \begin{bmatrix} 1&0&0 \\ 0&\phi _1&\phi _2 \\ 0&1&0 \end{bmatrix},\\ G= & {} \begin{bmatrix} 1&0 \\ 0&1 \\ 0&0\end{bmatrix}, \ \epsilon _t = \begin{bmatrix} u_{t} \\ e_{t} \end{bmatrix} \end{aligned}$$

In the above representation, the transition densities associated with Eq. (A.2) are degenerated because of the reduced rank covariance matrix of $G\epsilon _t$. We employ the disturbance smoothing sampler by Durbin and Koopman (2002) to circumvent the degeneracy problem and to sample the continuous latent variables $x_{1:T}$ and $z_{1:T}$ conditional on $s_{1:T}$. Since in this “Appendix” we focus on quantifying the uncertainty in break points, we do not provide detailed expositions for the smoothing algorithm here.

Unlike Wang and Zivot (2000), we suggest a simple alternative method of estimating the break date and constructing a corresponding Bayesian credible set using the smoothing algorithm of Hamilton (1989). The proposed algorithm is a multi-move sampler that samples the whole sequence of $s_t$ through one-time forward–backward recursion. Consequently, the proposed method has a good mixing property. The target distribution in sampling $s_{1:T}$ is given by:

$$\begin{aligned} \begin{aligned} f(s_{1:T}\vert \beta _{1:T}, y_{1:T})&= f(s_{1:T}\vert \beta _{1:T}) \\&= f(s_T \vert \beta _{T}) \prod _{t=1}^{T-1} f(s_t \vert s_{t+1:T},\beta _{1:T}), \end{aligned} \end{aligned}$$

(A.3)

where $s_{t+1:T}= \{s_{t+1}, s_{t+2}, \ldots ,s_T\}$. For the sake of notational simplicity, the model parameter $\theta$ is abbreviated throughout the derivation. The conditional density at time t is further decomposed as follows:

$$\begin{aligned} \begin{aligned} f(s_t \vert s_{t+1:T},\beta _{1:T}),&= f(s_t \vert s_{t+1:T},\beta _{1:t}, \beta _{t+1:T}), \\&= {f(s_t,\beta _{t+1:T} \vert s_{t+1:T},\beta _{1:t}) \over f(\beta _{t+1:T} \vert s_{t+1:T},\beta _{1:t})}, \\&\propto f(s_t \vert s_{t+1:T},\beta _{1:t}) \ f(\beta _{t+1:T} \vert s_{t:T},\beta _{1:t}), \\&\propto f(s_{t+1}\vert s_t) f(s_t\vert \beta _{1:t}). \end{aligned} \end{aligned}$$

(A.4)

In the last line, we drop the density of $\beta _k$ for $k = t+1,t+2,\ldots ,T$ because it is irrelevant conditional on $\beta _{t}$. The evaluation of the filtering density $f(s_t\vert \beta _{1:t})$ for $t = 1,\ldots ,T$ is performed by Hamilton (1989)’s filtering algorithm. In the prediction step, given $f(s_{t-1}\vert \beta _{1:t-1})$, the joint density for $s_t$ and $s_{t-1}$ is obtained by $f(s_t,s_{t-1}\vert \beta _{1:t-1}) = f(s_t\vert s_{t-1}) f(s_{t-1}\vert \beta _{1:t-1})$, where $f(s_t\vert s_{t-1})$ is the transition probability in Eq. (12) or (13). In the updating step, we compute the density of $s_t$ conditional on $\beta _{1:t}$ as follows:

$$\begin{aligned} f(s_t \vert \beta _{1:t}) = {f(\beta _t \vert s_t) f(s_t \vert \beta _{1:t-1})\over f(\beta _t \vert \beta _{1:t-1})}, \end{aligned}$$

(A.5)

where $f(s_t \vert \beta _{1:t-1}) = \sum _{s_{t-1}} f(s_t,s_{t-1}\vert \beta _{1:t-1})$. We obtain $f(\beta _{t}\vert \beta _{1:t-1})$ by integrating out $s_t$ as $f(\beta _{t}\vert \beta _{1:t-1}) = \sum _{s_t} f(\beta _t \vert s_t, \beta _{1:t-1})f(s_t \vert \beta _{1:t-1})$. Therefore, it is simple to obtain $f(s_t\vert \beta _{1:t})$ for $t=1,\ldots ,T$ by iterating the prediction and the updating steps. Based on Eq. (A.4), $s_t$ is drawn from the following posterior distribution:

$$\begin{aligned} pr(s_t = i \vert s_{t+1:T}, \beta _{1:T}) = {f(s_{t+1}\vert s_t = i) f(s_t = i\vert \beta _{1:t}) \over \sum _{s_t} f(s_{t+1}\vert s_t) f(s_t\vert \beta _{1:t})}, \end{aligned}$$

(A.6)

moving backward in time for $t = T,T-1,\ldots ,1$ and for $i=1,2$.