Large Shocks and the Business Cycle: The Effect of Outlier Adjustments

Abstract

This study examines the impact of outlier-adjusted data on business cycle inferences using coincident indicators of the composite index (CI) in Japan. To estimate the CI and business cycles, this study proposes a Markov switching dynamic factor model incorporating Student’s t-distribution in both the idiosyncratic noise and the factor equation. Furthermore, the model includes a stochastic volatility process to identify whether a large shock is associated with a business cycle. From the empirical analysis, both the factor and the idiosyncratic component have fat-tail error distributions, and the estimated CI and recession probabilities are close to those published by the Economic and Social Research Institute. Compared with the estimated CI using the adjusted data set, the outlier adjustment reduces the depth of the recession. Moreover, the results of the shock decomposition show that the financial crisis in mid-2008 was caused by increase of clustering shocks and large unexpected shocks. In contrast, the Great East Japan Earthquake in 2011 was derived from idiosyncratic noise and did not cause a recession. When analyzing whether to use a sample that includes outliers associated with the business cycle, it is not desirable to use the outlier-adjusted data set.

Sampling Algorithm for Parameters

1.1 Sampling $\gamma _i$

Let $\Delta (\psi _i ) = 1 - \psi _i L$, where L denotes a lag operator, and $\psi _i^* = (1 - \psi _i^2)^{\frac{1}{2}}$. Given c and s, the Eq. (3) can be rewritten as

$$\begin{aligned} \bar{y}_i = \bar{x}_i \gamma _i + \sqrt{\lambda _i} \epsilon _i, \end{aligned}$$

where $\bar{y}_i = \left( \psi _i^* y_{i1}, \Delta (\psi _i)y_{i2}, \ldots , \Delta (\psi _i) y_{iT} \right) '$, $\bar{x}_i = \left( \psi _i^* c_1, \Delta ( \psi _i ) c_{2}, \ldots , \Delta ( \psi _i ) c_T \right) '$, $\lambda _{i} = diag( \lambda _{i1}, \ldots , \lambda _{T} )$, $\epsilon _i = ( \epsilon _{1i}, \ldots , \epsilon _{Ti} )'$. Thus, the full conditional distribution of $\gamma _i$ is as follows:

$$\begin{aligned} \gamma _i | \theta _{- \gamma _i}, \vartheta , y \sim \mathcal {N}( \hat{\mu }_{\gamma _i}, \hat{\sigma }^2_{\gamma _i} ), \end{aligned}$$

where $\hat{\mu }_{\gamma _i} = \hat{\sigma }^2_{\gamma _i}( \sigma ^{-2}_i \bar{x}'_i \lambda _i^{-1}\bar{y} + \sigma _{\gamma 0}^{-2} \gamma _0 )$ and $\hat{\sigma }^2_{\gamma _i} = ( \sigma _i^{-2}\bar{x}_i' \lambda _i^{-1}\bar{x}_i + \sigma _{\gamma 0}^{-2} )^{-1}$.

1.2 Sampling $\psi _i$ and $\sigma ^2_i$

For sampling parameter $\psi _i$, we employ the Metropolis-Hastings algorithm proposed by Chib and Greenberg (1995). Let $\hat{y}_i = (z_{i2}, \ldots , z_{iT})'$ and $\hat{x}_i = (z_{i1}, \ldots , z_{i,T-1})'$. The full conditional distribution of $\psi _i$ is given by

$$\begin{aligned} \pi ( \psi _i | \theta _{- \psi _i}, \vartheta , y ) \propto \psi _i^* \exp \left[ - \frac{(\psi _i^* z_{i1})^2}{2 \lambda _{i1} \sigma ^2_i} - \frac{(\psi _i - \psi _{0i})^2}{2 \sigma ^2_{\psi _{i0}}} - \frac{1}{2 \sigma _i^2}\sum _{t=2}^T \frac{(z_{it} - \psi _i z_{i, t-1} )^2}{\lambda _{it}} \right] . \end{aligned}$$

It is difficult to directly draw the parameter. We generate the value from the following candidate distribution:

$$\begin{aligned} \psi _i | \theta _{- \psi _i}, \vartheta , y \sim \mathcal {TN}_{|\psi _i| < 1}( \hat{\mu }_{\psi _i}, \hat{\sigma }^2_{\psi _i} ), \end{aligned}$$

where

$$\begin{aligned} \hat{\mu }_{\psi _i} = \hat{\sigma }^{2}_{\psi _i} \left( \sum _{t=2}^T \frac{z_{it} z_{i, t-1}}{\lambda _{it}} + \frac{\psi _{0i}}{\sigma ^2_{\psi _{0i}}} \right) , \quad \text{ and }\quad \hat{\sigma }^2_{\psi _i} = \sigma ^2_i \left( \sum _{t=2}^T \frac{z_{i, t-1}^2}{\lambda _{it}} + \frac{1}{\sigma ^2_{\psi _{0i}}} \right) ^{-1} . \end{aligned}$$

Let $\psi _i^{old}$ be the previous value. Then, we draw a candidate $\psi ^{new}_i$ from $\mathcal {N}( \hat{\mu }_{\psi _i}, \hat{\sigma }^2_{\psi _i} )$, truncated on $(-1, 1)$, in order to satisfy the stationary condition, and accept it with probability

$$\begin{aligned} \alpha ( \psi _i^{old}, \psi ^{new}_i ) = \min \left[ \frac{\pi (\psi ^{new}_i | \theta _{- \psi _i}, \vartheta , y)}{\pi (\psi _i^{old}| \theta _{- \psi _i}, \vartheta , y)} ,\ 1 \right] . \end{aligned}$$

Next, the full conditional distribution of $\sigma ^2_i$ is as follows:

$$\begin{aligned} \sigma ^2_i | \theta _{- \sigma ^2_i}, \vartheta , y \sim \mathcal {IG}\left( \frac{\hat{\tau }_i}{2},\ \frac{\hat{\delta }_{i}}{2} \right) , \end{aligned}$$

where $\hat{\tau }_i = \tau _{0i} + T$ and $\hat{\lambda }_{i} = \frac{( \psi _i^* z_{i1} )^2}{\lambda _{i1}} + \sum _{t=2}^T \frac{(z_{it} - \psi _i z_{i, t-1})^2}{\lambda _{it}} + \delta _{0i}$.

1.3 Sampling c

We show the state space representation of the model for drawing the latent variables c. Let $\Delta y_{it} = y_{it} - \psi _i y_{i, t-1}$, for $i = 1, \ldots , n$, $\Delta y_t = ( \Delta y_{1t}, \ldots , \Delta y_{nt} )'$, $\Delta (\phi ) = 1 - \phi (L_q)$ and $\sigma ^2_{ht} = \exp ( h_t )$. Then, the model can be represented as

$$\begin{aligned} \Delta y_t&= {} \Gamma \alpha _t + \epsilon _t, \quad \epsilon _t \sim \mathcal {N}( 0_{n}, H_t ), \end{aligned}$$

(11)

$$\begin{aligned} \alpha _{t}&= {} m_{t} + T \alpha _{t - 1} + G_t \eta _t, \quad \eta _t \sim \mathcal {N}( 0_{q+1}, I_{q+1} ), \end{aligned}$$

(12)

where $I_{q+1}$ denotes $(q+1) \times (q+1)$ unit matrix, and $\Gamma$, $\alpha _t$, $H_t$, $m_t$, T, and $G_t$ are given by

$$\begin{aligned} {}&\Gamma = \left[ \begin{array}{ccc} \gamma _1 &{} - \gamma _1 \psi _1 &{} 0'_{q-1}\\ \vdots &{} \vdots &{} \vdots \\ \gamma _n &{} - \gamma _n \psi _n &{} 0'_{q-1} \end{array} \right] , \ \alpha _t = \left[ \begin{array}{c} c_t \\ \vdots \\ c_{t-q} \end{array} \right] , \ H_t = diag( \lambda _{1t} \sigma ^2_1, \ldots , \lambda _{nt}\sigma ^2_n) , \\ {}&m_t = \left[ \begin{array}{c} \Delta (\phi ) \mu _{t} \\ 0_{q} \end{array} \right] ,\ T = \left[ \begin{array}{cc} {\phi '}&{} \\ I_{q - 1} &{} 0_{q - 1} \end{array} \right] ,\ G_t = \left[ \begin{array}{cc} \sqrt{\omega _{t}} \sigma _{ht} &{} 0_{q}'\\ 0_{q} &{} 0_{q \times q} \end{array} \right] . \end{aligned}$$

Since Eqs. (11) and (12) constitute the linear Gaussian state space mode, we can sample c using the efficient simulation smoother (Durbin and Koopman 2002).

1.4 Sampling $\mu$

Let $\tilde{x}_t = ( 1 - s_t, s_t )$. Given c and s, the equation of (4) is rewritten as:

$$\begin{aligned} \tilde{y} = \tilde{x} \mu + \omega ^{\frac{1}{2}} \eta , \end{aligned}$$

where $\tilde{y} = ( \Delta (\phi )c_1, \Delta (\phi )c_2, \ldots , \Delta (\phi )c_T )'$, $\tilde{x} = ( \Delta (\phi ) x_1', \Delta (\phi )x_2', \ldots , \Delta (\phi )x_T' )'$, $\bar{\omega } = diag( \omega _1, \ldots , \omega _T )$, and $\eta = ( \eta _1, \ldots , \eta _T )'$. The full conditional distribution of $\mu$ can be obtained as

$$\begin{aligned} \mu | \theta _{- \mu }, \vartheta , y \sim \mathcal {N}( \hat{\mu }, \hat{\Sigma }_{\mu } )I[ \mu ^{(0)} < \mu ^{(1)} ], \end{aligned}$$

where $\hat{\mu } = \hat{\Sigma }_{\mu }( \tilde{x}'\bar{\omega }^{-1}\tilde{y} + \Sigma _{\mu 0}^{-1}\mu _0 )$ and $\hat{\Sigma }_{\mu } = ( \tilde{x}'\bar{\omega }^{-1}\tilde{x} + \Sigma _{\mu 0}^{-1} )^{-1}$. For sampling $\mu$, if it does not satisfy the inequality $\mu ^{(0)} < \mu ^{(1)}$, the generated values are rejected and then sampled again.

1.5 Sampling $\phi$

Let $\dot{c}_t = c_t - \mu _t$ and $\tilde{c}_{t-1} = ( \dot{c}_{t-1}, \ldots , \dot{c}_{t-q} )$. Then, the full conditional distribution of $\phi$ is given by

$$\begin{aligned} \phi | \theta _{- \phi }, \vartheta , y \sim \mathcal {N}( \hat{\mu }_{\phi }, \hat{\Sigma }_{\phi } )I[S(\phi )], \end{aligned}$$

where

$$\begin{aligned} \hat{\mu }_{\phi } = \hat{\Sigma }_{\phi } \left( \sum _{t=q}^{T} \frac{\tilde{c}'_{t-1} \dot{c}_t }{\omega _{t}\sigma ^{2}_{ht}} + \Sigma _{\phi _0}^{-1}\phi _0 \right) ,\quad \text{ and }\quad \hat{\Sigma }_{\phi } = \left( \sum _{t=q}^{T} \frac{\tilde{c}_{t-1} \tilde{c}_{t-1}'}{\omega _{t} \sigma ^{2}_{ht}} + \Sigma _{\phi _0}^{-1} \right) ^{-1} \end{aligned}$$

Let $\phi ^{old}$ be the previous value. Then, we draw a candidate $\phi ^{new}$ from $\mathcal {N}( \hat{\mu }_{\phi }, \hat{\sigma }^2_{\phi } )$, truncated on $S(\phi )$, in order to satisfy the stationary condition, and accept it with probability

$$\begin{aligned} \alpha ( \phi ^{old}, \phi ^{new}) = \min \left[ \frac{\pi (\phi ^{new} | \theta _{- \phi }, \vartheta , y)}{\pi (\phi ^{old}| \theta _{- \phi }, \vartheta , y)} ,\ 1 \right] . \end{aligned}$$

1.6 Sampling p

Following Watanabe (2014), we employ the acceptance rejection algorithm for sampling $p_{0}$ and $p_{1}$. The full conditional distribution of p is following as:

$$\begin{aligned} \pi ( p | \theta _{- p}, \vartheta , y )&\propto {} \frac{(1 - p_{0})^{s_1}( 1- p_{1})^{1 - s_1}}{2 - p_{0} - p_{1}} \\&\quad \times \, p_{0}^{\iota _{00} + n_{00}} (1 - p_{0})^{\iota _{01} + n_{01}} p_{1}^{\iota _{11} + n_{11}} (1 - p_{1})^{\iota _{10} + n_{10}} \\ {}&= {} g_{p}( p ) \times p_{0}^{\iota _{00} + n_{00}} (1 - p_{0})^{\iota _{01} + n_{01}} p_{1}^{\iota _{11} + n_{11}} (1 - p_{1})^{\iota _{10} + n_{10}}, \end{aligned}$$

where $n_{ij}$ means the number of transitions from state i to j. We sample a proposed value $p_{0}^{new}$ and $p_{1}^{new}$ from the following independent beta distribution:

$$\begin{aligned} p_{0}^{new} \sim Beta( \iota _{00} + n_{00}, \iota _{01} + n_{01} ) ,\quad p_{1}^{new} \sim Beta( \iota _{11} + n_{11}, \iota _{10} + n_{10} ) . \end{aligned}$$

Since $0< g_{p}( p^{new} ) <1$, we employ the MH step. Finally, we accept the proposed values with probability

$$\begin{aligned} \alpha ( p^{old}, p^{new} ) = \min \left[ \frac{g_{p}( p^{new} )}{g_{p}( p^{old} )}, 1 \right] . \end{aligned}$$

1.7 Sampling s

For sampling s, we employ the multi-move sampler (Kim and Nelson 1998, 1999). The joint conditional distribution of s is as follows:

$$\begin{aligned} f( s | \theta , \vartheta _{- s}, y ) = f( s_T | \theta , \vartheta _{- s_{T}}, y ) \prod _{t=1}^{T-1} f(s_t | s_{t+1}, \theta , \vartheta _{- s_t}, y), \end{aligned}$$

(13)

First, we sample $s_T$, which is the first term on the right-hand side of Eq. (13). Given $s_T$ sampled from $s_{T-1}$, we can proceed backwards in time. Then, $f(s_{t} | s_{t+1}, \theta , \vartheta _{- s_t}, y)$ includes the following:

$$\begin{aligned} f( s_{t} | s_{t+1}, \theta , \vartheta _{- s_t}, y) \propto f( s_{t+1} | s_{t} ) f(s_{t} | \theta , \vartheta _{- s_t}, y), \end{aligned}$$

where $f( s_{t+1} | s_{t} )$ means the transition probability. Next, $f(s_{t} | \theta , \vartheta _{- s_t}, y)$, for $t=1, \ldots , T$, is calculated using Hamilton (1989) filter. $s_{T}$ is sampled from $f(s_{T} | \theta , \vartheta _{- s_T}, y)$, and $s_{t}$ is sampled using $s_{T}$ and Eq. (14) backward in time.

1.8 Sampling $\lambda _{it}$ and $\nu _i$

Given c, s and $\omega$, the full conditional distribution of $\lambda _{it}$ is as follows:

$$\begin{aligned} \lambda _{it} | \theta , \vartheta _{- \lambda _{it}}, y \sim \mathcal {IG}\left( \frac{\hat{a}_{it}}{2}, \frac{\hat{b}_{it}}{2}\right) . \end{aligned}$$

where $\hat{a}_{it} = \nu _{i} + 1$ and $\hat{b}_{it} = \nu _i + \sigma _i^{-2} ( z_{it} - \psi _i z_{i, t-1})^2$.

Finally, the full conditional distribution of $\nu _i$ is given by

$$\begin{aligned} \pi ( \nu _i | \lambda _i )&\propto {} \nu ^{A_{i0} - 1} \exp ( - B_{i0} \nu _i ) \prod _{t=1}^{T} \frac{\frac{\nu _i}{2}^{\frac{\nu _i}{2}}}{\Gamma \left( \frac{\nu _i}{2} \right) } \lambda _{it}^{-\frac{\nu _i}{2} + 1} \exp \left( - \frac{\nu _i}{2\lambda _{it}} \right) , \end{aligned}$$

To sampling the degrees of freedom parameter, we employ the AR-MH algorithm extended by Watanabe (2001). The AR-MH algorithm was proposed by Tierney (1994) (see also Chib and Greenberg (1995) for details). This algorithm samples the parameter using the AR and MH step. Suppose there is a candidate function $h(\nu _i)$ which can be directly sampled, and $f(\nu _i)$, defined as the target distribution. Then, the AR step proceeds as follows:

• 1. Sample the candidate $\nu _i$ from $h(\nu _i)$ and u from the uniform distribution $\mathcal {U}[0, 1]$.

• 2. If $u \le \frac{f( \nu _i )}{h( \nu _i )}$, return $\nu _i^{new} = \nu _i$. Else, go to 1.

This step is repeated until the candidate draw is accepted. In this study, we utilize a normal distribution as the candidate function. Let $p^* ( \nu _i )$ denote $p( \nu _i | \omega )$ with the constant subtracted, and the log of $p^* ( \nu _i )$ is given by

$$\begin{aligned} \ln p^*( \nu _i ) = \frac{T}{2}\nu _i \ln \left( \frac{\nu _i}{2}\right) - T \ln \Gamma \left( \frac{\nu _i}{2}\right) - J_i \nu _i + (A_{i0} - 1)\ln ( \nu _i ), \end{aligned}$$

(14)

where

$$\begin{aligned} J_i = \frac{1}{2} \sum _{t =1}^{T} \left\{ \ln ( \lambda _{it} ) + \frac{1}{\lambda _{it}} \right\} + B_{i0}. \end{aligned}$$

We apply the second-order Taylor expansion around $\nu _i = \nu _i^*$ to (14), which yields

$$\begin{aligned} \ln p^*( \nu _i )&\approx {} \ln p^*( \nu _i^* ) + C_i' ( \nu _i - \nu _i^* ) + \frac{C_i''}{2}( \nu _i - \nu _i^* )^2\\&= {} h(\nu _i), \end{aligned}$$

where

$$\begin{aligned} C_i'&= {} \left. \frac{d \ln p^*(\nu _i)}{d\nu _i} \right| _{\nu _i = \nu _i^*} = \frac{T}{2} \left\{ \ln \left( \frac{\nu _i}{2}\right) + 1 - \psi \left( \frac{\nu _i^*}{2}\right) \right\} - J_i + \frac{A_{i0} - 1}{\nu _i^*},\\ C_i''&= {} \left. \frac{d^2 \ln p^*(\nu _i)}{d^2 \nu _f} \right| _{\nu _i = \nu _i^*} = \frac{T }{2} \left\{ \frac{1}{\nu _i} - \frac{1}{2} \psi ' \left( \frac{\nu _i^*}{2}\right) \right\} - \frac{A_{i0} - 1}{\nu _i^{*2}}, \end{aligned}$$

with $\psi ( \nu _i )$ and $\psi '( \nu _i )$ denoting a digamma function defined by $\psi ( \nu _i ) = \frac{d \ln \Gamma ( \nu _i )}{d \nu _i}$, and a trigamma function defined by $\psi '( \nu _i ) = \frac{d \psi ( \nu _i )}{d \nu _i}$. Then, the normalized version of $h(\nu _i )$ has a normal density with mean $\nu _i^* - \frac{C_i'}{C_i''}$ and variance $- \frac{1}{C_i''}$.

Next, let the previous sampled value of $\nu _i$ be $\bar{\nu _i}$. Then, the MH step proceeds as follows:

• 1. Calculate the acceptance probability $q(\nu _i)$

  • If $p^*( \bar{\nu _i} ) < \kappa _{\nu _i} h( \bar{\nu _i} )$, then set $q(\nu _i) = 1$;

  • If $p^*( \bar{\nu _i} ) \ge \kappa _{\nu _i} h( \bar{\nu _i} )$ and $p^*( \nu _i^{new} ) < \kappa _{\nu _i} h( \nu _i^{new} )$, then set $\displaystyle q(\nu _i) = \frac{\kappa _{\nu _i} h(\bar{\nu _i} )}{p^*( \bar{\nu _i} )}$;

  • If $p^*( \bar{\nu _i} ) \ge \kappa _{\nu _i} h( \bar{\nu _i} )$ and $p^*( \nu _i^{new} ) \ge \kappa _{\nu _i} h( \nu _i^{new} )$, then set $q(\nu _i) = \min \left[ \frac{p^*( \nu _i^{new} )h(\bar{\nu _i})}{p^*( \bar{\nu _i} )h(\nu _i^{new})}, 1 \right]$;

• 2. Sample a value u from the uniform distribution $\mathcal {U}[0, 1]$.

• 3. If $u \le q(\nu _i)$, return $\nu _i = \nu _i^{new}$. Else, return $\nu _i = \bar{\nu _i}$.

In this step, the candidate value is accepted with probability $q(\nu _i)$, and otherwise rejected. If a draw is rejected, the previously sampled value is sampled again. In the empirical analysis, we set $\kappa _{\nu _i} = 1$.

1.9 Sampling $\omega _t$ and $\nu _f$

Since the full conditional distribution of $\omega _t$ are mutually independent, it is straightforward to sample $\omega _t$. Thus, the full conditional distribution of $\omega _t$ are given as

$$\begin{aligned} \left. \frac{\eta _t^2 / \exp (h_t) + \nu _f - 2}{\omega _t}\right| \theta , \vartheta _{- \omega _t},y \sim \chi ^2 ( \nu _f + 1 ),\quad t =1, \ldots , T. \end{aligned}$$

Finally, we sample $\nu _f$ using the AR-MH algorithm, as in sampling $\nu _i$.

1.10 Sampling $\beta$ and $\xi ^2$

Given h, the full conditional distribution of $\beta$ is given by

$$\begin{aligned} \pi ( \beta | \theta , \vartheta _{- \beta }, y ) \propto (1 - \beta ^2) \exp \left[ - \frac{(1 - \beta ^2)h_{1}^2}{2\xi ^2} - \frac{(\beta - \mu _{\beta _0})^2}{2\sigma ^2_{\beta _0}} \right] \prod _{t=2}^{T} \exp \left[ - \frac{(h_{t} - \beta h_{t-1})^2}{2 \xi ^2 }\right] . \end{aligned}$$

It is difficult to directly draw the parameter. We generate the value from the following candidate distribution

$$\begin{aligned} \beta | \theta _{- \beta }, \vartheta , y \sim \mathcal {TN}_{|\beta | < 1}( \hat{\mu }_{\beta }, \hat{\sigma }^2_{\beta } ), \end{aligned}$$

where

$$\begin{aligned} \hat{\mu }_{\beta } = \hat{\sigma }^{-2}_{\beta } \left( \xi ^{-2} \sum _{t=2}^T h_{t} h_{t-1} + \sigma ^{-2}_{\beta _0}\mu _{\beta _0} \right) , \quad \text{ and }\quad \hat{\sigma }^2_{\beta } = \left( \xi ^{-2}\sum _{t=2}^T h_{t-1}^2 + \sigma ^{-2}_{\beta _0} \right) ^{-1}. \end{aligned}$$

Let $\beta ^{old}$ be the previous value. Then, we draw a candidate $\beta ^{new}$ from $\mathcal {N}( \hat{\mu }_{\beta }, \hat{\sigma }^2_{\beta } )$, truncated on $(-1, 1)$, in order to satisfy the stationary condition, and accept it with probability

$$\begin{aligned} \alpha ( \beta ^{old}, \beta ^{new} ) = \min \left[ \frac{\pi (\beta ^{new} | \theta _{- \beta }, \vartheta , y)}{\pi (\beta ^{old}| \theta _{- \beta }, \vartheta , y)},\ 1 \right] . \end{aligned}$$

Finally, the full conditional distribution of $\xi ^2$ is as follows:

$$\begin{aligned} \xi ^2 | \theta _{- \xi ^2}, \vartheta , y \sim \mathcal {IG}\left( \frac{\hat{\tau }_h}{2},\ \frac{\hat{\delta }_{h}}{2} \right) , \end{aligned}$$

where $\hat{\tau }_h = \tau _{0h} + T$ and $\hat{\delta }_{h} = (1 - \beta ^2) h_{1}^2 + \sum _{t=2}^T (h_{t} - \beta h_{t-1})^2 + \delta _{0h}$.

1.11 Sampling h

For sampling the latent variable h, we employ the multi-move sampler extended by Watanabe and Omori (2004). First, we divide h into $K + 1$ blocks, $(h_{k_{l-1}}, \ldots , h_{k_l})$ for $l = 1, \ldots , K+1$ with $k_0 = 0$ and $k_{K+1} = T$. The K knots ($k_{1}, \ldots , k_{K}$) are randomly drawn from

$$\begin{aligned} k_l = int \left[ T \times \frac{ l \times U_{l}}{K+2} \right] , \end{aligned}$$

where $U_l$ are independent uniforms in [0, 1] and “int” means the integer part. Following Pitt and Shephard (1999), we draw the error term ($\zeta _{k_{j-1}},\ldots , \zeta _{k_j - 1}$) instead of $(h_{k_{i-1} + 1}, \ldots , h_{k_i})$ from their full conditional distributions,

$$\begin{aligned} \pi \left( \zeta _{t-1}, \ldots , \zeta _{t+k-1} | h_{t-1}, h_{t+ k+1}, \theta , \vartheta _{- h_{k_{i-1} + 1}, \ldots , h_{k_i}} \right). \end{aligned}$$

(15)

Next, let $k_{l - 1} = t - 1$, $k_{l} = t + k$, and $h(k) = \{ h_j \}_{j=t-1}^{t+k}$. Then, we construct a candidate distribution in order to sampling the error vector. The log of the posterior density (15) is described as follows:

$$\begin{aligned} {}&\log \pi ( \zeta _{t - 1}, \ldots , \zeta _{t+k} | h_{t-1}, h_{t+k+1}, \theta , \vartheta _{- h(k)}, y)\\&\quad = \text{ const. } + \frac{1}{2\xi ^2}\sum _{j = t}^{t+k} \zeta _{j}^2 - \frac{1}{2} \sum _{j=t}^{t+k} \left\{ h_j + \frac{e^{*2}_{j} }{\omega _j} \exp ( - h_j )\right\} - \frac{1}{2 \xi ^2} (h_{t+k +1} - \beta h_{t+k})^2, \end{aligned}$$

where $e^*_j$ means the residual of Eq. (4). Then, we evaluate this logarithm of the posterior density using the Taylor expansion of the log-likelihood,

$$\begin{aligned} l (h_{j}) = - \frac{1}{2}h_j - \frac{1}{2} \frac{e^{*2}_{j}}{\omega _j} \exp ( - h_j ), \end{aligned}$$

around the mode $\hat{h}_j$, as follows:

$$\begin{aligned} {}&\log \pi (\zeta _{t-1}, \ldots , \zeta _{t+k - 1} | \theta , \vartheta _{- h(k)}, y)\\ {}&\quad \approx \text{ const. } + \sum _{j=t}^{t+k} \left\{ l (\hat{h}_j) + l' (\hat{h}_j) ( h_j - \hat{h}_j ) + \frac{1}{2}l''(\hat{h}_j)(h_j - \hat{h}_j)^2 \right\} \\ {}&\quad \quad - \frac{1}{2} \sum _{j=t}^{t+k} \left\{ h_j + \frac{e^{*2}_j}{\omega _j} \exp ( - h_j )\right\} - \frac{1}{2 \xi ^2} (h_{t+k +1} - \beta h_{t+k})^2\\ {}&\quad \equiv \log g_{h}(\zeta _{t-1}, \ldots , \zeta _{t+k-1}), \end{aligned}$$

where

$$\begin{aligned} l' (\hat{h}_j)&\equiv {} \frac{\partial l(\hat{h}_j)}{\partial h_{j}} = - \frac{1}{2} + \frac{1}{2}\frac{e_{j}^{*2}}{\omega _j} \exp ( - \hat{h}_j),\\ l'' (\hat{h}_j)&\equiv {} \frac{\partial ^2 l(\hat{h}_j)}{\partial ^2 h_{j}} = - \frac{1}{2}\frac{e_{j}^{*2}}{\omega _j}\exp ( - \hat{h}_j). \\ \end{aligned}$$

We sample the error term from the posterior distribution with the simulation smoother. Moreover, we employ the AR-MH algorithm. Finally, in order to select the posterior mode $\hat{h}_j$, we apply the Kalman filter and disturbance smoother (Watanabe and Omori 2004).