On the Effect of Bank of Japan’s Outright Purchase on the JGB Yield Curve

Abstract

This paper examines an impact of Bank of Japan (BOJ)’s outright purchase on the Japanese government bond (JGB) yield curve. Particularly, we develop a simple state space model, which incorporates new factors regarding the BOJ’s announcement for its outright purchase and the current market outstanding with standard level and spread factors. Based on the model with a filtering method, we also implement an empirical analysis with time series of the BOJ’s announcement records during 2014/10/22–2017/8/3 in the quantitative–qualitative easing period to estimate the sensitivities of interest rates against the changes in the market expectation for the net supply with each sector of JGB. We expect the current work provides a basis for considering quantitative effects on the term structure by BOJ’s policy changes such as termination or significant reduction of the BOJ’s outright purchase. For instance, our scenario analysis shows substantial increase in the 30 year yield with widening of 20–30 year spread.

Appendix: Derivation of Equation (12)

Recall that $x_{t,1}$ and $x_{t,2}$ follows stochastic processes (10) and (11), respectively. Then,

$$\begin{aligned} \int _t^T x_{s,1} ds = x_{t,1} \tau + \frac{\lambda }{2}\tau ^2 + \sigma _1 \int _0^\tau (\tau -u) dW_{u,1}^{\mathbb {Q}}, \end{aligned}$$

(16)

$$\begin{aligned} \int _t^T x_{s,2} ds = \frac{1-e^{-\kappa \tau }}{\kappa }x_{t,2} + \frac{\sigma _2}{\kappa } \int _0^\tau (1-e^{\kappa (u-\tau )}) dW_{u,2}^{\mathbb {Q}}. \end{aligned}$$

(17)

The above two equations derive the following equation with the assumption that the short rate $r_t$ is the sum of two variables, i.e. Eq. (9):

$$\begin{aligned} {\begin{matrix} &{}P(t,T) = {{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ e^{-\int _t^T r_s ds} | \mathcal {F}_t \right] = {{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ e^{-\int _t^T (x_{s,1} + x_{s,2}) ds} | \mathcal {F}_t \right] \\ &{}\quad = \exp \left\{ -x_{t,1} \tau - \frac{\lambda }{2}\tau ^2 - \frac{1-e^{-\kappa \tau }}{\kappa }x_{t,2} \right\} \times \\ &{}{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1 \int _0^\tau (\tau -u) dW_{u,1}^{\mathbb {Q}} - \frac{\sigma _2}{\kappa } \int _0^\tau (1-e^{\kappa (u-\tau )}) dW_{u,2}^{\mathbb {Q}} \right\} \right] \end{matrix}} \end{aligned}$$

(18)

Remark that

$$\begin{aligned} \int _0^\tau (\tau -u) dW_{u,1}^{\mathbb {Q}} = \int _0^\tau (\tau -u) \sqrt{1-\rho ^2} d\tilde{W}_{u,1}^{\mathbb {Q}} + \int _0^\tau (\tau -u) \rho dW_{u,2}^{\mathbb {Q}} \end{aligned}$$

(19)

where $\tilde{W}_{t,1}^{\mathbb {Q}}$ is a one dimensional Brownian motion independent of $W_{t,2}^{\mathbb {Q}}$, i.e. $\tilde{W}_{t,1}^{\mathbb {Q}} \mathop {\perp \!\!\!\!\perp }W_{t,2}^{\mathbb {Q}}$, under a risk-neutral measure $\mathbb {Q}$. Then,

$$\begin{aligned} {\begin{matrix} &{}{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1 \int _0^\tau (\tau -u) dW_{u,1}^{\mathbb {Q}} - \frac{\sigma _2}{\kappa } \int _0^\tau (1-e^{\kappa (u-\tau )}) dW_{u,2}^{\mathbb {Q}} \right\} \right] \\ &{}\quad = {{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1\int _0^\tau (\tau -u) \sqrt{1-\rho ^2} d\tilde{W}_{u,1}^{\mathbb {Q}} \right\} \right] \times \\ &{}{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1 \int _0^\tau (\tau -u) \rho dW_{u,2}^{\mathbb {Q}} - \frac{\sigma _2}{\kappa } \int _0^\tau (1-e^{\kappa (u-\tau )}) dW_{u,2}^{\mathbb {Q}} \right\} \right] \end{matrix}} \end{aligned}$$

(20)

Since Var$[\sigma _1\int _0^\tau (\tau -u) \sqrt{1-\rho ^2} d\tilde{W}_{u,1}^{\mathbb {Q}}] = \sigma _1^2(1-\rho ^2)\tau ^3/3$,

$$\begin{aligned}&{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1\int _0^\tau (\tau -u) \sqrt{1-\rho ^2} d\tilde{W}_{u,1}^{\mathbb {Q}} \right\} \right] = \exp \left\{ \frac{\sigma _1^2(1-\rho ^2)}{6}\tau ^3 \right\} \end{aligned}$$

(21)

$$\begin{aligned}&{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ -\sigma _1 \int _0^\tau (\tau -u) \rho dW_{u,2}^{\mathbb {Q}} - \frac{\sigma _2}{\kappa } \int _0^\tau (1-e^{\kappa (u-\tau )}) dW_{u,2}^{\mathbb {Q}} \right\} \right] \nonumber \\&\quad = {{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ - \int _0^\tau \left\{ \rho \sigma _1 (\tau -u) + \frac{\sigma _2}{\kappa } (1-e^{\kappa (u-\tau )}) \right\} dW_{u,2}^{\mathbb {Q}} \right\} \right] \nonumber \\&\quad = {{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ - \int _{-\tau }^0 \left\{ -\rho \sigma _1 z + \frac{\sigma _2}{\kappa } (1-e^{\kappa z})\right\} dW_{z,2}^{\mathbb {Q}} \right\} \right] ;~~z:=u-\tau \end{aligned}$$

(22)

Since

$$\begin{aligned}&Var\left[ \int _{-\tau }^0 \left\{ -\rho \sigma _1 z + \frac{\sigma _2}{\kappa } (1-e^{\kappa z}) \right\} dW_{z,2}^{\mathbb {Q}} \right] \nonumber \\&\quad = \int _{-\tau }^{0} \left\{ \rho ^2 \sigma _1^2z^2 + \left( \frac{\sigma _2}{\kappa } \right) ^2 (1-e^{\kappa z})^2- 2\frac{\rho \sigma _1 \sigma _2}{\kappa } z(1-e^{\kappa z}) \right\} dz \nonumber \\&\quad = \frac{\rho ^2 \sigma _1^2 }{3}\tau ^3 -2A(\tau ) - 2 \frac{\rho \sigma _1 \sigma _2}{\kappa } \int _{-\tau }^{0} (z-ze^{\kappa z}) dz = \frac{\rho ^2 \sigma _1^2 }{3}\tau ^3 -2A(\tau ) \nonumber \\&\qquad +\, 2 \frac{\rho \sigma _1 \sigma _2}{\kappa } \left\{ \frac{\tau ^2}{2} + \frac{\tau e^{-\kappa \tau } - B(\tau ) }{\kappa } \right\} , \end{aligned}$$

(23)

$$\begin{aligned}&{{\mathrm{\mathbb {E}}}}^{\mathbb {Q}}\left[ \exp \left\{ - \int _{-\tau }^0 \left\{ -\rho \sigma _1 z + \frac{\sigma _2}{\kappa } (1-e^{\kappa z}) dW_{z,2}^{\mathbb {Q}} \right\} \right\} \right] \nonumber \\&\quad = \exp \left[ \frac{\rho ^2 \sigma _1^2 }{6}\tau ^3 -A(\tau ) + \frac{\rho \sigma _1 \sigma _2}{\kappa } \left\{ \frac{\tau ^2}{2} + \frac{\tau e^{-\kappa \tau } - B(\tau ) }{\kappa } \right\} \right] \end{aligned}$$

(24)

Eq. (18), (20), (21) and (24) derive Eq. (12).