Appendix: Proofs
Before proving Proposition 2.1, several lemmas are presented.
Lemma 6.1
For \(i=1,2,3\) and any integer \(\tau >1\),
$$\begin{aligned} \mathcal {B}_{\tau }^{(i)}-1 &=\mathcal {B}_{\tau -1}^{(i)}\left( 1-\kappa ^{\mathbb {Q}}_{i,i}\right) -\mathbbm {1}_{ \{i=3\} } (1-\lambda )^{\tau -1} \\ &= \mathcal {B}_{\tau -1}^{(i)}(1-\lambda \mathbbm {1}_{ \{i>1\} } )-\mathbbm {1}_{ \{i=3\} } (1-\lambda )^{\tau -1}. \end{aligned}$$
Proof of Lemma 6.1
See Lemma A.2 of Eghbalzadeh et al. (2022).
Lemma 6.2
The following recursive relationship between the time-t and time-\(t+1\) zero-coupon bond prices presented in (2.3) holds for any integer \(\tau >0\):
$$\begin{aligned} P(t+1,t+\tau )&= P(t,t+\tau )e^{\Delta r_t} \exp \left[ \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) -\Delta \mathcal {B}_{\tau -1}^{\top }\left( \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \Sigma Z_{t+1}^\mathbb {Q}\right) \right] . \end{aligned}$$
Proof of Lemma 6.2
The case \(\tau =1\) is trivial. For \(\tau >1\), using (2.1) and (2.3) for the first equality, and then Lemma 6.1 for the third one,
$$\begin{aligned}&\log \left( \dfrac{P(t+1,t+\tau )}{P(t,t+\tau )} \right) - \Delta r_t = \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \sum _{i=1}^3 \left( \mathcal {B}_{\tau -1}^{(i)} X_{t+1}^{(i)}\right. \\&\qquad \left. -\left( \mathcal {B}_{\tau }^{(i)}-1\right) X_t^{(i)}\right) +\Delta \sum _{i=1}^3 X_t^{(i)} - \Delta (X_t^{(1)}+X_t^{(2)}) \\&\quad = \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \sum _{i=1}^2 \mathcal {B}_{\tau -1}^{(i)}\left( X_{t+1}^{(i)}-\left( \dfrac{\mathcal {B}_{\tau }^{(i)}-1}{\mathcal {B}_{\tau -1}^{(i)}}\right) X_t^{(i)}\right) + \Delta X_t^{(3)} \\&\quad \quad - \Delta \left( \mathcal {B}_{\tau -1}^{(3)} X_{t+1}^{(3)}-\left( \mathcal {B}_{\tau }^{(3)}-1\right) X_t^{(3)}\right) \\&\quad = \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \sum _{i=1}^2 \mathcal {B}_{\tau -1}^{(i)}\left( X_{t+1}^{(i)}-\left( 1-\kappa ^{\mathbb {Q}}_{i,i}\right) X_t^{(i)}\right) + \Delta X_t^{(3)} \\&\quad \quad - \left( \Delta \mathcal {B}_{\tau -1}^{(3)} X_{t+1}^{(3)}-\left( \Delta \mathcal {B}_{\tau -1}^{(3)}(1-\lambda )-(1-\lambda )^{\tau -1}\right) X_t^{(3)}\right) \\&\quad =\log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \sum _{i=1}^3 \mathcal {B}_{\tau -1}^{(i)}\left( X_{t+1}^{(i)}-\left( 1-\kappa ^{\mathbb {Q}}_{i,i}\right) X_t^{(i)}\right) \\&\quad \quad+ \Delta \left( 1-(1-\lambda )^{\tau - 1}\right) X_t^{(3)} \\&\quad = \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \mathcal {B}_{\tau -1}^{\top }\left( \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \Sigma Z_{t+1}^\mathbb {Q}\right) + \Delta \left( 1-(1-\lambda )^{\tau - 1} - \mathcal {B}_{\tau -1}^{(2)}\lambda \right) X_t^{(3)}\\&\quad = \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) - \Delta \mathcal {B}_{\tau -1}^{\top }\left( \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \Sigma Z_{t+1}^\mathbb {Q}\right) . \end{aligned}$$
Therefore,
$$\begin{aligned} P(t+1,t+\tau )&= P(t,t+\tau )e^{\Delta r_t} \exp \left[ \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) -\Delta \mathcal {B}_{\tau -1}^{\top }\left( \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \Sigma Z_{t+1}^\mathbb {Q}\right) \right] . \end{aligned}$$
\(\square\)
Lemma 6.3
Consider any integer \(\tau >0\) and any real number r. Using the convention \(0^0=0\), for functions \(\zeta _0(r,\tau ),\zeta _1(r,\tau )\) and \(\zeta _2(r,\tau )\) defined in (2.7)–(2.9),
$$\begin{aligned} \zeta _0(r,\tau -1)-\zeta _0(r,\tau )&= -r^{\tau -1} , \end{aligned}$$
(6.1)
$$\begin{aligned} \zeta _1(r,\tau -1)-\zeta _1(r,\tau )&= -r^{\tau -1}(\tau -1) , \end{aligned}$$
(6.2)
$$\begin{aligned} \zeta _2(r,\tau -1)-\zeta _2(r,\tau )&= -r^{\tau -1}(\tau -1)^2 . \end{aligned}$$
(6.3)
Proof of Lemma 6.3
This result is a direct consequence of the sum representations of \(\zeta _0(r,\tau ),\zeta _1(r,\tau )\) and \(\zeta _2(r,\tau )\) provided in (2.7)–(2.9). \(\square\)
Lemma 6.4
The following recursive connection holds for the quantity \(\upsilon _\tau\) defined in (2.6) for any integer \(\tau >0\):
$$\begin{aligned} \upsilon _{\tau } = \mathcal {B}_{\tau -1}^{\top } \Sigma \rho ( \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top + \upsilon _{\tau -1}. \end{aligned}$$
(6.4)
Proof of Lemma 6.4
The case \(\tau =1\) is trivial as it leads to \(0=0\). For \(\tau >1\), first,
$$\begin{aligned} \mathcal {B}_{\tau -1}^{\top } \Sigma \rho ( \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top&= \sum ^3_{i=1} \sum ^3_{j=1} \mathcal {B}_{\tau -1}^{(i)} \mathcal {B}_{\tau -1}^{(j)}\Sigma _{i,i}\Sigma _{j,j}\rho _{i,j}. \end{aligned}$$
Moreover, based on (2.6),
$$\begin{aligned} \upsilon _{\tau } - \upsilon _{\tau -1}&= \sum ^3_{i=1} \sum ^3_{j=1} a^{(i,j)}_\tau \Sigma _{i,i} \Sigma _{j,j} \rho _{i,j}, \end{aligned}$$
with \(a^{(i,j)}_\tau = (\upsilon ^{(i,j)}_{\tau } - \upsilon ^{(i,j)}_{\tau -1})/(\Sigma _{i,i} \Sigma _{j,j} \rho _{i,j})\).
To complete the proof, we now show that \(a^{(i,j)}_\tau =\mathcal {B}_{\tau -1}^{(i)} \mathcal {B}_{\tau -1}^{(j)}\) for any \(i,j=1,2,3\).
First, for \(i=j=1\),
$$\begin{aligned} a^{(1,1)}_\tau&= \dfrac{\tau (\tau -1)(2\tau -1)}{6} - \dfrac{(\tau -1)(\tau -2)(2\tau -3)}{6}. \\&=\dfrac{(\tau -1)(2\tau ^2-\tau -2\tau ^2+7\tau -6)}{6}\\&=(\tau -1)^2\\&=\mathcal {B}_{\tau -1}^{(1)} \mathcal {B}_{\tau -1}^{(1)}. \end{aligned}$$
Secondly, for \(i=1\) and \(j=2\), using (6.2),
$$\begin{aligned} a^{(1,2)}_\tau&= \frac{1}{\lambda }\left( \frac{\tau (\tau -1)}{2} - \zeta _1 (1-\lambda ,\tau ) - \frac{(\tau -1)(\tau -2)}{2} + \zeta _1 (1-\lambda ,\tau -1)\right) \\&= \dfrac{(\tau -1)-(1-\lambda )^{\tau -1}(\tau -1)}{\lambda }\\ &=\mathcal {B}_{\tau -1}^{(1)}\mathcal {B}_{\tau -1}^{(2)}. \end{aligned}$$
Thirdly, consider \(i=1\) and \(j=3\). For \(\tau =2\), \(a^{(1,3)}_2= \mathcal {B}_{1}^{(1)}\mathcal {B}_{1}^{(3)}=0\) since \(\upsilon ^{(1,3)}_{2}=\upsilon ^{(1,3)}_{1}=0\). For \(\tau >2\), using (6.1), (6.2), and (6.3),
$$\begin{aligned} a^{(1,3)}_\tau&= \frac{1}{\lambda } \bigg [\frac{\tau (\tau -1)}{2}-1 -\zeta _0\left( 1-\lambda ,\tau -1\right) -(1+\lambda )\zeta _1\left( 1-\lambda ,\tau -1\right) \\&\quad -\lambda \zeta _2\left( 1-\lambda ,\tau -1\right)- \frac{(\tau -1)(\tau -2)}{2}+1 +\zeta _0\left( 1-\lambda ,\tau -2\right) \\&\quad +(1+\lambda )\zeta _1\left( 1-\lambda ,\tau -2\right) +\lambda \zeta _2\left( 1-\lambda ,\tau -2\right) \bigg ]\\&=\dfrac{1}{\lambda }\bigg [\tau - 1 - (1-\lambda )^{\tau -2}-(1-\lambda )^{\tau - 2}(\tau -2)(1+\lambda )-\lambda (1-\lambda )^{\tau -2 }(\tau - 2)^2 \bigg ]\\&=\dfrac{1}{\lambda }\bigg [\tau - 1 - (1-\lambda )^{\tau -2}-(1-\lambda )^{\tau - 2}(\tau -1 - 1)(1+\lambda )\\&\quad -\lambda (1-\lambda )^{\tau -2 }(\tau - 2)(\tau - 1 -1) \bigg ]\\&=\dfrac{1}{\lambda }\bigg [\tau - 1 -(1-\lambda )^{\tau - 2}(\tau -1 )(1+\lambda )\\&\quad +\lambda (1-\lambda )^{\tau - 2}-\lambda (1-\lambda )^{\tau -2 }(\tau - 2)(\tau - 1 )+\lambda (1-\lambda )^{\tau -2 }(\tau - 1-1) \bigg ]\\&=\dfrac{\tau -1}{\lambda }\bigg [1 -(1-\lambda )^{\tau - 2}-\lambda (1-\lambda )^{\tau -2 }(\tau - 2) \bigg ]\\&=\mathcal {B}_{\tau -1}^{(1)}\mathcal {B}_{\tau -1}^{(3)}. \end{aligned}$$
Fourthly, for \(i=j=2\),
$$\begin{aligned} a^{(2,2)}_\tau&=\frac{1}{\lambda ^2} \left( \tau - 2 \left[ \frac{1-(1-\lambda )^\tau }{\lambda }\right] + \frac{1-(1-\lambda )^{2 \tau }}{1-(1-\lambda )^2} - \tau + 1 + 2 \left[ \frac{1-(1-\lambda )^{\tau -1} }{\lambda }\right] \right. \\&\quad \left. -\frac{1-(1-\lambda )^{2 (\tau -1)}}{1-(1-\lambda )^2}\right) \\&=\frac{1}{\lambda ^2} \left( 1 -2(1-\lambda )^{\tau -1}+(1-\lambda )^{2(\tau -1)}\right) \\&=\mathcal {B}_{\tau -1}^{(2)} \mathcal {B}_{\tau -1}^{(2)}. \end{aligned}$$
Fifthly, consider \(i=2\) and \(j=3\). For \(\tau =2\), \(a^{(2,3)}_2= \mathcal {B}_{1}^{(2)}\mathcal {B}_{1}^{(3)}=0\) since \(\upsilon ^{(2,3)}_{2}=\upsilon ^{(2,3)}_{1}=0\). For \(\tau >2\), using (6.1) and (6.2),
$$\begin{aligned} a^{(2,3)}_\tau&= \frac{\tau -2- (2-\lambda )\zeta _0\left( 1-\lambda ,\tau -1\right) + (1-\lambda )\zeta _0\left( (1-\lambda )^2,\tau -1\right) }{\lambda ^2}\\&\quad + \frac{- \zeta _1\left( 1-\lambda ,\tau -1\right) + (1-\lambda )\zeta _1\left( (1-\lambda )^2,\tau -1\right) }{\lambda }\\&\quad - \frac{\tau -3- (2-\lambda )\zeta _0\left( 1-\lambda ,\tau -2\right) + (1-\lambda )\zeta _0\left( (1-\lambda )^2,\tau -2\right) }{\lambda ^2}\\&\quad - \frac{- \zeta _1\left( 1-\lambda ,\tau -2\right) + (1-\lambda )\zeta _1\left( (1-\lambda )^2,\tau -2\right) }{\lambda }\\&= \frac{1- (2-\lambda )(1-\lambda )^{\tau -2} + (1-\lambda )(1-\lambda )^{2(\tau -2)} }{\lambda ^2} \\&\quad +\frac{-(1-\lambda )^{\tau -2}(\tau -2)+(1-\lambda )(1-\lambda )^{2(\tau -2)}(\tau -2)}{\lambda }\\&= \frac{\left( 1-(1-\lambda )^{\tau -2}\right) ^2 +\lambda (1-\lambda )^{\tau -2}\left( 1- (1-\lambda )^{\tau -2}\right) }{\lambda ^2} \\&\quad +\frac{(1-\lambda )^{\tau -2}(\tau -2)\left( (1-\lambda )^{\tau -1}-1\right) }{\lambda }\\&= \frac{\left( 1-(1-\lambda )^{\tau -2}\right) \left( 1-(1-\lambda )^{\tau -2}(1-\lambda )\right) }{\lambda ^2} \\&\quad+ \frac{(1-\lambda )^{\tau -2}(\tau -2)\left( (1-\lambda )^{\tau -1}-1\right) }{\lambda }\\&= \frac{1-(1-\lambda )^{\tau -1}}{\lambda }\left( \frac{1-(1-\lambda )^{\tau -2}}{\lambda }-(1-\lambda )^{\tau -2}(\tau -2)\right) \\&=\mathcal {B}_{\tau -1}^{(2)}\mathcal {B}_{\tau -1}^{(3)}. \end{aligned}$$
Lastly, consider \(i=j=3\). For \(\tau =2\), \(a^{(3,3)}_2= \mathcal {B}_{1}^{(3)}\mathcal {B}_{1}^{(3)}=0\) since \(\upsilon ^{(3,3)}_{2}=\upsilon ^{(3,3)}_{1}=0\). For \(\tau >2\), using (6.1), (6.2), and (6.3),
$$\begin{aligned} a^{(3,3)}_\tau&= \frac{1}{\lambda ^2} \bigg [ \tau -2 + \zeta _0\left( (1-\lambda )^2,\tau -1\right) +\lambda ^2 \zeta _2\left( (1-\lambda )^2,\tau -1\right) \\&\quad - 2\zeta _0\left( 1-\lambda ,\tau -1\right) - 2\lambda \zeta _1\left( 1-\lambda ,\tau -1\right) + 2\lambda \zeta _1\left( (1-\lambda )^2,\tau -1\right) \\&\quad - \tau +3 - \zeta _0\left( (1-\lambda )^2,\tau -2\right) -\lambda ^2 \zeta _2\left( (1-\lambda )^2,\tau -2\right) \\&\quad + 2\zeta _0\left( 1-\lambda ,\tau -2\right) + 2\lambda \zeta _1\left( 1-\lambda ,\tau -2\right) - 2\lambda \zeta _1\left( (1-\lambda )^2,\tau -2\right) \bigg ]\\&=\frac{1}{\lambda ^2} \bigg [1+(1-\lambda )^{2(\tau -2)}+\lambda ^2(1-\lambda )^{2(\tau -2)}(\tau -2)^2-2(1-\lambda )^{\tau -2}\\&\quad -2\lambda (1-\lambda )^{\tau -2}(\tau -2)+2\lambda (1-\lambda )^{2(\tau -2)}(\tau -2)\bigg ]\\&=\frac{1}{\lambda ^2} \bigg [\left( 1-(1-\lambda )^{\tau -2}\right) ^2+\lambda ^2(1-\lambda )^{2(\tau -2)}(\tau -2)^2\\&\quad -2\lambda (1-\lambda )^{\tau -2}(\tau -2)\left( 1-(1-\lambda )^{\tau -2}\right) \bigg ]\\&=\frac{1}{\lambda ^2} \bigg [\left( 1-(1-\lambda )^{\tau -2} -\lambda (1-\lambda )^{\tau -2}(\tau -2)\right) ^2\bigg ]\\&=\left( \dfrac{1-(1-\lambda )^{\tau -2}}{\lambda } -(1-\lambda )^{\tau -2}(\tau -2)\right) ^2\\&=\mathcal {B}_{\tau -1}^{(3)} \mathcal {B}_{\tau -1}^{(3)}. \end{aligned}$$
\(\square\)
Lemma 6.5
For \(i=1,2,3\) and any integer \(\tau >0\), the following recursive relationships between \(\mathcal {B}^{(i)}_{\tau }\) and \(\mathcal {B}^{(i)}_{\tau -1}\) hold:
$$\begin{aligned} \mathcal {B}^{(1)}_{\tau }= & {} \mathcal {B}^{(1)}_{\tau -1} + 1,\\ \mathcal {B}^{(2)}_{\tau }= & {} \mathcal {B}^{(2)}_{\tau -1}+(1-\lambda )^{\tau -1},\\ \mathcal {B}^{(3)}_{\tau }= & {} \mathcal {B}^{(3)}_{\tau -1}+(1-\lambda )^{\tau -2}\lambda (\tau -1). \end{aligned}$$
Proof of Lemma 6.5
The case \(\tau =1\) is trivial. For \(\tau >1\), based on (2.4), \(\mathcal {B}^{(1)}_{\tau } =\tau -1+1=\mathcal {B}^{(1)}_{\tau -1}+1.\) Furthermore,
$$\begin{aligned} \mathcal {B}^{(2)}_{\tau }&= \dfrac{1-(1-\lambda )^{\tau }}{\lambda }-1+1 = (1-\lambda )\mathcal {B}^{(2)}_{\tau -1}+1\\&=\mathcal {B}^{(2)}_{\tau -1}-1+(1-\lambda )^{\tau -1}+1=\mathcal {B}^{(2)}_{\tau -1}+(1-\lambda )^{\tau -1}. \end{aligned}$$
Lastly,
$$\begin{aligned} \mathcal {B}^{(3)}_{\tau }&=\frac{1-(1-\lambda )^{\tau -1}}{\lambda } -1+1 - (\tau -2+1) (1-\lambda )^{\tau -1}\\&=(1-\lambda )\left( \frac{1-(1-\lambda )^{\tau -2}}{\lambda } +\dfrac{1}{1-\lambda } - (\tau -2) (1-\lambda )^{\tau -2}-(1-\lambda )^{\tau -2}\right) \\&=\mathcal {B}^{(3)}_{\tau -1} +\dfrac{1}{1-\lambda } -(1-\lambda )^{\tau -2} \\&\quad -1+(1-\lambda )^{\tau -2}-\dfrac{\lambda }{1-\lambda }+ \lambda (\tau -2) (1-\lambda )^{\tau -2}+\lambda (1-\lambda )^{\tau -2} \\&=\mathcal {B}^{(3)}_{\tau -1}+(1-\lambda )^{\tau -2}\lambda (\tau -1). \end{aligned}$$
\(\square\)
Lemma 6.6
The following recursive relationship holds for quantity \(A_{\tau }\) defined in (2.3) and any integer \(\tau >0\):
$$\begin{aligned} A_{\tau }= A_{\tau -1} \exp \left( \frac{1}{2} \Delta ^2\mathcal {B}_{\tau -1}^{\top } \Sigma \rho ( \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top - \Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q}\right) . \end{aligned}$$
Proof of Lemma 6.6
For \(\tau =1\), the proof is trivial. For \(\tau >1\), using Lemma 6.4, Lemma 6.5 and (2.4) to substitute into (2.5),
$$\begin{aligned}&\log \left( \dfrac{A_{\tau -1}}{A_{\tau }}\right) -\Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q} = -\Delta \theta _2^\mathbb {Q}\left( \mathcal {B}_{\tau -1}^{(1)} - \mathcal {B}_{\tau }^{(1)} + \mathcal {B}_{\tau }^{(2)} - \mathcal {B}_{\tau -1}^{(2)} \right) \\&\qquad + \Delta \theta _3^\mathbb {Q}\left( \mathcal {B}_{\tau -1}^{(3)}-\mathcal {B}_{\tau }^{(3)}\right)-\Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q}- \frac{1}{2} \Delta ^2\left( \upsilon _\tau -\upsilon _{\tau -1}\right) \\&\quad = -\Delta \theta _2^\mathbb {Q}\left( -1 + (1-\lambda )^{\tau -1} \right) - \Delta \theta _3^\mathbb {Q}(1-\lambda )^{\tau -2}\lambda (\tau -1) \\&\quad \quad -\Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q}- \frac{1}{2} \Delta ^2\left( \upsilon _\tau -\upsilon _{\tau -1}\right) \\&\quad =\Delta \theta _2^\mathbb {Q}\lambda \mathcal {B}_{\tau -1}^{(2)} - \Delta \theta _3^\mathbb {Q}\lambda (1-\lambda )^{\tau -2}(\tau -1)-\Delta \mathcal {B}_{\tau -1}^{(2)} \lambda (\theta _2^\mathbb {Q}- \theta _3^\mathbb {Q}) -\Delta \mathcal {B}_{\tau -1}^{(3)} \lambda \theta _3^\mathbb {Q}\\&\quad \quad - \frac{1}{2} \Delta ^2\left( \upsilon _\tau -\upsilon _{\tau -1}\right) \\&\quad = - \Delta \theta _3^\mathbb {Q}\lambda (1-\lambda )^{\tau -2}(\tau -1) + \Delta \lambda \theta _3^\mathbb {Q}( \mathcal {B}_{\tau -1}^{(2)} - \mathcal {B}_{\tau -1}^{(3)}) - \frac{1}{2} \Delta ^2\mathcal {B}_{\tau -1}^{\top } \Sigma \rho ( \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top \\&\quad =- \frac{1}{2} \Delta ^2\mathcal {B}_{\tau -1}^{\top } \Sigma \rho ( \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top . \end{aligned}$$
\(\square\)
Proof of Proposition 2.1
The proof relies on calculating the moments generating function of innovations under the \(\mathcal {T}\)-forward measure. Consider the row vector \(\Gamma =[\Gamma _1,\Gamma _2,\Gamma _3]\) with \(\Gamma _i \in \mathbb {R}\) for \(i=1,2,3\). Then
$$\begin{aligned}&\mathbb {E}^{\mathcal {T}}\left[ \exp (\Gamma Z^{\mathcal {T}}_{t+1})\bigg |\mathcal {F}_t\right] =\mathbb {E}^{\mathcal {T}}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1}+\Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1}) \bigg |\mathcal {F}_t\right] \\&\quad = \frac{\mathbb {E}^{{\mathbb {Q}}}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1}+\Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1}) \dfrac{d\mathbb {Q}^{\mathcal {T}}}{d\mathbb {Q}} \bigg |\mathcal {F}_t\right] }{\mathbb {E}^{{\mathbb {Q}}}\left[ \dfrac{d\mathbb {Q}^{\mathcal {T}}}{d\mathbb {Q}} \bigg |\mathcal {F}_t\right] }\\&\quad = \exp \left[ \Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1}\right] \dfrac{\mathbb {E^Q}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1})\dfrac{P(t+\tau ,t+\tau )B(0)}{P(0,t+\tau )B(t+\tau )} \bigg |\mathcal {F}_t \right] }{\mathbb {E^Q}\left[ \dfrac{P(t+\tau ,t+\tau )B(0)}{P(0,t+\tau )B(t+\tau )} \bigg |\mathcal {F}_t \right] }\\&\quad = \exp \left[ \Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1} \right] \frac{1}{P(0,t+\tau )}\dfrac{\mathbb {E^Q}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1})\dfrac{P(t+\tau ,t+\tau )}{B(t+\tau )} \bigg |\mathcal {F}_t \right] }{\dfrac{P(t,t+\tau )}{P(0,t+\tau )B(t)}}\\&\quad =\exp \left[ \Gamma \rho \Delta \Sigma \mathcal {B}_{\tau -1} \right] \dfrac{B(t)}{P(t,t+\tau )} \mathbb {E^Q}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1})\mathbb {E^Q}\left[ \dfrac{P(t+\tau ,t+\tau )}{B(t+\tau )} \bigg |\mathcal {F}_{t+1} \right] \bigg |\mathcal {F}_t \right] \\&\quad = \exp \left[ \Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1} \right] \mathbb {E^Q}\left[ \exp (\Gamma Z^{\mathbb {Q}}_{t+1}) \dfrac{B(t)}{P(t,t+\tau )}\dfrac{P(t+1,t+\tau )}{B(t+1)} \bigg |\mathcal {F}_t \right] , \end{aligned}$$
where the fourth and fifth equalities rely on the fact that \(\frac{P(\cdot ,\mathcal {T})}{B(\cdot )}\) is a \(\mathbb {Q}\)-martingale.
Define
$$\begin{aligned} Y(Z_{t+1}^\mathbb {Q}) \equiv \Gamma Z^{\mathbb {Q}}_{t+1}+\log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) -\Delta \mathcal {B}_{\tau -1}^{\top }\left( \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \Sigma Z_{t+1}^\mathbb {Q}\right) . \end{aligned}$$
Using Lemma 6.2 therefore leads to
$$\begin{aligned} \mathbb {E}^{\mathcal {T}}\left[ \exp \left( \Gamma Z^{\mathcal {T}}_{t+1}\right) \bigg |\mathcal {F}_t\right] =\exp \left[ \Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1} \right] \mathbb {E^Q}\bigg [ \exp \left( Y(Z_{t+1}^\mathbb {Q}) \right) \bigg |\mathcal {F}_t \bigg ], \end{aligned}$$
where, given \(\mathcal {F}_t\), \(Y(Z_{t+1}^\mathbb {Q})\) follows the Gaussian distribution with conditional mean and variance
$$\begin{aligned} \mathbb {E^Q}\left[ Y(Z_{t+1}^\mathbb {Q}) \bigg |\mathcal {F}_t \right]&= \log \left( \dfrac{A_{\tau -1}}{A_\tau }\right) -\Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q},\\ \text {Var}^{\mathbb {Q}} \left[ Y(Z_{t+1}^\mathbb {Q}) \bigg |\mathcal {F}_t \right]&= \Gamma \rho \Gamma ^\top + \Delta \mathcal {B}_{\tau -1}^{\top } \Sigma \rho (\Delta \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top - 2 \Gamma \Delta \rho \Sigma \mathcal {B}_{\tau -1}. \end{aligned}$$
Thus,
$$\begin{aligned} \mathbb {E}^{\mathcal {T}}\left[ \exp (\Gamma Z^{\mathcal {T}}_{t+1})\bigg |\mathcal {F}_t\right]&=\dfrac{A_{\tau -1}}{A_\tau } \exp \left( -\Delta \mathcal {B}_{\tau -1}^{\top } \kappa ^\mathbb {Q}\theta ^\mathbb {Q}+ \dfrac{1}{2} \Delta \mathcal {B}_{\tau -1}^{\top } \Sigma \rho (\Delta \mathcal {B}_{\tau -1}^{\top } \Sigma )^\top \right)\times \\&\quad \exp \left( \dfrac{1}{2}\Gamma \rho \Gamma ^\top \right) . \end{aligned}$$
Therefore, using Lemma 6.6 leads to
$$\begin{aligned} \mathbb {E}^{\mathcal {T}}\left[ \exp (\Gamma Z^{\mathcal {T}}_{t+1})\bigg |\mathcal {F}_t\right] =\exp \left( \dfrac{1}{2}\Gamma \rho \Gamma ^\top \right) . \end{aligned}$$
\(\square\)
Lemma 6.7
Assume (2.13) holds. For \(t=0,\ldots ,\mathcal {T}\) and \(n=0,\ldots ,\mathcal {T}-t\),
$$\begin{aligned} X^{(i)}_{t+n}&=X^{(i)}_{t}(1-\kappa ^{\mathcal {T}}_{i,i})^{n}+\kappa ^{\mathcal {T}}_{i,i} \theta ^{\mathcal {T}}_i \sum _{l=1}^{n}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n-l)}+\Sigma _{i,i}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,i}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n-l)} \\&\quad +\sum _{l=1}^{n}\sum _{j\ne i}^3 \kappa ^{\mathcal {T}}_{i,j} (\theta ^{\mathcal {T}}_j-X^{(j)}_{t+l-1}) (1-\kappa ^{\mathcal {T}}_{i,i})^{(n-l)} - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(i)}_{t+l} (1- \kappa ^{\mathcal {T}}_{i,i})^{n-1-l}. \end{aligned}$$
(6.5)
Proof of Lemma 6.7
This proof is analogous to that of Lemma A.1 from Eghbalzadeh et al. (2022), which is based on induction. We apply the convention \(\sum _{l=0}^{-1} x_l= \sum _{l=1}^{0} x_l \equiv 0\). The case \(n=0\) is therefore trivial. Then, assume (6.5) holds for some \(n \le \mathcal {T}-t-1\). Using (2.13), for any \(i=1,2,3\),
$$\begin{aligned} X^{(i)}_{t+n+1}&=X^{(i)}_{t+n}+\sum _{j=1}^{3}\kappa ^{\mathcal {T}}_{i,j}(\theta ^{\mathcal {T}}_j-X^{(j)}_{t+n})+\Sigma _{i,i}Z^{\mathcal {T}}_{t+n+1,i} - \eta _{t+n}^{\mathcal {T},(i)}\\&=X^{(i)}_{t+n}(1-\kappa ^{\mathcal {T}}_{i,i})+ \kappa ^{\mathcal {T}}_{i,i}\theta ^{\mathcal {T}}_i +\sum _{j\ne i}^{3}\kappa ^{\mathcal {T}}_{i,j}(\theta ^{\mathcal {T}}_j-X^{(j)}_{t+n})+\Sigma _{i,i}Z^{\mathcal {T}}_{t+n+1,i} - \eta _{t+n}^{\mathcal {T},(i)}. \end{aligned}$$
Applying (6.5) in the latter equality yields
$$\begin{aligned} X^{(i)}_{t+n+1}&= X^{(i)}_{t}(1-\kappa ^{\mathcal {T}}_{i,i})^{n+1}+\kappa ^{\mathcal {T}}_{i,i}\theta ^{\mathcal {T}}_i \sum _{l=1}^{n}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)}+\Sigma _{i,i}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,i}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)}\\&\quad +\sum _{l=1}^{n}\sum _{j\ne i}^{3}\kappa ^{\mathcal {T}}_{i,j}(\theta ^{\mathcal {T}}_j-X^{(j)}_{t+l-1})(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)} + \kappa ^{\mathcal {T}}_{i,i}\theta ^{\mathcal {T}}_i - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(i)}_{t+l} (1- \kappa ^{\mathcal {T}}_{i,i} )^{n-l}\\&\quad +\sum _{j\ne i}^{3}\kappa ^{\mathcal {T}}_{i,j}(\theta ^{\mathcal {T}}_j-X^{(j)}_{t+n})+\Sigma _{i,i}Z^{\mathcal {T}}_{t+n+1,i} - \eta ^{\mathcal {T},(i)}_{t+n}\\&= X^{(i)}_{t}(1-\kappa ^{\mathcal {T}}_{i,i})^{n+1}+\kappa ^{\mathcal {T}}_{i,i}\theta ^{\mathcal {T}}_i \sum _{l=1}^{n+1}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)}+\Sigma _{i,i}\sum _{l=1}^{n+1}Z^{\mathcal {T}}_{t+l,i}(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)}\\&\quad +\sum _{l=1}^{n+1}\sum _{j\ne i}^{3}\kappa ^{\mathcal {T}}_{i,j}(\theta ^{\mathcal {T}}_j-X^{(j)}_{t+l-1})(1-\kappa ^{\mathcal {T}}_{i,i})^{(n+1-l)} - \sum _{l=0}^{n}\eta ^{\mathcal {T},(i)}_{t+l} (1- \kappa ^{\mathcal {T}}_{i,i} )^{n-l} , \end{aligned}$$
thereby finishing the induction. \(\square\)
Lemma 6.8
For \(n=0,\ldots ,\mathcal {T}-t\), the factors \(X^{(i)}_{t+n}\) can be expressed in terms of \(X_{t}\) and innovations \(\{ Z^{\mathcal {T}}_{t+l} \}^{\mathcal {T}-t}_{l=1}\) as follows:
$$\begin{aligned} X^{(1)}_{t+n}=& X^{(1)}_{t}+\Sigma _{1,1}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,1} - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(1)}_{t+l},\\ X^{(2)}_{t+n}=& X^{(2)}_{t}(1-\lambda )^{n}+(\theta ^\mathcal {T}_2-\theta ^\mathcal {T}_3)\left( 1-(1-\lambda )^n\right) +\Sigma _{2,2}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,2}(1-\lambda )^{(n-l)}\\&\quad + \lambda \bigg (nX^{(3)}_{t} (1-\lambda )^{n-1} + \theta ^{\mathcal {T}}_3 \left( \frac{\left( 1-(1-\lambda )^n\right) }{\lambda }-n(1-\lambda )^{n-1} \right) \\&\quad + \Sigma _{3,3} \sum _{k=1}^{n-1} (n-k)(1-\lambda )^{n-k-1} Z^{\mathcal {T}}_{t+k,3} -\sum _{k=0}^{n-1}(n-k-1) \eta ^{\mathcal {T},(3)}_{t+k} (1-\lambda )^{n-k-2}\bigg )\\&\quad -\sum _{l=0}^{n-1}\eta ^{\mathcal {T},(2)}_{t+l} (1- \lambda )^{n-1-l},\\ X^{(3)}_{t+n}=& X^{(3)}_{t}(1-\lambda )^{n}+ \theta ^\mathcal {T}_3\left( 1-(1-\lambda )^n\right) +\Sigma _{3,3}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,3}(1-\lambda )^{(n-l)}\\&\quad - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(3)}_{t+l} (1- \lambda )^{n-1-l}. \end{aligned}$$
Proof of Lemma 6.8
From (2.12), \(\kappa ^{\mathcal {T}}_{1,1}=\kappa ^{\mathcal {T}}_{1,2}=\kappa ^{\mathcal {T}}_{1,3}=\kappa ^{\mathcal {T}}_{2,1}=\kappa ^{\mathcal {T}}_{3,1}=\kappa ^{\mathcal {T}}_{3,2}=0\), \(\kappa ^{\mathcal {T}}_{2,2}=\kappa ^{\mathcal {T}}_{3,3}=\lambda\) and \(\kappa ^{\mathcal {T}}_{2,3}=-\lambda\). When placed into (6.5), this leads to
$$\begin{aligned} X^{(1)}_{t+n}= & {} X^{(1)}_{t}+\Sigma _{1,1}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,1} - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(1)}_{t+l}, \\ X^{(2)}_{t+n}= & {} X^{(2)}_{t}(1-\lambda )^{n}+\lambda \theta ^\mathcal {T}_2 \sum _{l=1}^{n}(1-\lambda )^{(n-l)}+\Sigma _{2,2}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,2}(1-\lambda )^{(n-l)} \\{} & {} \quad -\lambda \sum _{l=1}^{n} (\theta ^\mathcal {T}_3-X^{(3)}_{t+l-1}) (1-\lambda )^{(n-l)} - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(2)}_{t+l} (1- \lambda )^{n-1-l}, \end{aligned}$$
(6.6)
$$\begin{aligned} X^{(3)}_{t+n}= & {} X^{(3)}_{t}(1-\lambda )^{n}+\lambda \theta ^\mathcal {T}_3 \sum _{l=1}^{n}(1-\lambda )^{(n-l)}+\Sigma _{3,3}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,3}(1-\lambda )^{(n-l)} \\{} & {} \quad - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(3)}_{t+l} (1- \lambda )^{n-1-l}. \end{aligned}$$
(6.7)
Furthermore,
$$\begin{aligned}\sum _{l=1}^{n} (1-\lambda )^{n-l} X^{(3)}_{t+l-1} &= \sum _{l=0}^{n-1}(1-\lambda )^{n-l-1} X^{(3)}_{t+l} \\& = \sum _{l=0}^{n-1} (1-\lambda )^{n-l-1} \bigg [ X^{(3)}_{t}(1-\lambda )^{l}+\lambda \theta ^{\mathcal {T}}_3 \sum _{k=1}^{l}(1-\lambda )^{(l-k)} \\&\quad \quad +\Sigma _{3,3}\sum _{k=1}^{l}Z^{\mathcal {T}}_{t+k,3}(1-\lambda )^{(l-k)}- \sum _{k=0}^{l-1}(1-\lambda )^{l-1-k}\eta ^{\mathcal {T},(3)}_{t+k}\bigg ] \\& = nX^{(3)}_{t} (1-\lambda )^{n-1} + \lambda \theta ^{\mathcal {T}}_3 \sum _{l=0}^{n-1} \frac{(1-\lambda )^{n-l-1}-(1-\lambda )^{n-1}}{\lambda } \\&\quad \quad + \Sigma _{3,3} \sum _{l=0}^{n-1} \sum _{k=1}^{n-1} \mathbbm {1}_{\{k \le l\}} Z^{\mathcal {T}}_{t+k,3} (1-\lambda )^{n-k-1} - \sum _{l=0}^{n-1} \sum _{k=0}^{n-1}\mathbbm {1}_{\{k \le l-1\}}(1-\lambda )^{n-k-2} \eta ^{\mathcal {T},(3)}_{t+k} \\ &= nX^{(3)}_{t} (1-\lambda )^{n-1} + \theta ^{\mathcal {T}}_3 \left( \dfrac{1-(1-\lambda )^n}{\lambda }-n(1-\lambda )^{n-1} \right) \\&\quad \quad + \Sigma _{3,3} \sum _{k=1}^{n-1} Z^{\mathcal {T}}_{t+k,3} \sum _{l=k}^{n-1} (1-\lambda )^{n-k-1}-\sum _{k=0}^{n-1} \eta ^{\mathcal {T},(3)}_{t+k} \sum _{l=k+1}^{n-1}(1-\lambda )^{n-k-2} \\& = nX^{(3)}_{t} (1-\lambda )^{n-1} + \theta ^{\mathcal {T}}_3 \left( \dfrac{1-(1-\lambda )^n}{\lambda }-n(1-\lambda )^{n-1} \right) \\&\quad \quad + \Sigma _{3,3} \sum _{k=1}^{n-1} (n-k)(1-\lambda )^{n-k-1} Z^{\mathcal {T}}_{t+k,3} -\sum _{k=0}^{n-1}(n-k-1) \eta ^{\mathcal {T},(3)}_{t+k} (1-\lambda )^{n-k-2}. \end{aligned}$$
(6.8)
Using (6.8) in (6.6),
$$\begin{aligned} X^{(2)}_{t+n}&=X^{(2)}_{t}(1-\lambda )^{n}+\theta ^\mathcal {T}_2\left( 1-(1-\lambda )^n\right) +\Sigma _{2,2}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,2}(1-\lambda )^{(n-l)}\\&\quad - \theta ^\mathcal {T}_3\left( 1-(1-\lambda )^n\right) \\&\quad + \lambda \bigg (n X^{(3)}_{t} (1-\lambda )^{n-1} + \theta ^{\mathcal {T}}_3 \left( \dfrac{1-(1-\lambda )^n}{\lambda }-n(1-\lambda )^{n-1} \right) \\&\quad + \Sigma _{3,3} \sum _{k=1}^{n-1} (n-k)(1-\lambda )^{n-k-1} Z^{\mathcal {T}}_{t+k,3} -\sum _{k=0}^{n-1}(n-k-1) \eta ^{\mathcal {T},(3)}_{t+k} (1-\lambda )^{n-k-2}\bigg ) \\&\quad - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(2)}_{t+l} (1- \lambda )^{n-1-l}. \end{aligned}$$
Moreover, using (2.7) in (6.7) leads to
$$\begin{aligned} X^{(3)}_{t+n}&=X^{(3)}_{t}(1-\lambda )^{n}+ \theta ^\mathcal {T}_3\left( 1-(1-\lambda )^n\right) +\Sigma _{3,3}\sum _{l=1}^{n}Z^{\mathcal {T}}_{t+l,3}(1-\lambda )^{(n-l)}\\&\quad - \sum _{l=0}^{n-1}\eta ^{\mathcal {T},(3)}_{t+l} (1- \lambda )^{n-1-l}. \end{aligned}$$
\(\square\)
Proof of Proposition 2.2
The joint normality of \(X_{t+n}\) given \(\mathcal {F}_t\) is a direct consequence of Lemma 6.8, which expresses \(X_{t+n}\) as a linear combination of the \(\mathcal {F}_t\)-measurable elements of \(X_t\) and of jointly Gaussian innovations \(Z^\mathcal {T}_{t+1},\ldots , Z^\mathcal {T}_{t+n}\) that are independent of \(\mathcal {F}_t\). The composition of \(\mathcal {M}_{t,n}\) is also a direct consequence of Lemma 6.8, and of the null expectation of innovations \(Z^\mathcal {T}_{t+1},\ldots , Z^\mathcal {T}_{t+n}\) given \(\mathcal {F}_t\).
Components of \(\mathcal {V}_{n}\) are also obtained through Lemma 6.8 and (2.7)–(2.9):
$$\begin{aligned} \mathcal {V}^{(1,1)}_{n}&=\text {Var}^\mathcal {T}(X^{(1)}_{t+n} |\mathcal {F}_t)= \Sigma _{1,1}^2 \sum _{l=1}^{n}\text {Var}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,1}) =n \Sigma _{1,1}^2,\\ \mathcal {V}^{(2,2)}_{n}&=\text {Var}^\mathcal {T}(X^{(2)}_{t+n}|\mathcal {F}_t) = \Sigma ^2_{2,2} \sum _{l=1}^{n}(1-\lambda )^{2(n-l)} \text {Var}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,2}) \\&\quad + \lambda ^2 \Sigma ^2_{3,3} \sum _{l=1}^{n-1} (n-l)^2(1-\lambda )^{2(n-l-1)} \text {Var}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,3})\\&\quad +2\Sigma _{2,2} \lambda \Sigma _{3,3}\sum _{l=1}^{n-1} (n-l)(1-\lambda )^{2(n-l)-1} \text {Cov}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,2},Z^{\mathcal {T}}_{t+l,3})\\&= \Sigma _{2,2}^2 \left( 1+\zeta _0((1-\lambda )^2,n)\right) + \lambda ^2\Sigma _{3,3}^2(1-\lambda )^{-2}\zeta _2((1-\lambda )^2,n)\\&\quad +2\Sigma _{2,2} \lambda \Sigma _{3,3}\rho _{2,3}(1-\lambda )^{-1}\zeta _1\left( \left( 1-\lambda \right) ^2,n\right) ,\\ \mathcal {V}^{(3,3)}_{n}&=\text {Var}^\mathcal {T}(X^{(3)}_{t+n}|\mathcal {F}_t)=\Sigma _{3,3}^2\sum _{l=1}^n(1-\lambda )^{2(n-l)}\text {Var}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,3}) \\&=\Sigma _{3,3}^2 \left( 1+\zeta _0((1-\lambda )^2,n)\right) \end{aligned}$$
and
$$\begin{aligned} \mathcal {V}^{(1,2)}_{n}=\mathcal {V}^{(2,1)}_{n}&=\text {Cov}^\mathcal {T}(X^{(1)}_{t+n},X^{(2)}_{t+n}|\mathcal {F}_t)=\Sigma _{1,1}\Sigma _{2,2}\sum _{l=1}^{n}(1-\lambda )^{n-l}\text {Cov}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,1},Z^{\mathcal {T}}_{t+l,2})\\&\quad +\lambda \Sigma _{1,1}\Sigma _{3,3}\sum _{l=1}^{n-1}(n-l)(1-\lambda )^{n-l-1}\text {Cov}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,1},Z^{\mathcal {T}}_{t+l,3})\\&=\Sigma _{1,1}\Sigma _{2,2}\rho _{1,2}\left( 1+\zeta _0(1-\lambda ,n)\right) +\lambda \Sigma _{1,1}\Sigma _{3,3}\rho _{1,3}\dfrac{\zeta _1(1-\lambda ,n)}{1-\lambda },\\ \mathcal {V}^{(1,3)}_{n}=\mathcal {V}^{(3,1)}_{n}&=\text {Cov}^\mathcal {T}(X^{(1)}_{t+n},X^{(3)}_{t+n}|\mathcal {F}_t)=\Sigma _{1,1}\Sigma _{3,3}\sum _{l=1}^{n}(1-\lambda )^{n-l}\text {Cov}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,1},Z^{\mathcal {T}}_{t+l,3})\\&=\Sigma _{1,1}\Sigma _{3,3}\rho _{1,3} \left( 1+\zeta _0(1-\lambda ,n)\right) ,\\ \mathcal {V}^{(2,3)}_{n}=\mathcal {V}^{(3,2)}_{n}&=\text {Cov}^\mathcal {T}(X^{(2)}_{t+n},X^{(3)}_{t+n}|\mathcal {F}_t)\\&=\Sigma _{2,2}\Sigma _{3,3}\sum _{l=1}^{n}(1-\lambda )^{2(n-l)}\text {Cov}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,2},Z^{\mathcal {T}}_{t+l,3})\\&\quad +\lambda \Sigma _{3,3}^2\sum _{l=1}^{n-1}(n-l)(1-\lambda )^{2(n-l)-1}\text {Var}^\mathcal {T}(Z^{\mathcal {T}}_{t+l,3})\\&=\Sigma _{2,2}\Sigma _{3,3} \rho _{2,3}\left( 1+\zeta _0((1-\lambda )^2,n)\right) +\lambda \Sigma _{3,3}^2\dfrac{\zeta _1((1-\lambda )^2,n)}{1-\lambda }. \end{aligned}$$
\(\square\)
Proof of Lemma 2.1
From, (2.12), for \(i=1,2,3\), \(\eta ^{\mathcal {T},(i)}_t = \Delta \Sigma _{i,i}\sum ^3_{j=1} \Sigma _{j,j} \rho _{i,j} \mathcal {B}^{(j)}_{\mathcal {T}-t-1}\). Therefore,
$$\begin{aligned} \sum _{l=0}^{\mathcal {T}-t-1} \eta ^{\mathcal {T},(1)}_{t+l}&= \Delta \Sigma _{1,1}\sum ^3_{j=1} \Sigma _{j,j} \rho _{1,j} \sum _{l=0}^{\mathcal {T}-t-1} \mathcal {B}^{(j)}_{\mathcal {T}-t-l-1}\\&= \Delta \Sigma _{1,1}\sum ^3_{j=1} \Sigma _{j,j} \rho _{1,j} \sum _{l=0}^{\mathcal {T}-t-1} \mathcal {B}^{(j)}_{l}\\&= \Delta \Sigma _{1,1}\sum ^3_{j=1} \Sigma _{j,j} \rho _{1,j} \sum _{l=1}^{\mathcal {T}-t-1} \mathcal {B}^{(j)}_{l}\\&=\Delta \Sigma _{1,1}\bigg [\Sigma _{1,1} \frac{(\mathcal {T}-t-1)(\mathcal {T}-t)}{2} +\dfrac{\Sigma _{2,2} \rho _{1,2}}{\lambda }\left( \mathcal {T}-t-1 -\zeta _0\left( 1-\lambda ,\mathcal {T}-t\right) \right) \\&\quad +\Sigma _{3,3} \rho _{1,3}\bigg ( \dfrac{\mathcal {T}-t-1-\left( 1+\zeta _0\left( 1-\lambda ,\mathcal {T}-t-1\right) \right) }{\lambda }\\&\quad -\zeta _1\left( 1-\lambda , \mathcal {T}-t-1\right) \bigg )\bigg ]. \end{aligned}$$
Moreover, for \(i=2,3\),
$$\begin{aligned}&\sum _{l=0}^{\mathcal {T}-t-1}\eta ^{\mathcal {T},(i)}_{t+l} (1- \lambda )^{\mathcal {T}-t-1-l}\\&\quad = \Delta \Sigma _{i,i}\sum _{l=0}^{\mathcal {T}-t-1}(1- \lambda )^{\mathcal {T}-t-1-l}\left( \Sigma _{1,1} \rho _{i,1} \mathcal {B}^{(1)}_{\mathcal {T}-t-l-1} \!+\! \Sigma _{2,2} \rho _{i,2} \mathcal {B}^{(2)}_{\mathcal {T}-t-l-1} \!\right. \\&\quad \quad \left. +\! \Sigma _{3,3} \rho _{i,3} \mathcal {B}^{(3)}_{\mathcal {T}-t-l-1} \right) \\&\quad = \Delta \Sigma _{i,i}\sum _{l=1}^{\mathcal {T}-t-1}(1- \lambda )^{l}\left( \Sigma _{1,1} \rho _{i,1} \mathcal {B}^{(1)}_{l} + \Sigma _{2,2} \rho _{i,2} \mathcal {B}^{(2)}_{l} + \Sigma _{3,3} \rho _{i,3} \mathcal {B}^{(3)}_{l} \right) \\&\quad =\Delta \Sigma _{i,i}\bigg [\Sigma _{1,1} \rho _{i,1} \zeta _1\left( 1-\lambda ,\mathcal {T}-t\right) \\&\quad \quad+\dfrac{\Sigma _{2,2} \rho _{i,2}}{\lambda }\left( \zeta _0\left( 1-\lambda ,\mathcal {T}-t\right) -\zeta _0\left( \left( 1-\lambda \right) ^{2},\mathcal {T}-t\right) \right) \\&\quad \quad +\Sigma _{3,3} \rho _{i,3}\bigg (\dfrac{\zeta _0\left( 1-\lambda ,\mathcal {T}-t\right) -(1-\lambda )^{-1}\zeta _0\left( \left( 1-\lambda \right) ^{2},\mathcal {T}-t\right) }{\lambda } \\&\qquad -(1+\lambda )\zeta _1\left( \left( 1-\lambda \right) ^{2},\mathcal {T}-t-1\right) \bigg )\bigg ]. \end{aligned}$$
Lastly,
$$\begin{aligned}&\sum _{k=0}^{\mathcal {T}-t-1}(\mathcal {T}-t-k-1) \eta ^{\mathcal {T},(3)}_{t+k} (1-\lambda )^{\mathcal {T}-t-k-2}\\&\quad = \sum _{k=0}^{\mathcal {T}-t-1}(\mathcal {T}-t-k-1) \Delta \Sigma _{3,3}\sum ^3_{j=1} \Sigma _{j,j} \rho _{3,j} \mathcal {B}^{(j)}_{\mathcal {T}-t-k-1} (1-\lambda )^{\mathcal {T}-t-k-2}\\&\quad = \Delta \Sigma _{3,3} \sum _{k=0}^{\mathcal {T}-t-1} k \sum ^3_{j=1} \Sigma _{j,j} \rho _{3,j} \mathcal {B}^{(j)}_{k} (1-\lambda )^{k-1}\\&\quad = \Delta \Sigma _{3,3} \sum _{k=1}^{\mathcal {T}-t-1} \sum ^3_{j=1} \Sigma _{j,j} \rho _{3,j} k \mathcal {B}^{(j)}_{k} (1-\lambda )^{k-1}\\&\quad = \frac{\Delta \Sigma _{3,3}}{1-\lambda } \bigg ( \Sigma _{1,1} \rho _{3,1} \zeta _2\left( 1-\lambda ,\mathcal {T}-t\right) + \frac{\Sigma _{2,2} \rho _{2,1}}{\lambda } \left[ \zeta _1\left( 1-\lambda ,\mathcal {T}-t\right) \right. \\&\qquad \left. - \zeta _1\left( (1-\lambda )^2,\mathcal {T}-t\right) \right] \\&\quad \quad + \Sigma _{3,3} \rho _{3,1} \left[ \frac{\zeta _1\left( 1-\lambda ,\mathcal {T}-t\right) }{\lambda } - \frac{1}{\lambda }\zeta _1\left( (1-\lambda )^2,\mathcal {T}-t\right) \right. \\&\qquad \left. - (1-\lambda )^{-1}\zeta _2\left( (1-\lambda )^2,\mathcal {T}-t\right) \right] \bigg ). \end{aligned}$$
Lemma 6.9
Under the risk-neutral measure \(\mathbb {Q}\), conditionally on \(\mathcal {F}_t\), factors \(X_{t+n}\) follow the multivariate Gaussian distribution with mean vector \(\tilde{\mathcal {M}}_{t,n}=\left[ \tilde{\mathcal {M}}^{(i)}_{t,n}\right] ^3_{i=1}\) and covariance matrix \(\mathcal {V}_{n}=\left[ \mathcal {V}^{(i,j)}_{n}\right] ^3_{i,j=1}\), where
$$\begin{aligned} \tilde{\mathcal {M}}^{(1)}_{t,n}&=X^{(1)}_{t}, \\ \tilde{\mathcal {M}}^{(2)}_{t,n}&=X^{(2)}_{t}(1-\lambda )^{n}+(\theta ^\mathbb {Q}_2-\theta ^\mathbb {Q}_3)\left( 1-(1-\lambda )^n\right) \\&\quad + \lambda \bigg (nX^{(3)}_{t} (1-\lambda )^{n-1} + \theta ^{\mathbb {Q}}_3 \left( \dfrac{\zeta _0(1-\lambda ,n+1)}{1-\lambda }-n(1-\lambda )^{n-1} \right) \bigg ),\\ \tilde{\mathcal {M}}^{(3)}_{t,n}&=X^{(3)}_{t}(1-\lambda )^{n}+ \theta ^\mathbb {Q}_3\left( 1-(1-\lambda )^n\right) . \end{aligned}$$
Proof of Lemma 6.9
The proof is analogous to that of Proposition 2.2, replacing forward measure superscripts \(\mathcal {T}\) by \(\mathbb {Q}\) and setting the process \(\eta\) to zero. \(\square\)
Proof of Proposition 2.3
From (2.14), For \(t=0,\ldots ,T\) and \(n=0,\ldots ,T-t\),
$$\begin{aligned} X^{(i)}_{t+n}&=X^{(i)}_{t}(1-\kappa ^{\mathbb {P}}_{i,i})^{n}+\kappa ^{\mathbb {P}}_{i,i} \theta ^{\mathbb {P}}_i \sum _{l=1}^{n}(1-\kappa ^{\mathbb {P}}_{i,i})^{(n-l)}+\Sigma _{i,i}\sum _{l=1}^{n}Z^{\mathbb {P}}_{t+l,i}(1-\kappa ^{\mathbb {P}}_{i,i})^{(n-l)} \\&\quad +\sum _{l=1}^{n}\sum _{j\ne i}^3 \kappa ^{\mathbb {P}}_{i,j} (\theta ^{\mathbb {P}}_j-X^{(j)}_{t+l-1}) (1-\kappa ^{\mathbb {P}}_{i,i})^{(n-l)}. \end{aligned}$$
(6.9)
Indeed, steps for the proof of (6.9) are identical to these for the proof of Lemma 6.7, setting the process \(\eta\) to zero. This leads to
$$\begin{aligned} X^{(i)}_{t+n}&=X^{(i)}_{t}(1-\kappa ^{\mathbb {P}}_{i,i})^{n}+ \theta ^\mathbb {P}_i\left( 1-(1-\kappa ^{\mathbb {P}}_{i,i})^n\right) +\Sigma _{i,i}\sum _{l=1}^{n}Z^{\mathbb {P}}_{t+l,i}(1-\kappa ^{\mathbb {P}}_{i,i})^{(n-l)}, \quad i=1,3. \end{aligned}$$
(6.10)
and
$$\begin{aligned} X^{(2)}_{t+n}= & {} X^{(2)}_{t}(1-\kappa ^{\mathbb {P}}_{2,2})^{n}+\kappa ^{\mathbb {P}}_{2,2} \theta ^\mathbb {P}_2 \sum _{l=1}^{n}(1-\kappa ^{\mathbb {P}}_{2,2})^{(n-l)}+\Sigma _{2,2}\sum _{l=1}^{n}Z^{\mathbb {P}}_{t+l,2}(1-\kappa ^{\mathbb {P}}_{2,2})^{(n-l)} \\{} & {} \quad -\lambda \sum _{l=1}^{n} (\theta ^\mathbb {P}_3-X^{(3)}_{t+l-1}) (1-\kappa ^{\mathbb {P}}_{2,2})^{(n-l)}. \end{aligned}$$
(6.11)
Furthermore, assuming \(\kappa ^{\mathbb {P}}_{2,2} \ne \kappa ^{\mathbb {P}}_{3,3}\) and denoting \(\omega =\frac{1-\kappa ^{\mathbb {P}}_{3,3}}{1-\kappa ^{\mathbb {P}}_{2,2}}\),
$$\begin{aligned} -\sum _{l=1}^{n} (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-l} \left( \theta ^\mathbb {P}_3- X^{(3)}_{t+l-1}\right)&= -\sum _{l=0}^{n-1}(1-\kappa ^{\mathbb {P}}_{2,2} )^{n-l-1} \left( \theta ^\mathbb {P}_3- X^{(3)}_{t+l} \right) \\&= \sum _{l=0}^{n-1} (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-l-1} \bigg [ X^{(3)}_{t}(1-\kappa ^{\mathbb {P}}_{3,3} )^{l}- \theta ^{\mathbb {P}}_3 (1-\kappa ^{\mathbb {P}}_{3,3})^l \\&\quad +\Sigma _{3,3}\sum _{k=1}^{l}Z^{\mathbb {P}}_{t+k,3}(1-\kappa ^{\mathbb {P}}_{3,3} )^{(l-k)} \bigg ] \\&= (X^{(3)}_{t}-\theta ^{\mathbb {P}}_3)\frac{1-\omega ^n}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1} \\&\quad + \Sigma _{3,3} \sum _{l=0}^{n-1} \sum _{k=1}^{n-1} \mathbbm {1}_{\{k \le l\}} Z^{\mathbb {P}}_{t+k,3} \omega ^l\frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{k}} \\&= (X^{(3)}_{t}-\theta ^{\mathbb {P}}_3)\frac{1-\omega ^n}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1} \\&\quad + \Sigma _{3,3} \sum _{k=1}^{n-1} Z^{\mathbb {P}}_{t+k,3} \frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{k}} \sum _{l=k}^{n-1} \omega ^{l} \\&= (X^{(3)}_{t}-\theta ^{\mathbb {P}}_3)\frac{1-\omega ^n}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1} \\&\quad + \Sigma _{3,3} \sum _{k=1}^{n-1} Z^{\mathbb {P}}_{t+k,3} \frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{k}} \frac{\omega ^{k}-\omega ^{n}}{1-\omega } . \end{aligned}$$
(6.12)
Substituting (6.12) into (6.11) leads to
$$\begin{aligned} X^{(2)}_{t+n}&= X^{(2)}_{t}(1-\kappa ^{\mathbb {P}}_{2,2})^{n}+ \theta ^\mathbb {P}_2 \left( 1-(1-\kappa ^{\mathbb {P}}_{2,2})^{n} \right) +\Sigma _{2,2}\sum _{l=1}^{n}Z^{\mathbb {P}}_{t+l,2}(1-\kappa ^{\mathbb {P}}_{2,2})^{(n-l)} \\&\quad + \lambda \bigg ( (X^{(3)}_{t}-\theta ^{\mathbb {P}}_3)\frac{1-\omega ^n}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1} + \Sigma _{3,3} \sum _{l=1}^{n-1} Z^{\mathbb {P}}_{t+l,3} \frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{l}} \frac{\omega ^{l}-\omega ^{n}}{1-\omega } \bigg ). \end{aligned}$$
(6.13)
Combining (6.10) and (6.13) directly yields expressions for \(\mathcal {M}^{\mathbb {P},(i)}_{t,n}\). Additionally,
$$\begin{aligned} \mathcal {V}^{\mathbb {P},(1,1)}_{n}&=\text {Var}^{\mathbb{P}}(X^{(1)}_{t+n}|\mathcal {F}_t)=\Sigma _{1,1}^2\sum _{l=1}^n (1-\kappa ^{\mathbb {P}}_{1,1})^{2(n-l)}\text {Var}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,1})\\&= {\left\{ \begin{array}{ll} \Sigma _{1,1}^2 \left( 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{1,1})^2,n)\right) \quad \text { if } \kappa ^{\mathbb {P}}_{1,1}\in (0,1),\\ n \Sigma _{1,1}^2 \quad \text { if } \kappa ^{\mathbb {P}}_{1,1}=0, \end{array}\right. }\\ \mathcal {V}^{\mathbb {P},(2,2)}_{n}&=\text {Var}^{\mathbb{P}}(X^{(2)}_{t+n}|\mathcal {F}_t) = \Sigma ^2_{2,2} \sum _{l=1}^{n}(1-\kappa ^{\mathbb {P}}_{2,2})^{2(n-l)} \text {Var}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,2})\\&\quad + \lambda ^2 \Sigma ^2_{3,3} \sum _{l=1}^{n-1} \left( \frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{l}} \frac{\omega ^{l}-\omega ^{n}}{1-\omega }\right) ^2 \text {Var}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,3})\\&\quad +2\Sigma _{2,2} \lambda \Sigma _{3,3}\sum _{l=1}^{n-1} (1-\kappa ^{\mathbb {P}}_{2,2} )^{(2n-l-1)} \left( \frac{1}{1-\omega }\left( \frac{\omega }{1-\kappa ^{\mathbb {P}}_{3,3}} \right) ^l \right. \\&\quad \left. - \frac{\omega ^n}{1-\omega }\left( \frac{1}{1-\kappa ^{\mathbb {P}}_{3,3}} \right) ^l\right) \text {Cov}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,2},Z^{\mathbb {P}}_{t+l,3})\\&= \Sigma _{2,2}^2 \left( 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{2,2})^2,n)\right) + \lambda ^2 \Sigma ^2_{3,3} \frac{(1-\kappa ^{\mathbb {P}}_{2,2} )^{2n-2}}{(1-\omega )^2} \sum _{l=1}^{n-1} \left( \frac{ \omega ^{2l}-2\omega ^{l+n}+\omega ^{2n}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{2l}} \right) \\&\quad +2 \frac{\rho _{2,3} \lambda \Sigma _{2,2} \Sigma _{3,3}}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{(2n-1)}\sum _{l=1}^{n-1} \left( \left( \frac{\omega }{(1-\kappa ^{\mathbb {P}}_{2,2})(1-\kappa ^{\mathbb {P}}_{3,3})} \right) ^l \right. \\&\quad \left. -\omega ^n\left( \frac{1}{(1-\kappa ^{\mathbb {P}}_{2,2})(1-\kappa ^{\mathbb {P}}_{3,3})} \right) ^l\right) \\&= \Sigma _{2,2}^2 \left( 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{2,2})^2,n)\right) \\&\quad + \lambda ^2 \Sigma ^2_{3,3} \frac{(1-\kappa ^{\mathbb {P}}_{2,2} )^{2n-2}}{(1-\omega )^2} \left[ \zeta _0\left( (1-\kappa ^{\mathbb {P}}_{2,2})^{-2},n\right) -2\omega ^n \zeta _0\left( \omega (1-\kappa ^{\mathbb {P}}_{3,3})^{-2},n\right) \right. \\&\quad \left. + \omega ^{2n}\zeta _0\left( (1-\kappa ^{\mathbb {P}}_{3,3})^{-2},n\right) \right] \\&\quad +2 \frac{\rho _{2,3} \lambda \Sigma _{2,2} \Sigma _{3,3}}{1-\omega } (1-\kappa ^{\mathbb {P}}_{2,2} )^{(2n-1)} \left[ \zeta _0\left( \frac{\omega }{(1-\kappa ^{\mathbb {P}}_{2,2})(1-\kappa ^{\mathbb {P}}_{3,3})},n\right) \right. \\&\quad \left. -\omega ^n \zeta _0\left( \frac{1}{(1-\kappa ^{\mathbb {P}}_{2,2})(1-\kappa ^{\mathbb {P}}_{3,3})},n\right) \right] ,\\ \mathcal {V}^{\mathbb {P},(3,3)}_{n}&=\text {Var}^{\mathbb{P}}(X^{(3)}_{t+n}|\mathcal {F}_t)=\Sigma _{3,3}^2\sum _{l=1}^n(1-\kappa ^{\mathbb {P}}_{3,3})^{2(n-l)}\text {Var}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,3})\\&=\Sigma _{3,3}^2 \left( 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{3,3})^2,n)\right) \end{aligned}$$
and
$$\begin{aligned} \mathcal {V}^{\mathbb {P},(1,2)}_{n}&=\mathcal {V}^{\mathbb {P},(2,1)}_{n}=\text {Cov}^\mathbb {P}(X^{(1)}_{t+n},X^{(2)}_{t+n}|\mathcal {F}_t)\\&=\Sigma _{1,1}\Sigma _{2,2}\sum _{l=1}^{n}(1-\kappa ^{\mathbb {P}}_{1,1})^{n-l} (1-\kappa ^{\mathbb {P}}_{2,2})^{n-l} \text {Cov}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,1},Z^{\mathbb {P}}_{t+l,2})\\&\quad +\lambda \Sigma _{1,1}\Sigma _{3,3}\sum _{l=1}^{n-1} (1-\kappa ^{\mathbb {P}}_{1,1})^{n-l} \frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{l}} \frac{\omega ^{l}-\omega ^{n}}{1-\omega } \text {Cov}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,1},Z^{\mathbb {P}}_{t+l,3})\\&=\Sigma _{1,1}\Sigma _{2,2}\rho _{1,2} \left[ 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{1,1})(1-\kappa ^{\mathbb {P}}_{2,2}),n) \right] \\&\quad +\lambda \Sigma _{1,1}\Sigma _{3,3}\rho _{1,3} \frac{(1-\kappa ^{\mathbb {P}}_{1,1})^n (1-\kappa ^{\mathbb {P}}_{2,2})^{n-1}}{1-\omega } \times \\&\quad \left( \zeta _0 \left( \frac{\omega }{(1-\kappa ^{\mathbb {P}}_{1,1})(1-\kappa ^{\mathbb {P}}_{3,3})},n\right) - \omega ^n \zeta _0 \left( \frac{1}{(1-\kappa ^{\mathbb {P}}_{1,1})(1-\kappa ^{\mathbb {P}}_{3,3})},n\right) \right) ,\\ \mathcal {V}^{\mathbb {P},(1,3)}_{n}&=\mathcal {V}^{\mathbb {P},(3,1)}_{n}=\text {Cov}^\mathbb {P}(X^{(1)}_{t+n},X^{(3)}_{t+n}|\mathcal {F}_t)\\&=\Sigma _{1,1}\Sigma _{3,3}\sum _{l=1}^{n} (1-\kappa ^{\mathbb {P}}_{1,1})^{(n-l)} (1-\kappa ^{\mathbb {P}}_{3,3})^{(n-l)} \text {Cov}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,1},Z^{\mathbb {P}}_{t+l,3})\\&=\Sigma _{1,1}\Sigma _{3,3}\rho _{1,3}\left[ 1 +\zeta _0( (1-\kappa ^{\mathbb {P}}_{1,1})(1-\kappa ^{\mathbb {P}}_{3,3}),n)\right] ,\\ \mathcal {V}^{\mathbb {P},(2,3)}_{n}&=\mathcal {V}^{\mathbb {P},(3,2)}_{n}=\text {Cov}^\mathbb {P}(X^{(2)}_{t+n},X^{(3)}_{t+n}|\mathcal {F}_t)\\&=\Sigma _{2,2}\Sigma _{3,3}\sum _{l=1}^{n}(1-\kappa ^{\mathbb {P}}_{2,2})^{n-l} (1-\kappa ^{\mathbb {P}}_{3,3})^{n-l} \text {Cov}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,2},Z^{\mathbb {P}}_{t+l,3})\\&\quad +\lambda \Sigma _{3,3}^2\sum _{l=1}^{n-1}\frac{ (1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1}}{(1-\kappa ^{\mathbb {P}}_{3,3} )^{l}} \frac{\omega ^{l}-\omega ^{n}}{1-\omega } (1-\kappa ^{\mathbb {P}}_{3,3})^{n-l}\text {Var}^\mathbb {P}(Z^{\mathbb {P}}_{t+l,3})\\&=\Sigma _{2,2}\Sigma _{3,3} \rho _{2,3} \left[ 1+\zeta _0((1-\kappa ^{\mathbb {P}}_{2,2})(1-\kappa ^{\mathbb {P}}_{3,3}),n)\right] \\&\quad +\lambda \Sigma _{3,3}^2 \frac{(1-\kappa ^{\mathbb {P}}_{2,2} )^{n-1} (1-\kappa ^{\mathbb {P}}_{3,3} )^{n}}{1-\omega } \left( \zeta _0\left( \frac{1}{(1-\kappa ^{\mathbb {P}}_{2,2} )(1-\kappa ^{\mathbb {P}}_{3,3} )},n\right) \right. \\&\quad - \omega ^n \zeta _0\left( (1-\kappa ^{\mathbb {P}}_{3,3})^{-2} ,n\right) \bigg) \!. \end{aligned}$$
\(\square\)
Proof of Theorem 4.1
Using Lemma (6.9), the futures prices can be obtained from the moment generating function of the normal distribution:
$$\begin{aligned} F_{\mathcal {T}_1,\mathcal {T}_2,\mathcal {T}_3}= & {} \mathbb {E^Q}\left[ P(\mathcal {T}_2,\mathcal {T}_3)\bigg | \mathcal {F}_{\mathcal {T}_1}\right] ,\\= & {} \mathbb {E^Q}\left[ A_{\tau _3}\exp \left[ -\Delta \mathcal {B}_{\tau _3}^\top X_{\mathcal {T}_2}\right] \bigg | \mathcal {F}_{\mathcal {T}_1}\right] \\= & {} A_{\tau _3} \exp \left[ -\Delta \sum ^3_{i=1} \mathcal {B}^{(i)}_{\tau _3} \tilde{\mathcal {M}}^{(i)}_{\mathcal {T}_1,\tau _2} +\frac{\Delta ^2}{2} \mathcal {B}^\top _{\tau _3} \mathcal {V}_{\tau _2} \mathcal {B}_{\tau _3} \right] . \end{aligned}$$
This implies
$$\begin{aligned} F_{\mathcal {T}_1,\mathcal {T}_2,\mathcal {T}_3} = \tilde{A}_{\tau _2,\tau _3} \exp \left[ -\Delta \sum ^3_{i=1} \tilde{\mathcal {B}}^{(i)}_{\tau _3} X^{(i)}_{\mathcal {T}_1} \right] \end{aligned}$$
where
$$\begin{aligned} \tilde{A}_{\tau _2,\tau _3}&= A_{\tau _3} \exp \bigg [ \frac{\Delta ^2}{2} \mathcal {B}^\top _{\tau _3} \mathcal {V}_{\tau _2} \mathcal {B}_{\tau _3} -\Delta \mathcal {B}^{(2)}_{\tau _3} (\theta ^\mathbb {Q}_2-\theta ^\mathbb {Q}_3)\left( 1-(1-\lambda )^{\tau _2}\right) \\&\quad \quad -\Delta \mathcal {B}^{(2)}_{\tau _3} \lambda \theta ^{\mathbb {Q}}_3 \left( \dfrac{\zeta _0(1-\lambda ,\tau _2+1)}{1-\lambda }-\tau _2(1-\lambda )^{\tau _2-1} \right) \bigg ) \\&\qquad -\Delta \mathcal {B}^{(3)}_{\tau _3} \theta ^\mathbb {Q}_3\left( 1-(1-\lambda )^{\tau _2}\right) \bigg ],\\ \tilde{\mathcal {B}}^{(1)}_{n}&= \mathcal {B}^{(1)}_{n}, \quad \tilde{\mathcal {B}}^{(2)}_{n} = \mathcal {B}^{(2)}_{n} (1-\lambda )^{n}, \quad \tilde{\mathcal {B}}^{(3)}_{n} = \mathcal {B}^{(3)}_{n} (1-\lambda )^{n} + \mathcal {B}^{(2)}_{n} \lambda n (1-\lambda )^{n-1}. \end{aligned}$$
\(\square\)
