A least-squares Monte Carlo approach to the estimation of enterprise risk

Abstract

The estimation of enterprise risk for financial institutions entails a re-evaluation of the company’s economic balance sheet at a future time for a (large) number of stochastic scenarios. The current paper discusses tackling this nested valuation problem based on least-squares Monte Carlo techniques familiar from American option pricing. We formalise the algorithm in an operator setting and discuss the choice of the regressors (“basis functions”). In particular, we show that the left singular functions of the corresponding conditional expectation operator present robust basis functions. Our numerical examples demonstrate that the algorithm can produce accurate results at relatively low computational costs.

Appendix: Proofs and technical material

Proof of Lemma 2.1

1) Let $A \in \mathcal{F}_{t}$, $0 \leq t \leq \tau$. Then, by “taking out what is known”,

$$\begin{aligned} \tilde{\mathbb{P}}[A] =& {\mathbb{E} }^{\tilde{\mathbb{P}}}\left [ { \textbf{1}}_{A} \right ] = {\mathbb{E} }^{\mathbb{P}}\bigg[ \frac{d \tilde{\mathbb{P}}}{d {\mathbb{P}}}{\textbf{1}}_{A} \bigg] = { \mathbb{E} }^{\mathbb{P}}\bigg[ {\mathbb{E} }^{\mathbb{P}}\Big[ \frac{\frac{d {\mathbb{Q}}}{d {\mathbb{P}}}}{{\mathbb{E} }^{\mathbb{P}}[ \frac{d {\mathbb{Q}}}{d {\mathbb{P}}} | \mathcal{F}_{\tau} ] }{ \textbf{1}}_{A}\Big| \mathcal{F}_{\tau} \Big] \bigg] = {\mathbb{P}}[A]. \end{aligned}$$

2) Let $X:\Omega \rightarrow \mathbb{R}$ be a random variable. Then similarly as above,

$$\begin{aligned} {\mathbb{E} }^{\tilde{\mathbb{P}}}[X | \mathcal{F}_{\tau} ] =& \frac{{\mathbb{E} }^{\mathbb{P}}[ \frac{d \tilde{\mathbb{P}}}{d {\mathbb{P}}} X | \mathcal{F}_{\tau} ] }{{\mathbb{E} }^{\mathbb{P}}[ \frac{d \tilde{\mathbb{P}}}{d {\mathbb{P}}} | \mathcal{F}_{\tau} ]} = {\mathbb{E} }^{\mathbb{P}}\bigg[ \frac{X \frac{d {\mathbb{Q}}}{d {\mathbb{P}}}}{{\mathbb{E} }^{\mathbb{P}}[ \frac{d {\mathbb{Q}}}{d {\mathbb{P}}} | \mathcal{F}_{\tau} ]} \bigg| \mathcal{F}_{\tau} \bigg] = {\mathbb{E} }^{{\mathbb{Q}}} [ X | \mathcal{F}_{\tau} ]. \end{aligned}$$

 □

Proof of Lemma 2.3

Linearity is obvious. For continuity, take $h^{(n)} \rightarrow h$. Then

$$\begin{aligned} & {\mathbb{E} }^{\mathbb{P}}[ (\mathcal{V}\,h^{(n)} - \mathcal{V}\,h )^{2}] \\ &= {\mathbb{E} }^{\mathbb{P}}\bigg[ \sum _{s,t = \tau}^{T} {\mathbb{E} }^{ \tilde{\mathbb{P}}} [ (h^{(n)}_{s} - h_{s} ) (Y_{s}) | Y_{\tau} ] { \mathbb{E} }^{\tilde{\mathbb{P}}} [ (h^{(n)}_{t} - h_{t} ) (Y_{t}) | Y_{ \tau} ] \bigg] \\ &\leq \sum _{s,t = \tau}^{T} \sqrt{ {\mathbb{E} }^{\mathbb{P}}\big[ \big({\mathbb{E} }^{\tilde{\mathbb{P}}} [ (h^{(n)}_{s} - h_{s} ) (Y_{s}) | Y_{\tau} ] \big)^{2}\big]} \, \sqrt{ {\mathbb{E} }^{\mathbb{P}}\big[ \big({\mathbb{E} }^{\tilde{\mathbb{P}}} [ (h^{(n)}_{t} - h_{t} ) (Y_{t}) | Y_{\tau} ] \big)^{2}\big]} \\ &\leq \sum _{s,t = \tau}^{T} \sqrt{ {\mathbb{E} }^{\tilde{\mathbb{P}}} [ (h^{(n)}_{s} - h_{s} )^{2} (Y_{s}) ]} \, \sqrt{ {\mathbb{E} }^{ \tilde{\mathbb{P}}} [ (h^{(n)}_{t} - h_{t} )^{2} (Y_{t}) ]} \longrightarrow 0 \qquad \text{as $n\rightarrow \infty $}. \end{aligned}$$

 □

Proof of Lemma 2.4

Let $\mathcal{V}^{(t)}$, $t=\tau ,\dots ,T$, be the component mappings of $\mathcal{V}$ introduced in (2.3), such that $\mathcal{V}^{(t)}: L^{2}(\mathbb{R}^{d},\mathcal{B}, \tilde{\mathcal{P}}_{Y_{t}}) \rightarrow L^{2}(\mathbb{R}^{d}, \mathcal{B},\tilde{\mathcal{P}}_{Y_{\tau}})$ is the conditional expectation. Under the assumption that there exists a joint density, $\mathcal{V}^{(t)}$ can be represented as

$$\begin{aligned} \mathcal{V}^{(t)}\,\mathbf{x}&=\int _{\mathbb{R}^{d}}\mathbf{x}(y)\, \pi _{Y_{t}|Y_{\tau}}(y|x)\,dy=\int _{\mathbb{R}^{d}}\mathbf{x}(y) \frac{\pi _{Y_{\tau},Y_{t}}(x,y)\,}{\pi _{Y_{\tau}}(x)}\,dy \\ &=\int _{\mathbb{R}^{d}}\mathbf{x}(y) \frac{\pi _{Y_{\tau},Y_{t}}(x,y)}{\pi _{Y_{t}}(y)\pi _{Y_{\tau}}(x)} \pi _{Y_{t}}(y)\,dy=\int _{\mathbb{R}^{d}}\mathbf{x}(y)\,k(x,y)\,\pi _{Y_{t}}(y) \,dy, \end{aligned}$$

where $\mathbf{x} \in {L}^{2}(\mathbb{R}^{d},\mathcal{B},\tilde{\mathbb{P}}_{Y_{t}})$, $\pi _{Y_{t}}(y)$ and $\pi _{Y_{\tau}}(x)$ are the marginal density functions for $Y_{t}$ and $Y_{\tau}$ in ${L}^{2}(\mathbb{R}^{d},\mathcal{B},\tilde{\mathbb{P}}_{Y_{t}})$ and ${L}^{2}(\mathbb{R}^{d},\mathcal{B}, {\mathbb{P}}_{Y_{\tau}})$, and $k(x,y)= \frac{\pi _{Y_{\tau},Y_{t}}(x,y)}{\pi _{Y_{t}}(y)\pi _{Y_{\tau}}(x)}$. Thus $\mathcal{V}^{(t)}$ is an integral operator with kernel $k(x,y)$, and

$$ \int _{\mathbb{R}^{d}}\int _{\mathbb{R}^{d}}|k(x,y)|^{2}\,\pi _{Y_{t}}(y) \pi _{Y_{\tau}}(x)\,dy\,dx= \int _{\mathbb{R}^{d}}\int _{\mathbb{R}^{d}} \pi _{Y_{t}|Y_{\tau}}(y|x)\, \pi _{Y_{\tau}|Y_{t}}(x|y)\,dy\,dx< \infty . $$

Thus $\mathcal{V}^{(t)}$ is a Hilbert–Schmidt operator (e.g. Alt [2, Proposition 10.15]) and therefore compact. Finally, $\mathcal{V}$ is the sum of the $\mathcal{V}^{(t)}$ and therefore also compact. □

Proof of Proposition 3.1

${\mathbb{P}}_{Y_{\tau}}$ is a finite Borel measure, and hence ${L}^{2}\left (\mathbb{R}^{d},\mathcal{B},{\mathbb{P}}_{Y_{\tau}} \right )$ is separable (see Cohn [16, Proposition 3.4.5]). Now if $e_{k},\;k=1,\dots ,M$, are independent, we can find by Gram–Schmidt an orthonormal system $S = \{f_{k}, k=1,\dots ,M\}$ with ${\mathrm{lin}}\{e_{k}, k=1,\dots ,M\} = {\mathrm{lin}}\, S$. For $S$, in turn, we find an orthonormal basis $\{f_{k}, k \in \mathbb{N}\} = S' \supseteq S$. Hence

$$ \widehat{C}_{\tau}^{(M)} = \sum _{k=1}^{M} \alpha _{k}\,e_{k} = \sum _{k=1}^{M} \underbrace{\tilde{\alpha}_{k}}_{\displaystyle{=\langle C_{\tau},f_{k} \rangle}}\,f_{k} \longrightarrow \sum _{k=1}^{\infty} \tilde{\alpha}_{k} \,f_{k} = C_{\tau} \qquad \text{as $M\rightarrow \infty $,} $$

where

$$ \| \widehat{C}_{\tau}^{(M)} - C_{\tau} \|^{2} = \sum _{k=M+1}^{\infty} | \langle C_{\tau},f_{k} \rangle |^{2} \longrightarrow 0 \qquad \text{as $M\rightarrow \infty $} $$

by Parseval’s identity. For the second part, we note that

$$ (\hat{\alpha}_{1}^{(N)},\dots ,\hat{\alpha}_{M}^{(N)})' = \hat{\alpha}^{(N)} = (A^{(M,N)} )^{-1} \frac{1}{N} \sum _{i=1}^{N} e (Y_{ \tau}^{(i)} )\,V_{\tau}^{(i)}, $$

where $e(\, \cdot \, ) = (e_{1}(\, \cdot \, ),\dots , e_{M}(\, \cdot \, ))'$ and

$$ A^{(M,N)} = \bigg( \frac{1}{N} \sum _{i=1}^{N} e_{k}(Y_{\tau}^{(i)}) \,e_{\ell}(Y_{\tau}^{(i)}) \bigg)_{1\leq k,\ell \leq M} $$

is invertible for large enough $N$ since we assume linear independence. Hence

$$ \hat{\alpha}^{(N)} \longrightarrow \alpha = (\alpha _{1},\dots , \alpha _{M})' = (A^{(M)} )^{-1} {\mathbb{E} }^{\tilde{\mathbb{P}}} \bigg[e (Y_{\tau} )\,\bigg(\sum _{t=\tau}^{T} \,x_{t} \bigg)\bigg] $$

$\tilde{\mathbb{P}}$-a.s. by the law of large numbers, where $A^{(M)} = ({\mathbb{E} }^{\tilde{\mathbb{P}}} [e_{k} (Y_{\tau} )\,e_{\ell}(Y_{\tau} ) ] )_{1\leq k,\ell \leq M}$. Thus

$$ \widehat{C}_{\tau}^{(M,N)} = e'\,\hat{\alpha}^{(N)} \longrightarrow e' \alpha = \widehat{C}_{\tau}^{(M)} $$

$\tilde{\mathbb{P}}$-a.s. Finally, for the third part, write

$$\begin{aligned} V_{\tau}^{(i)}&=\sum _{t=\tau}^{T}x_{t} (Y_{t}^{(i)} )=\sum _{k=1}^{M} \alpha _{k}\,e_{k} (Y_{\tau}^{(i)} )+\epsilon _{j}, \end{aligned}$$

where $\mathbb{E} [\epsilon _{j}|Y_{\tau} ]=0$, ${\mathrm{Var}}[\epsilon _{j}|Y_{ \tau} ]=\Sigma (Y_{\tau})$ and ${\mathrm{Cov}}[\epsilon _{i},\,\epsilon _{j}|Y_{\tau} ]=0$. Thus (see e.g. Amemiya [3, Chap. 1])

$$ \sqrt{N}(\alpha -\hat{\alpha}^{(N)}) \stackrel{\mathcal {D}}{\longrightarrow }{\mathcal{N}}\Big(0, \underbrace{ (A^{(M)} )^{-1}\big(\mathbb{E}^{\mathbb{P}} [e_{k}(Y_{\tau})e_{\ell}(Y_{\tau})\Sigma (Y_{\tau}) ]\big)_{1\le k, \,\ell \le M} (A^{(M)} )^{-1}}_{ \displaystyle{=\tilde{\xi}}}\Big) $$

so that

$$ \sqrt{N} (\widehat{C}_{\tau}^{(M)}-\widehat{C}_{\tau}^{(M,N)} )=e'( \alpha -\hat{\alpha}^{(N)})\sqrt{N} \stackrel{\mathcal {D}}{\longrightarrow} {\mathcal{N}}(0,\xi ^{(M)}), $$

where $\xi ^{(M)}=e'\,\tilde{\xi}\,e$. □

Proof of Proposition 3.2

Since $(V_{\tau}^{(i)},Y_{\tau}^{(i)})$ are i.i.d. as Monte Carlo trials, the first part of Newey [34, Assumption 1] is automatically satisfied. The conditions in the proposition are then exactly Assumptions 1 (part 2), 2 and 3 in Newey [34] for $d=0$. Thus the claim follows by the first part of Newey [34, Theorem 1]. □

Proof of Proposition 3.3

Analogously to the proof of Proposition 3.2, the first part of Newey [34, Assumption 1] is automatically satisfied. The conditions in the proposition are taken from the second part of Assumption 1, Assumption 8, the discussion following Assumption 8, and Assumption 9 in Newey [34]. Thus the claim follows by the first part of Newey [34, Theorem 4]. □

Proof of Proposition 4.2

We approximate $\mathcal{V}$ by an arbitrary rank-$M$ operator $\mathcal{V}_{F}^{(M,e)}$, which can be written as $\mathcal{V}_{F}^{(M,e)}=\sum _{k=1}^{M}\alpha _{k}\,\langle \,\cdot \,, u_{k}\rangle \,e_{k}$, where $(\alpha _{k})_{k=1}^{M} \subseteq \mathbb{R}_{+}$, $(u_{k})^{M}_{k=1}$ are orthonormal in ℋ and $(e_{k})_{k=1}^{M}$ are orthonormal in ${L}^{2} (\mathbb{R}^{d},\mathcal{B},{\mathbb{P}}_{Y_{\tau}} )$. Denote by $\mathcal{V}_{F}^{(M,e^{*})}$ the operator we obtain when choosing $(\alpha _{k}, u_{k}, e_{k})=(\omega _{k}, s_{k}, \varphi _{k})$. Then

$$\begin{aligned} \inf _{\mathcal{V}_{F}^{(M,e)}}\|\mathcal{V}-\mathcal{V}_{F}^{(M,e)} \|^{2}&\le \sup _{\|\mathbf{x}\|=1}\|\mathcal{V}\mathbf{x}- \mathcal{V}_{F}^{(M,e^{*})}\mathbf{x}\|^{2} \\ &=\sup _{\|\mathbf{x}\|=1}\bigg\| \sum _{k=M+1}^{\infty}\omega _{k} \langle \mathbf{x}, s_{k}\rangle \varphi _{j}\bigg\| ^{2} \\ &=\sup _{\|\mathbf{x}\|=1}\sum _{k=M+1}^{\infty}\omega _{k}^{2}\, \langle \mathbf{x}, s_{k}\rangle ^{2} \; =\; \omega _{M+1}^{2}. \end{aligned}$$

On the other hand, consider any alternative system $(\alpha _{k}, u_{k}, e_{k})$ for an arbitrary finite-rank operator $\mathcal{V}_{F}^{(M,e)}$. Then choose a non-zero $\mathbf{x}_{0}$ such that

$$ \mathbf{x}_{0}\in {\mathrm{lin}}\{s_{1},\dots ,s_{M+1}\}\, \cap \, {\mathrm{lin}} \{u_{1},\dots ,u_{M}\}^{\bot} \neq \{0\}. $$

Note that $\mathcal{V}-\mathcal{V}_{F}^{(M,e)}$ is compact and bounded. Therefore

$$\begin{aligned} \|\mathcal{V}-\mathcal{V}_{F}^{(M,e)}\|^{2}&\ge \frac{\|\mathcal{V}\mathbf{x}_{0}-\mathcal{V}_{F}^{(M,e)}\,\mathbf{x}_{0}\|^{2}}{\|\mathbf{x}_{0}\|^{2}} \\ &=\frac{\|\mathcal{V}\mathbf{x}_{0}\|^{2}}{\|\mathbf{x}_{0}\|^{2}}= \frac{\sum _{k=1}^{M+1}\omega _{k}^{2}|\langle \mathbf{x}_{0}, s_{k}\rangle |^{2}}{\sum _{k=1}^{M+1}|\langle \mathbf{x}_{0},s_{k}\rangle |^{2}} \\ & \ge \omega _{M+1}^{2}. \end{aligned}$$

Hence

$$ \inf _{\mathcal{V}_{F}^{(M,e)}}\|\mathcal{V}-\mathcal{V}_{F}^{(M,e)} \|^{2}=\omega _{M+1}^{2}=\|\mathcal{V}-\mathcal{V}_{F}^{(M,e^{*})}\|. $$

Now since $\inf _{\mathcal{V}_{F}^{(M,e)}}\|\mathcal{V}-\mathcal{V}_{F}^{(M,e)} \|^{2}=\inf _{\{e_{1},\dots ,e_{M}\}}\|\mathcal{V}-P^{(M,e)} \mathcal{V}\|^{2}$, the claim follows by (4.1). □

Proof of Proposition 4.3

Proceeding as in (4.3) and with (4.1), we obtain

$$\begin{aligned} & \inf _{\alpha _{M}} \sup _{y\in \mathcal{Y}} \bigg|C_{\tau}(y)- \sum _{k=1}^{M}\alpha _{M,k}\,e_{k}(y)\bigg| \\ &\le \sup _{y\in \mathcal{Y}} \bigg|C_{\tau}(y)-\widehat{C}_{\tau}^{(M)}(y) \bigg| =\sup _{y\in \mathcal{Y}}\bigg|\sum _{k=M+1}^{\infty}\omega _{k} \,{\langle \mathbf{x}, s_{k}\rangle}\,\varphi _{k}(y)\bigg| \\ &\le \sum _{k=M+1}^{\infty}\omega _{k} \, \|\mathbf{x}\|\,\|s_{k}\|\, \sup _{y\in \mathcal{Y}}|\varphi _{k}(y)| =\sum _{k=M+1}^{\infty} \omega _{k}\,\|\mathbf{x}\|\, \sup _{y\in \mathcal{Y}}|\varphi _{k}(y)|=O \left (\omega _{M}\right ) \end{aligned}$$

for a fixed $\mathbf{x}$ since the $\varphi _{k}$ are uniformly bounded, where the second inequality follows from the triangle and Cauchy–Schwarz inequalities.

Then, going through the assumptions of Proposition 3.2 with the choices $B=I$ and $e^{(M)}=(e_{1},\dots ,e_{M})'$, we obtain

$$ \mathbb{E}^{\tilde{\mathbb{P}}} [\tilde{e}^{(M)}(Y_{\tau})\tilde{e}^{(M)}(Y_{\tau})' ]=I $$

due to the orthonormality of the singular functions. Therefore, the smallest eigenvalue is bounded away from zero uniformly for every $M$. Moreover, for fixed $y\in \mathcal{Y}$, $|\tilde{e}^{(M)}(y)|=\sqrt{\varphi _{1}(y)^{2}+\cdots + \varphi _{M}(y)^{2}}$ so that

$$\begin{aligned} \sup _{y\in \mathcal{Y}}|\tilde{e}^{(M)}(y)| \le &\sqrt{\sum _{k=1}^{M} \,\sup _{y\in \mathcal{Y}}\varphi _{k}(y)^{2}}\le \sqrt{M \max _{1 \le k\le M} \sup _{y\in \mathcal{Y}}\varphi _{k}(y)^{2}}=C\sqrt{M}= \zeta _{0}(M) \end{aligned}$$

as the $\varphi _{k}$ are uniformly bounded. Thus the claim follows by Proposition 3.2. □

Proof of Lemma 4.4

The assertions on the conditional distributions are standard. In order to show that $\mathcal{V}$ is compact, we check that the transition and the reverse transition density functions satisfy the condition in Lemma 2.4. Note that the transition density function can be written as

$$\begin{aligned} \pi _{Y_{T}|Y_{\tau}}(y|x) &= g\big(y; \mu _{T}+\Gamma '\Sigma _{\tau}^{-1}(x- \mu _{\tau}), \Sigma _{T|\tau}\big) \\ &= \frac{1}{(2\pi )^{d/2}|\Sigma _{T|\tau}|^{1/2}} \frac{ |\Sigma _{\tau}(\Gamma ')^{-1}\Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{\tau} |^{1/2}}{ |\Sigma _{\tau}(\Gamma ')^{-1}\Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{\tau} |^{1/2}} \\ & \phantom{=:} \times \exp \bigg(-\frac{1}{2}\big(x-\mu _{\tau}-\Sigma _{\tau}( \Gamma ')^{-1}(y-\mu _{T})\big)'\Sigma _{\tau}^{-1}\Gamma \Sigma _{T| \tau}^{-1} \\ & \phantom{=\times \exp \Bigg(-\frac{1}{2}\big(} \times \Gamma '\Sigma _{\tau}^{-1}\big(x- \mu _{\tau}-\Sigma _{\tau}(\Gamma ')^{-1}(y-\mu _{T})\big)\bigg) \\ &= \frac{|\Sigma _{\tau}|}{|\Gamma |} g\big(x;\mu _{\tau}+\Sigma _{ \tau}(\Gamma ')^{-1}(y-\mu _{T}), \Sigma _{\tau}(\Gamma ')^{-1} \Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{\tau}\big). \end{aligned}$$

We evaluate the integral

$$\begin{aligned} &\int _{\mathbb{R}^{d}}\pi _{Y_{T}|Y_{\tau}}(y|x)\pi _{Y_{\tau}|Y_{T}}(x|y)dx \\ &=\frac{|\Sigma _{\tau}|}{|\Gamma |} \int _{\mathbb{R}^{d}}g\big(x; \mu _{\tau}+\Sigma _{\tau}(\Gamma ')^{-1}(y-\mu _{T}), \Sigma _{\tau}( \Gamma ')^{-1}\Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{\tau}\big)\, \\ & \phantom{=\frac{|\Sigma _{\tau}|}{|\Gamma |} \int _{\mathbb{R}^{d}}} \times g\big(x;\mu _{\tau}+\Gamma \Sigma _{\tau}^{-1}(y- \mu _{T}), \Sigma _{\tau |T}\big)\,dx \\ &=\frac{|\Sigma _{\tau}|}{|\Gamma |(2\pi )^{d/2}} \frac{1}{ |\Sigma _{\tau}(\Gamma ')^{-1}\Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{\tau}+\Sigma _{\tau |T} |^{1/2}} \\ & \phantom{=:} \times \exp \bigg(-\frac{1}{2}(y-\mu _{T})'V^{-1}(y-\mu _{T})\bigg) \\ &=C_{1}\, g(y;\mu _{T}, V), \end{aligned}$$

where

$$\begin{aligned} V^{-1} =& (\Gamma ^{-1}\Sigma _{\tau}-\Sigma _{T}^{-1}\Gamma ' ) \big(\Sigma _{\tau}(\Gamma ')^{-1}\Sigma _{T|\tau}\Gamma ^{-1}\Sigma _{ \tau}+\Sigma _{\tau |T}\big)^{-1} \big(\Sigma _{\tau}(\Gamma ')^{-1}- \Gamma \Sigma _{T}^{-1}\big) \end{aligned}$$

and $C_{1}$ is an appropriate constant to obtain $g(y; \mu _{T}, V)$ (see the results from Vinga [42] on the product of Gaussian densities). Therefore,

$$ \int _{\mathbb{R}^{d}}\int _{\mathbb{R}^{d}}\pi _{Y_{T}|Y_{\tau}}(y|x) \pi _{Y_{\tau}|Y_{T}}(x|y)\,dx\,dy=\int _{\mathbb{R}^{d}}C_{1}g(y;\mu _{T}, V)\,dy=C_{1} < \infty . $$

 □

Proof of Lemma 4.5

The operator $\mathcal{V}^{*}$ can be found via

$$\begin{aligned} {\langle \mathcal{V}h,m\rangle}_{\pi _{Y_{\tau}}}&=\int _{\mathbb{R}^{d}} (\mathcal{V}h)(x)\,m(x)\,\pi _{Y_{\tau}}(x)\,dx \\ &=\int _{\mathbb{R}^{d}} \bigg(\int _{\mathbb{R}^{d}} h(y)\pi _{Y_{T}|Y_{ \tau}}(y|x)\,dy\bigg)\,m(x)\,\pi _{Y_{\tau}}(x)\,dx \\ &=\int _{\mathbb{R}^{d}} h(y)\bigg(\int _{\mathbb{R}^{d}} m(x)\pi _{Y_{ \tau}|Y_{T}}(x|y)\,dx\bigg)\pi _{Y_{T}}(y)\,dy= {\langle h, \mathcal{V}^{*}m\rangle}_{\pi _{Y_{T}}}, \end{aligned}$$

where $(\mathcal{V}^{*}m)(y)=\int _{\mathbb{R}^{d}} m(x)\pi _{Y_{\tau}|Y_{T}}(x|y) \,dx$. We obtain for $\mathcal{V}\mathcal{V}^{*}$ that

$$\begin{aligned} (\mathcal{VV^{*}}\varphi )(x) &=\int _{\mathbb{R}^{d}} (\mathcal{V}^{*} \varphi )(s)\pi _{Y_{T}|Y_{\tau}}(s|x)\,ds \\ & =\int _{\mathbb{R}^{d}}\bigg[\int _{\mathbb{R}^{d}} \varphi (y) \pi _{Y_{\tau}|Y_{T}}(y|s)\,dy \bigg]\pi _{Y_{T}|Y_{\tau}}(s|x)\,ds \\ &=\int _{\mathbb{R}^{d}} \varphi (y) \underbrace{\int _{\mathbb{R}^{d}} \pi _{Y_{\tau}|Y_{T}}(y|s)\pi _{Y_{T}|Y_{\tau}}(s|x)\,ds}_{ \displaystyle{=K_{A}(x,y)}}\,dy. \end{aligned}$$

It is useful to express the reverse density as in the proof of Lemma 4.4 as

$$\begin{aligned} &g(y; \mu _{Y_{\tau}|s}, \Sigma _{{\tau}|T})= \frac{|\Sigma _{T}|}{|\Gamma |}g\big(s; \mu _{T}+\Sigma _{T}\Gamma ^{-1}(y- \mu _{\tau}),\,\, \Sigma _{T}\Gamma ^{-1}\Sigma _{{\tau}|T}(\Gamma ')^{-1} \Sigma _{T}\big). \end{aligned}$$

Hence we get

$$\begin{aligned} K_{A}(x,y)&=\int _{\mathbb{R}^{d}} \pi _{Y_{\tau}|Y_{T}}(y|s) \pi _{Y_{T}|Y_{ \tau}}(s|x)\,ds \\ &= \frac{1}{(2\pi )^{d/2} |\Gamma \Sigma _{T}^{-1} (\Sigma _{T}\Gamma ^{-1}\Sigma _{{\tau}|T}(\Gamma ')^{-1}\Sigma _{T}+\Sigma _{T|{\tau}} )\Sigma _{T}^{-1}\Gamma ' |^{1/2}} \\ & \phantom{=:} \times \exp \bigg(-\frac{1}{2}\big(y-\mu _{\tau}-\Gamma \Sigma _{T}^{-1} \Gamma '\Sigma _{{\tau}}^{-1}(x-\mu _{{\tau}})\big)'(\Gamma ^{-1})' \Sigma _{T}\, \\ & \phantom{= \times \exp \bigg(-\frac{1}{2}\big(} \times \big(\Sigma _{T}\Gamma ^{-1}\Sigma _{{ \tau}|T}(\Gamma ')^{-1}\Sigma _{T}+\Sigma _{T|{\tau}}\big)^{-1} \Sigma _{T}\Gamma ^{-1} \\ & \phantom{= \times \exp \bigg(-\frac{1}{2}\big(} \times \big(y-\mu _{\tau}-\Gamma \Sigma _{T}^{-1} \Gamma '\Sigma _{{\tau}}^{-1}(x-\mu _{{\tau}})\big)\bigg) \\ &=g \big(y; \mu _{\tau}+ \underbrace{\Gamma \Sigma _{T}^{-1}\Gamma '\Sigma _{{\tau}}^{-1}}_{ \displaystyle{=A}}(x-\mu _{{\tau}}),\, \Sigma _{\tau}-\Gamma \Sigma _{T}^{-1} \Gamma '\Sigma _{\tau}^{-1}\Gamma \Sigma _{T}^{-1}\Gamma '\big) \\ &=g\big(y; \underbrace{\mu _{\tau}+A(x-\mu _{{\tau}})}_{ \displaystyle{=\mu _{A}(x)}},\, \underbrace{\Sigma _{\tau}-A\Sigma _{\tau} A'}_{\displaystyle{= \Sigma _{A}}}\big) =g\big(y; \mu _{A}(x), \Sigma _{A}\big), \end{aligned}$$

where we again rely on results from Vinga [42]. The formula for $\mathcal{V}^{*}\mathcal{V}$ can be derived analogously. □

Proof of Lemma 4.6

We start by recalling the considerations from Khare and Zhou [28]. Let $(X_{t})$ with values in $\mathbb{R}^{d}$ be an MAR(1) process satisfying the equation

$$ X_{t}=\Phi X_{t-1}+\eta _{t},\qquad t\ge 1, $$

(7.1)

where $\Phi \in \mathbb{R}^{d\times d}$ and $(\eta _{t})_{t\ge 1}$ are i.i.d. $\sim N(0, H)$. The process $(X_{t})$ has a unique stationary distribution $N(0, \Sigma )$ if and only if $H=\Sigma -\Phi \Sigma \Phi '$, and it is reversible if and only if $\Phi \Sigma =\Sigma \Phi '$. [28] show that if these assumptions are satisfied, the transformed Markov operator for (7.1) has eigenvalues which are products of eigenvalues of $\Phi$ and the corresponding eigenfunctions are products of Hermite polynomials.

Now note that for $Y \sim K_{A}(x,\,\cdot \,)$, we can write

$$ Y-\mu _{\tau}=A(x-\mu _{\tau})+\zeta _{A}, $$

where $\zeta _{A}\sim N(0, \Sigma _{A})$. Since from Lemma 4.5, we have $\Sigma _{A}=\Sigma _{\tau}-A\Sigma _{\tau}A'$ and

$$ A\Sigma _{\tau}= \Gamma \Sigma _{T}^{-1}\Gamma ' = \Sigma _{\tau}\, A', $$

the operator $\mathcal{V}\,\mathcal{V}^{*}$ has for $\Sigma = \Sigma _{\tau}$ the same structure as the Markov operator for (7.1) which is reversible and stationary.

Following [28], denote by $\Sigma _{\tau}^{1/2}$ the square root matrix of $\Sigma _{\tau}$. Then

$$ \Sigma _{\tau}^{-1/2}A\Sigma _{\tau}^{1/2} = \Sigma _{\tau}^{-1/2}\, \Gamma \Sigma _{T}^{-1}\Gamma '\Sigma _{{\tau}}^{-1/2} $$

is symmetric and thus orthogonally diagonalisable as

$$ \Sigma _{\tau}^{-1/2} A \Sigma _{\tau}^{1/2} = P \Lambda P' $$

for an orthogonal matrix $P$. Then

$$ A=(\Sigma _{\tau}^{1/2} P)\Lambda (P'\Sigma _{\tau}^{-1/2}), $$

and the entries of the diagonal matrix $\Lambda$ are the eigenvalues of $A$.

Now for the transformation $z_{P}(Y)$ defined in (4.4), we obtain

$$\begin{aligned} \mathbb{E}_{K_{A}} [z^{P}(Y)|x ] =&P'\Sigma _{\tau}^{-1/2}A(x-\mu _{ \tau}) \\ =&P'\Sigma _{\tau}^{-1/2}\Sigma _{\tau}^{1/2}P\Lambda P'\Sigma _{ \tau}^{-1/2}(x-\mu _{\tau}) =\Lambda z^{P}(x) \end{aligned}$$

and

$$\begin{aligned} {\mathrm{Var}}_{K_{A}} [z^{P}(Y)|x ] =& P'\Sigma _{\tau}^{-1/2}\Sigma _{A} \Sigma _{\tau}^{-1/2}P \\ =& P'\Sigma _{\tau}^{-1/2}(\Sigma _{\tau}-A\Sigma _{\tau}A') \Sigma _{ \tau}^{-1/2}P =I-\Lambda ^{2}. \end{aligned}$$

Moreover,

$$\begin{aligned} \mathbb{E}_{\pi _{Y_{\tau}}} [z^{P}({Y_{\tau}}) ] =&P'\Sigma _{\tau}^{-1/2} \mathbb{E}_{\pi _{Y_{\tau}}}\left [Y_{\tau}-\mu _{\tau}\right ]=0, \\ {\mathrm{Var}}_{\pi _{Y_{\tau}}} [z^{P}({Y_{\tau}}) ] =& P'\Sigma _{\tau}^{-1/2} \Sigma _{\tau}\Sigma _{\tau}^{-1/2}P = I. \end{aligned}$$

The second part follows analogously. □

Proof of Proposition 4.7

For fixed $z_{i}^{P}(Y)$, from Carrasco and Florens [13], the univariate orthonormal Hermite polynomial of order $n_{i}$ is an eigenfunction under $K_{A}$, so that

$$ \mathbb{E}_{K_{A}}\big[h_{n_{i}}\big(z_{i}^{P}(Y)\big)\big|x\big]= \lambda _{i}^{n_{i}} h_{n_{i}}\big(z_{i}^{P}(x)\big). $$

Moreover, the products of these polynomials are also eigenfunctions since

$$ \mathbb{E}_{K_{A}}\bigg[\prod _{i=1}^{d}h_{n_{i}}\big(z_{i}^{P}(Y) \big)\bigg|x\bigg]=\prod _{i=1}^{d}\mathbb{E}_{K_{A}}\big[h_{n_{i}} \big(z_{i}^{P}(Y)\big)\big|x\big]. $$

The orthogonality of the eigenfunctions is proved in Khare and Zhou [28]. Note that the product of normalised Hermite polynomials is already normalised since

$$\begin{aligned} \mathbb{E}_{\pi _{Y_{\tau}}}\bigg[\bigg(\prod _{i=1}^{d}h_{n_{i}}\big(z_{i}^{P}(Y) \big)\bigg)^{2}\bigg] =\prod _{i=1}^{d}\mathbb{E}_{\pi _{Y_{\tau}}} \big[h_{n_{i}}\big(z_{i}^{P}(Y)\big)^{2}\big]=1. \end{aligned}$$

The right singular functions are obtained similarly from $z_{i}^{Q}(X)$. □

Proof of (5.5)

We have $B_{r}(t,T)=\frac{1-e^{-\alpha (T-t)}}{\alpha}$, $B_{\mu}(t,T)=\frac{e^{\kappa (T-t)}-1}{\kappa}$ and

$$\begin{aligned} A&\!(t,T) \\ =&\exp \bigg(\bar{\gamma}\big(B_{r}(t,T)-T+t\big) \\ &\hphantom{\exp \bigg(}{} +\frac{1}{2}\bigg( \frac{\sigma _{r}^{2}}{\alpha ^{2}}\Big(T-t-2B_{r}(t,T) + \frac{1-e^{-2\alpha (T-t)}}{2\alpha}\Big) \\ & \hphantom{\exp \bigg( +\frac{1}{2}\bigg(} +\frac{\delta ^{2}}{\kappa ^{2}}\Big(T-t-2B_{ \mu}(t,T)+\frac{e^{2\kappa (T-t)}-1}{2\kappa}\Big) \\ &\hphantom{\exp \bigg( +\frac{1}{2}\bigg(} + \frac{2\rho _{23}\sigma _{r}\delta}{\alpha \kappa}\Big(B_{\mu}(t,T)-T+t+B_{r}(t,T)- \frac{1-e^{-(\alpha -\kappa )(T-t)}}{\alpha -\kappa}\Big)\bigg)\bigg). \end{aligned}$$

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Proof of Lemma 5.1

Under ℙ, the solutions of (5.1)–(5.3) are

$$\begin{aligned} q_{\tau}&=q_{0}+\bigg(m-\frac{1}{2}\sigma _{S}^{2}\bigg)\tau +\sigma _{S} \int _{0}^{\tau}\,dW_{s}^{S}, \\ r_{\tau}&=r_{0}e^{-\alpha \tau}+\gamma (1-e^{-\alpha \tau} )+\sigma _{r} \int _{0}^{\tau}e^{-\alpha (\tau -t)}dW_{t}^{r}, \\ \mu _{x+\tau}&=\mu _{x} e^{\kappa \tau}+\delta \int _{0}^{\tau}e^{ \kappa (\tau -u)}dW_{u}^{\mu}. \end{aligned}$$

Thus $Y_{\tau}= (q_{\tau},r_{\tau},\mu _{x+\tau})'$ is distributed according to ${\mathcal{N}}(\mu _{\tau},\Sigma _{\tau})$ with

$$\begin{array}{rcl}{\mu }_{\tau }& =& \left(\begin{array}{c}{q}_0+(m-\frac{1}{2}{\sigma }_S^2)\tau \\ r_0{e}^{-\alpha \tau }+\gamma (1-e^{-\alpha \tau })\\ {\mu }_x{e}^{\kappa \tau }\end{array}\right),\\ {\mathrm{\Sigma }}_{\tau }& =& \left(\begin{array}{ccc}{\sigma }_S^2\tau & {\rho }_{12}{\sigma }_S{\sigma }_r{B}_r(0,\tau )& {\rho }_{13}{\sigma }_S\delta B_{\mu }(0,\tau )\\ {\rho }_{12}{\sigma }_S{\sigma }_r{B}_r(0,\tau )& {\sigma }_r^2\frac{1-e^{-2\alpha \tau }}{2\alpha }& {\rho }_{23}{\sigma }_r\delta \frac{1-e^{-(\alpha -\kappa )\tau }}{\alpha -\kappa }\\ {\rho }_{13}{\sigma }_S\delta B_{\mu }(0,\tau )& {\rho }_{23}{\sigma }_r\delta \frac{1-e^{-(\alpha -\kappa )\tau }}{\alpha -\kappa }& {\delta }^2\frac{e^{2\kappa \tau }-1}{2\kappa }\end{array}\right).\end{array}$$

To derive the distribution under $\mathbb{Q}_{E}$, first note that for $\tau \le s< T$,

$$\begin{aligned} r_{s}&=e^{-\alpha (s-\tau )}r_{\tau}+\bigg(\bar{\gamma} - \frac{\sigma _{r}^{2}}{\alpha ^{2}}\bigg) (1-e^{-\alpha (s-\tau )} ) \\ & \phantom{=:} +\frac{\sigma _{r}^{2}}{2\alpha ^{2}} (e^{-\alpha (T-s)}-e^{-\alpha (T+s-2 \tau )} ) \\ & \phantom{=:} -\frac{\rho _{23}\sigma _{r}\delta}{\kappa} \bigg( \frac{e^{\kappa (T-s)}-e^{-\alpha (s-\tau )+\kappa (T-\tau )}}{\alpha -\kappa}- \frac{1-e^{-\alpha (s-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} +\sigma _{r}\int _{\tau}^{s}e^{-\alpha (s-y)}dZ_{y}^{r}, \end{aligned}$$

so that $\int _{\tau}^{T} r_{s}\,ds$ can be evaluated, using the stochastic Fubini theorem, as

$$\begin{aligned} \int _{\tau}^{T} r_{s} ds &=\frac{1-e^{-\alpha (T-\tau )}}{\alpha} r_{\tau}+\bigg(\bar{\gamma}-\frac{\sigma _{r}^{2}}{\alpha ^{2}}\bigg) \bigg(T-\tau -\frac{1-e^{-\alpha (T-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} +\frac{\sigma _{r}^{2}}{2\alpha ^{2}}\bigg( \frac{1-e^{-\alpha (T-\tau )}}{\alpha}- \frac{e^{-\alpha (T-\tau )}-e^{-2\alpha (T-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} -\frac{\rho _{23}\sigma _{r}\delta}{\kappa}\bigg( \frac{e^{\kappa (T-\tau )}-1}{\kappa (\alpha -\kappa )}- \frac{e^{\kappa (T-\tau )}-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha (\alpha -\kappa )} \\ & \phantom{=:} \quad \qquad \qquad -\frac{1}{\alpha}\Big(T-\tau - \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\Big)\bigg) \\ & \phantom{=:} +\sigma _{r}\int _{\tau}^{T}\frac{1-e^{-\alpha (T-y)}}{\alpha}dZ_{y}^{r}. \end{aligned}$$

Thus under $\mathbb{Q}_{E}$ with known $Y_{\tau}$, the solutions of (5.6)–(5.8) are

$$\begin{aligned} q_{T}&= \mu _{q_{T}|q_{\tau}} +\sigma _{S}\int _{\tau}^{T}dZ_{s}^{S}+ \sigma _{r}\int _{\tau}^{T}\frac{1-e^{-\alpha (T-y)}}{\alpha}dZ_{y}^{r}, \\ r_{T}&=\mu _{r_{T}|r_{\tau}} +\sigma _{r}\int _{\tau}^{T}e^{-\alpha (T-y)}dZ_{y}^{r}, \\ \mu _{x+T}&=\mu _{\mu _{x+T}|\mu _{x+\tau}}+\delta \int _{\tau}^{T}e^{ \kappa (T-t)}dZ_{t}^{\mu}, \end{aligned}$$

so that the conditional distribution of $Y_{T}|Y_{\tau}$ is Gaussian with parameters

$$\begin{aligned} \mu _{r_{T}|r_{\tau}}& =e^{-\alpha (T-\tau )}r_{\tau}+\bigg(\bar{\gamma}- \frac{\sigma _{r}^{2}}{\alpha ^{2}}\bigg) (1-e^{-\alpha (T-\tau )} ) \\ & \phantom{=:} +\frac{\sigma _{r}^{2}}{2\alpha ^{2}} (1-e^{-2\alpha (T-\tau )} ), \\ \sigma _{r_{T}|r_{\tau}}^{2}&= \sigma _{r}^{2} \frac{1-e^{-2\alpha (T-\tau )}}{2\alpha} \\ & \phantom{=:} -\frac{\rho _{23}\sigma _{r}\delta}{\kappa}\bigg( \frac{1-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha -\kappa}- \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\bigg), \\ \mu _{\mu _{x+T}|\mu _{x+\tau}}&= \mu _{x+\tau}e^{\kappa (T-\tau )}- \frac{\rho _{23}\sigma _{r}\delta}{\alpha}\bigg( \frac{e^{\kappa (T-\tau )}-1}{\kappa}- \frac{1-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha -\kappa}\bigg), \\ \sigma _{\mu _{x+T}|\mu _{x+\tau}}^{2}&= \delta ^{2} \frac{e^{2\kappa (T-\tau )}-1}{2\kappa}, \\ \sigma _{r_{T}, \mu _{x+T}|r_{\tau}, \mu _{x+\tau}}& = \rho _{23} \sigma _{r}\delta \frac{1-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha -\kappa}, \\ \mu _{q_{T}|q_{\tau}} &= q_{\tau}+B_{r}(\tau ,T)r_{\tau}+\bigg( \bar{\gamma}-\frac{\sigma _{r}^{2}}{\alpha ^{2}}\bigg)\bigg(T-\tau - \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} +\frac{\sigma _{r}^{2}}{2\alpha ^{2}}\bigg( \frac{1-e^{-\alpha (T-\tau )}}{\alpha}- \frac{e^{-\alpha (T-\tau )}-e^{-2\alpha (T-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} -\frac{\rho _{23}\sigma _{r}\delta}{\kappa}\bigg( \frac{e^{\kappa (T-\tau )}-1}{\kappa (\alpha -\kappa )}- \frac{e^{\kappa (T-\tau )}-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha (\alpha -\kappa )} \\ & \phantom{=} \quad \qquad \qquad -\frac{1}{\alpha}\Big(T-\tau - \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\Big)\bigg)-\frac{1}{2}\sigma _{S}^{2}(T- \tau ) \\ & \phantom{=:} -\frac{\rho _{12}\sigma _{S}\sigma _{r}}{\alpha}\bigg(T-\tau - \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\bigg) \\ & \phantom{=:} -\frac{\rho _{13}\sigma _{S}\delta}{\kappa}\bigg( \frac{e^{\kappa (T-\tau )}-1}{\kappa}-T+\tau \bigg) \\ & \phantom{=:} -\frac{\delta ^{2}}{\kappa}\bigg( \frac{e^{2\kappa (T-\tau )}-1}{2\kappa}- \frac{e^{\kappa (T-\tau )}-1}{\kappa}\bigg), \\ \sigma _{q_{T}|q_{\tau}}^{2}&=\sigma _{S}^{2}(T-\tau )+ \frac{\sigma _{r}^{2}}{\alpha ^{2}}\bigg(T-\tau -2 \frac{1-e^{-\alpha (T-\tau )}}{\alpha}+ \frac{1-e^{-2\alpha (T-\tau )}}{2\alpha}\bigg) \\ & \phantom{=:} +\frac{2\rho _{12}\sigma _{S}\sigma _{r}}{\alpha}\bigg(T-\tau - \frac{1-e^{-\alpha (T-\tau )}}{\alpha}\bigg), \\ \sigma _{q_{T}, r_{T}|q_{\tau}, r_{\tau}} &=\rho _{12}\sigma _{S}\sigma _{r} \frac{1-e^{-\alpha (T-\tau )}}{\alpha}+\frac{\sigma _{r}^{2}}{\alpha} \frac{1-2e^{-\alpha (T-\tau )}+e^{-2\alpha (T-\tau )}}{2\alpha}, \\ \sigma _{q_{T}, \mu _{x+T}|q_{\tau}, \mu _{x+\tau}}& = \rho _{13} \sigma _{S}\delta \frac{e^{\kappa (T-\tau )}-1}{\kappa} \\ & \phantom{=:} +\frac{\rho _{23}\sigma _{r}\delta}{\alpha}\bigg( \frac{e^{\kappa (T-\tau )}-1}{\kappa}- \frac{1-e^{-(\alpha -\kappa )(T-\tau )}}{\alpha -\kappa}\bigg). \end{aligned}$$

We can write the conditional mean of $Y_{T}$ given $Y_{\tau}$ in the affine form

$$\left(\begin{array}{c}{\mu }_{q_T|q_{\tau }}\\ {\mu }_{r_T|r_{\tau }}\\ {\mu }_{{\mu }_{x+T}|{\mu }_{x+\tau }}\end{array}\right)=\underset{=H}{\underbrace{\left(\begin{array}{ccc}1& \frac{1-e^{-\alpha (T-\tau )}}{\alpha }& 0\\ 0& e^{-\alpha (T-\tau )}& 0\\ 0& 0& e^{\kappa (T-\tau )}\end{array}\right)}}\left(\begin{array}{c}{q}_{\tau }\\ r_{\tau }\\ {\mu }_{x+\tau }\end{array}\right)+C_{\tau }=H{Y}_{\tau }+C_{\tau },$$

where $C_{\tau}$ is a constant matrix defined by the remaining terms of the mean vector of $Y_{T}|Y_{\tau}$ after defining $HY_{\tau}$. The unconditional distribution of $Y_{T}$ under $\tilde{\mathbb{P}}$ is also Gaussian since $Y_{\tau}$ and $Y_{T}|Y_{\tau}$ follow Gaussian distributions. Thus it suffices to specify the mean vector and the covariance matrix of $Y_{T}$ under $\tilde{\mathbb{P}}$ to specify its distribution, and we have

$$\begin{aligned} \mu _{T}=\mathbb{E}^{\tilde{\mathbb{P}}}[Y_{T}]&=\mathbb{E}^{ \mathbb{P}}\big[\mathbb{E}^{\mathbb{Q}_{E}}[Y_{T}|Y_{\tau}]\big]= \mathbb{E}^{\mathbb{P}} [HY_{\tau}+C_{\tau} ]=H\mu _{\tau}+C_{\tau}, \\ \Sigma _{T} = {\mathrm{Cov}}^{\tilde{\mathbb{P}}}[Y_{T}]&={\mathrm{Cov}}^{ \mathbb{P}}\big[\mathbb{E}^{\mathbb{Q}_{E}}[Y_{T}|Y_{\tau}]\big]+ \mathbb{E}^{\mathbb{P}}\big[{\mathrm{Cov}}^{\mathbb{Q}_{E}}[Y_{T}|Y_{\tau}] \big] \\ &={\mathrm{Cov}}^{\mathbb{P}}\left [HY_{\tau}+C_{\tau}\right ]+\mathbb{E}^{ \mathbb{P}} [\Sigma _{T|\tau} ] =H\Sigma _{{\tau}}H'+\Sigma _{T|\tau}. \end{aligned}$$

The final step is to specify the joint distribution of $Y_{\tau}$ and $Y_{T}$ by finding ${\mathrm{Cov}}(Y_{\tau}, Y_{T})$. Note that

$$\begin{aligned} \Gamma = {\mathrm{Cov}}(Y_{\tau}, Y_{T})&=\mathbb{E}^{\tilde{\mathbb{P}}}[Y_{\tau}Y_{T}']-\mathbb{E}^{\tilde{\mathbb{P}}}[Y_{\tau}] \,\mathbb{E}^{ \tilde{\mathbb{P}}}[Y_{T}']=\mathbb{E}^{\mathbb{P}}\big[\mathbb{E}^{ \mathbb{Q}_{E}}[Y_{\tau}Y_{T}'|Y_{\tau}]\big]-\mu _{\tau}\mu _{T}' \\ &=\mathbb{E}^{\mathbb{P}} [Y_{\tau}(Y_{\tau}'H'+C_{\tau}') ]-\mu _{\tau} \mu _{T}'=\Sigma _{\tau}H'. \end{aligned}$$

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