**Proof.** Proof of Proposition 2.

We start by defining new processes *dSik*,*<sup>t</sup>* and *dSijt* with *i*, *j*, *k* = 1, 2, . . . , *n* and *t* ≥ 0.

$$\begin{split} dZ\_{i}(t) &= \sum\_{j=1}^{n} \left( \epsilon^{\delta t} \right)\_{ij} \left[ \left( \tilde{\mathbf{c}}\_{j} - \frac{1}{2} \right) \left( \sqrt{\tilde{\mathbf{v}}\_{j}(t)} + \frac{\tilde{b}\_{j}}{\sqrt{\tilde{\mathbf{v}}\_{j}(t)}} \right) dt + \left( \sqrt{\tilde{\mathbf{v}}\_{j}(t)} + \frac{\tilde{b}\_{j}}{\sqrt{\tilde{\mathbf{v}}\_{j}(t)}} \right) d\tilde{\mathbf{W}}\_{j} \right] \\ &+ \sum\_{k=1}^{n} \left[ \left( \sum\_{j=1}^{n} \left( \epsilon^{\delta t} \right)\_{ij} \frac{L\_{j}}{n} \right) dt + \left( \sum\_{j=1}^{n} \left( \epsilon^{\delta t} \right)\_{ij} \left( c\_{k} - \frac{1}{2} \right) a\_{jk}^{2} \right) \left( \sqrt{v\_{k}(t)} + \frac{b\_{k}}{\sqrt{v\_{k}(t)}} \right)^{2} dt \right] \\ &+ \sum\_{k=1}^{n} \left( \sum\_{j=1}^{n} \left( \epsilon^{\delta t} \right)\_{ij} a\_{jk} \right) \left( \sqrt{v\_{k}(t)} + \frac{b\_{k}}{\sqrt{v\_{k}(t)}} \right) d\mathsf{W}\_{k} \\ &= \sum\_{j=1}^{n} dS\_{t}^{ij} + \sum\_{k=1}^{n} dS\_{ik,t} \end{split}$$

By the dependence structure implied by the model, it follows that all *S* are independent for a fix *i*, hence we can transform the characteristic function using the processes *S* as follows:

$$\Phi\_{Z(t),\boldsymbol{\nu}(t)}(T,\omega) = \prod\_{k=1}^{n} E\left[\exp\{i\omega^{\prime}\left(\mathcal{S}\_{k,T} - \mathcal{S}\_{k,t}\right)\} \mid \mathcal{S}\_{t},\boldsymbol{\nu}(t)\right] \prod\_{j=1}^{n} E\left[\exp\{i\omega\_{l}\left(\mathcal{S}\_{T}^{j} - \mathcal{S}\_{t}^{j}\right)\} \mid \mathcal{S}\_{t},\boldsymbol{\nu}(t)\right]$$

For each factor *j* = 1, 2, ... , *n* we define *<sup>S</sup>*<sup>∗</sup>*k*,*<sup>t</sup>* = *<sup>ω</sup> <sup>S</sup>*·*k*,*<sup>t</sup>* = ∑*ni*=<sup>1</sup> *<sup>ω</sup>iSik*,*t*; the dynamics of *<sup>S</sup>*<sup>∗</sup>*k*,*<sup>t</sup>* can be expressed as

$$\begin{aligned} dS\_{k,t}^\* &= \omega^\prime d\mathcal{S}\_{k,t} \\ &= \left[ L(\omega, t) + h\_k(\omega, t) \left( \sqrt{v\_k(t)} + \frac{b\_k}{\sqrt{v\_k(t)}} \right)^2 \right] dt + \mathcal{g}\_k \left( \omega, t \right) \left( \sqrt{v\_k(t)} + \frac{b\_k}{\sqrt{v\_k(t)}} \right) d\mathcal{W}\_{k,t} \end{aligned}$$

where *hk*(*<sup>ω</sup>*, *t*) = ∑*nj*=<sup>1</sup> *ck* − 12 *a*2*jk fj*(*<sup>ω</sup>*, *t*), *<sup>L</sup>*(*<sup>ω</sup>*, *t*) = ∑*nj*=<sup>1</sup> *Lj n fj*(*<sup>ω</sup>*, *t*) and *gk* (*<sup>ω</sup>*, *t*) = ∑*nj*=<sup>1</sup> *ajk fj*(*<sup>ω</sup>*, *t*) and *fj*(*<sup>ω</sup>*, *t*) = ∑*ni*=<sup>1</sup> *ωi eβ<sup>t</sup>ij*. These three functions are deterministic, linear combinations of *fj*(*<sup>ω</sup>*, *t*).

Next, we find the characteristic function for the increments of *<sup>S</sup>*<sup>∗</sup>*k*,*t*:

$$E\left[\exp\left\{i\phi\left(S\_{\mathbf{k},T}^{\*}-S\_{\mathbf{k},t}^{\*}\right)\right\}\mid S\_{\mathbf{k},t}^{\*},v\_{\mathbf{k}}(t)=v\_{\mathbf{k},t}\right]=\Phi\_{\mathbf{G}G}\left(T,\phi;L(\omega),h\_{\mathbf{k}}(\omega),\varrho\_{\mathbf{k}}(\omega),\mathbf{x}\_{\mathbf{k}},\theta\_{\mathbf{k}},\mathbb{Z}\_{\mathbf{k}},\rho\_{\mathbf{k},\mathbf{r}}b\_{\mathbf{k}},c\_{\mathbf{k},\mathbf{r}}v\_{\mathbf{k},\mathbf{r}},S\_{\mathbf{k},t}^{\*}\right).$$

The generic function Φ*GG* is provided in Lemma A1.It follows similarly for idiosyncratic factors:

$$\begin{split} dS\_{t}^{\gamma^{j}} &= \omega^{\prime} dS\_{t}^{\bar{j}} \\ &= \left(\sum\_{i=1}^{n} \omega\_{i} \left(\sum\_{j=1}^{n} \left(\varepsilon^{\delta t}\right)\_{ij} \left(\tilde{c}\_{j} - \frac{1}{2}\right)\right)\right) \left(\sqrt{\bar{v}\_{j}(t)} + \frac{\tilde{b}\_{j}}{\sqrt{\bar{v}\_{j}(t)}}\right) dt \\ &+ \left(\sum\_{i=1}^{n} \omega\_{i} \sum\_{j=1}^{n} \left(\varepsilon^{\delta t}\right)\_{ij}\right) \left(\sqrt{\bar{v}\_{j}(t)} + \frac{\tilde{b}\_{j}}{\sqrt{\bar{v}\_{j}(t)}}\right) \, d\tilde{W}\_{j,t} \\ &= h\_{\bar{j}}(\omega, t) \left(\sqrt{\bar{v}\_{\bar{j}}(t)} + \frac{\tilde{b}\_{\bar{j}}}{\sqrt{\bar{v}\_{\bar{j}}(t)}}\right)^{2} dt + g\_{\bar{j}}\left(\omega, t\right) \left(\sqrt{\bar{v}\_{\bar{j}}(t)} + \frac{\tilde{b}\_{\bar{j}}}{\sqrt{\bar{v}\_{\bar{j}}(t)}}\right) \, d\tilde{W}\_{\bar{j},t}, \end{split}$$

where *hj*(*<sup>ω</sup>*, *t*) = ∑*ni*=<sup>1</sup> *ωi* ∑*nj*=<sup>1</sup> *eβ<sup>t</sup>ij c*#*j* − 12 and *gj* (*<sup>ω</sup>*, *t*) = ∑*ni*=<sup>1</sup> *ωi* ∑*nj*=<sup>1</sup> *eβ<sup>t</sup>ij*.

> Combining all pieces together, we obtain:

$$\begin{split} \Phi\_{Z(t),x(t)}(T,\omega) &= \prod\_{k=1}^{n} E\left[\exp\{i\omega' \left(S\_{k,T} - S\_{k,t}\right)\} \mid S\_{l\prime}, v(t)\right] \prod\_{j=1}^{n} E\left[\exp\{i\omega' \left(S\_{T}^{j} - S\_{l}^{j}\right)\} \mid S\_{l\prime}, v(t)\right] \\ &= \prod\_{k=1}^{n} \Phi\_{GG}\left(T, 1; L\_{k}(\omega), h\_{k}(\omega), g\_{k}(\omega), \kappa\_{k}, \theta\_{k}, \tilde{\xi}\_{k\prime}\rho\_{k\prime}b\_{k\prime}c\_{k\prime}v\_{k\prime}, S\_{k,t}^{\*}\right) \\ &\times \prod\_{j=1}^{n} \Phi\_{GG}\left(T, 1; 0, h\_{j}(\omega), g\_{j}(\omega), \tilde{\xi}\_{j\prime}\tilde{\theta}\_{j\prime}\tilde{\xi}\_{j\prime}\tilde{\rho}\_{j\prime}\tilde{b}\_{j\prime}\tilde{\omega}\_{j\prime}\tilde{v}\_{j\prime}, \tilde{v}\_{j\prime}^{\*}\right) \end{split}$$

**Lemma A1.** *Let the generic process be:*

$$\begin{aligned} dZ(t) &= \left[ L(t) + h(t) \left( \sqrt{v(t)} + \frac{b}{\sqrt{v(t)}} \right)^2 \right] dt + g'(t) \left( \sqrt{v(t)} + \frac{b}{\sqrt{v(t)}} \right) dW\_t \\ d\upsilon(t) &= a(\theta - \upsilon(t)) dt + \xi \sqrt{v(t)} dB(t) \\ \langle dB(t), dW(t) \rangle &= \rho dt \end{aligned}$$

*with g* (*t*) *differentiable, then*

$$\begin{split} \Phi\_{\mathsf{G},\mathsf{G}}\left(T,\emptyset;\mathrm{L},h,\mathrm{g},\kappa,\theta,\mathsf{E},\rho,b,\nu,\upsilon,\mathcal{Z}\_{l}\right) &= \exp\left\{u\int\_{t}^{T}A\left(s\right)ds\right\}\nu(t)^{\frac{b\nu}{\xi}g(t)}\exp\left\{-u\rho\frac{\mathcal{G}\left(t\right)\nu(t)}{\xi}\right\} \\ \times\mathbbm{E}\left[\nu(T)^{u\frac{b\nu}{\xi}g(T)}\exp\left\{u\left(\int\_{t}^{T}B(s)\nu(s)ds + \int\_{t}^{T}\mathsf{C}(s)\frac{1}{\nu(s)}ds + \int\_{t}^{T}D(s)\ln(\nu(s))ds + \rho\frac{\mathcal{G}\left(T\right)\nu(T)}{\xi}\right)\right\}\mid\mathcal{F}\_{\mathsf{I}}\right] \end{split}$$

*where A, B, C and D are provided in the proof.*

### **Proof.** Proof of Lemma A1.

Let the generic process be:

$$\begin{aligned} dZ(t) &= \left[ L(t) + h(t) \left( \sqrt{v(t)} + \frac{b}{\sqrt{v(t)}} \right)^2 \right] dt + \mathfrak{g}'(t) \left( \sqrt{v(t)} + \frac{b}{\sqrt{v(t)}} \right) dW\_t \\ dv(t) &= \mathfrak{a}(\theta - v(t)) dt + \mathfrak{f} \sqrt{v(t)} dB(t) \\ \langle dB(t), dW(t) \rangle &= \rho dt \end{aligned}$$

We want to find

$$\mathbb{E}\left[e^{\mathfrak{u}Z(T)}|\mathcal{F}\_{\mathfrak{t}}\right] = e^{\mathfrak{u}Z(\mathfrak{t})}\Phi\_{\text{GG}}\left(T,\mathfrak{q};\mathcal{L},\mathfrak{h},\mathfrak{g},\mathfrak{k},\mathfrak{e},\mathfrak{e},\mathfrak{e},\mathfrak{v},\mathfrak{v},\mathfrak{v},\mathfrak{Z}\_{\mathfrak{t}}\right)$$

Letting *ν*˜(*t*) = *g* (*t*) *ν*(*t*) and *ν*<sup>ˆ</sup>(*t*) = *g* (*t*)ln(*ν*(*t*)), we have following:

$$d\ddot{\boldsymbol{v}}(t) = a\theta \boldsymbol{g}\left(t\right)dt + \left(\boldsymbol{g}'\left(t\right) - \boldsymbol{a}\boldsymbol{g}\left(t\right)\right)\boldsymbol{v}\left(t\right)dt + \boldsymbol{g}\left(t\right)\boldsymbol{\xi}\sqrt{\boldsymbol{v}\left(t\right)}dB(t),\tag{A9}$$

and

$$\begin{split} d\psi(t) &= \frac{g'(t)}{\operatorname{g}\left(t\right)}\vartheta(t)dt + \frac{\partial\vartheta(t)}{\partial\ln\left(\nu(t)\right)}d\ln(\nu(t)) + \frac{1}{2}\frac{\partial^2\vartheta(t)}{\partial\ln\left(\nu(t)\right)^2} < d\ln(\nu(t)) > \\ &= \frac{g'(t)}{\operatorname{g}\left(t\right)}\vartheta(t)dt + \frac{\operatorname{g}\left(t\right)\frac{\pi}{\operatorname{\nu}}}{\sqrt{\nu(t)}}dB(t) + \operatorname{g}\left(t\right)\left(\frac{a\theta}{\nu(t)} - a\right)dt - \operatorname{g}\left(t\right)\frac{\frac{\mathcal{E}^2}{\operatorname{\nu}}}{2\nu(t)}dt \end{split} \tag{A10}$$

From Equations (A9) and (A10), we solve for 1 *Tt g* (*s*) *ν*(*s*)*dB*(*s*) and 1 *Tt* √*g*(*s*) *ν*(*s*) *dB*(*s*):

$$\int\_{t}^{T} \mathbf{g}\left(\mathbf{s}\right) \sqrt{\mathbf{v}(\mathbf{s})} dB(\mathbf{s}) = \frac{\underline{g}\left(\underline{T}\right)\mathbf{v}\left(\underline{T}\right) - \underline{g}\left(\underline{t}\right)\mathbf{v}\left(\underline{t}\right)}{\frac{1}{\underline{\delta}}} - \frac{a\theta}{\frac{1}{\underline{\delta}}} \int\_{t}^{T} \mathbf{g}\left(\mathbf{s}\right) ds - \frac{1}{\underline{\delta}} \int\_{t}^{T} \left(\mathbf{g}'\left(\mathbf{s}\right) - a\mathbf{g}\left(\mathbf{s}\right)\right) \mathbf{v}\left(\mathbf{s}\right) ds,\tag{A11}$$

$$\int\_{I}^{T} \frac{\varrho(s)}{\sqrt{\nu(s)}} dB(s) = \frac{1}{\xi} \ln \frac{\nu(T)\delta^{(T)}}{\nu(\mathfrak{l})^{\mathfrak{gl}(I)}} + \frac{a}{\xi} \int\_{I}^{T} \mathfrak{g}'(s) \, ds + \frac{1}{\xi} \left(\frac{\xi^{2}}{2} - a\theta\right) \int\_{I}^{T} \frac{\varrho(s)}{\nu(s)} ds - \frac{1}{\xi} \int\_{I}^{T} \mathfrak{g}'(s) \ln(\nu(s)) ds.\tag{A12}$$

Split *W*(*t*) into *B*(*t*) and its orthogonal part *<sup>B</sup>*(*t*)<sup>⊥</sup>:

$$\begin{aligned} Z(T) &= Z(t) + \int\_t^T L\left(s\right) ds + \int\_t^T h(s) \left(\sqrt{\nu(s)} + \frac{b}{\sqrt{\nu(s)}}\right)^2 ds \\ &+ \int\_t^T g\left(s\right) \left(\sqrt{\nu(s)} + \frac{b}{\sqrt{\nu(s)}}\right) \left(\rho dB(s) + \sqrt{1 - \rho^2} dB(s)^\perp\right), \end{aligned}$$

then substitute Equation (A11) and (A12) to eliminate *dB*(*t*). *Z*(*t*) can be rewritten now as:

$$\begin{split} Z(T) &= Z(t) + \int\_{t}^{T} L\left(s\right) ds + \int\_{t}^{T} h(s) \left(\nu(s) + 2b + \frac{b^{2}}{\nu(s)}\right) ds + \sqrt{1 - \rho^{2}} \int\_{t}^{T} g(s) \left(a\sqrt{\nu(s)} + \frac{b}{\sqrt{\nu(s)}}\right) dB(s)^{\perp} \\ &+ \rho \frac{\mathcal{g}\left(T\right)\nu\left(T\right) - \mathcal{g}\left(t\right)\nu\left(t\right)}{\xi} - \frac{a\theta\rho}{\overline{\xi}} \int\_{t}^{T} g\left(s\right) ds - \frac{\rho}{\overline{\xi}} \int\_{t}^{T} \left(\mathcal{g}'\left(s\right) - a\mathcal{g}\left(s\right)\right) \nu\left(s\right) ds \\ &+ \frac{b\rho}{\overline{\xi}} \ln \frac{\nu\left(T\right)\mathcal{g}^{\left(T\right)}}{\nu\left(t\right)\mathcal{g}^{\left(t\right)}} + \frac{ab\rho}{\overline{\xi}} \int\_{t}^{T} \mathcal{g}\left(s\right) ds + \frac{b\rho}{\overline{\xi}} \left(\frac{\xi^{2}}{2} - a\theta\right) \int\_{t}^{T} \frac{\mathcal{g}\left(s\right)}{\nu\left(s\right)} ds - \frac{b\rho}{\overline{\xi}} \int\_{t}^{T} \mathcal{g}'\left(s\right) \ln(\nu\left(s\right)) ds \end{split}$$

Grouping conveniently, we obtain:

$$\begin{split} Z(T) &= Z(t) + \int\_{t}^{T} \overline{A}(s) \, ds + \int\_{t}^{T} \overline{B}(s) \boldsymbol{\nu}(\mathbf{s}) \, ds + \int\_{t}^{T} \overline{C}(s) \frac{1}{\boldsymbol{\nu}(\mathbf{s})} \, ds + \int\_{t}^{T} D(s) \ln(\boldsymbol{\nu}(\mathbf{s})) \, ds \\ &+ \frac{b\rho}{\overline{\mathfrak{s}}} \ln \frac{\boldsymbol{\nu}(T)\boldsymbol{\nu}^{(T)}}{\boldsymbol{\nu}(\mathbf{t})^{\otimes(t)}} + \rho \frac{\operatorname{g}\left(T\right)\boldsymbol{\nu}(T) - \operatorname{g}\left(t\right)\boldsymbol{\nu}(t)}{\overline{\mathfrak{s}}} + \sqrt{1-\rho^{2}} \int\_{t}^{T} \operatorname{g}(\boldsymbol{s}) \left( a\sqrt{\boldsymbol{\nu}(\mathbf{s})} + \frac{b}{\sqrt{\boldsymbol{\nu}(\mathbf{s})}} \right) \, dB(\mathbf{s})^{\perp} .\end{split}$$

where

$$\begin{array}{rcl} \overline{A}\left(s\right) &=& L\left(s\right) + 2bh(s) + \left(\frac{a b \rho}{\overline{\xi}} - \frac{a \theta \rho}{\overline{\xi}}\right)g\left(s\right) \\\\ \overline{B}\left(s\right) &=& h(s) - \frac{\rho}{\overline{\xi}}\left(\mathcal{g}'\left(s\right) - a \mathcal{g}\left(s\right)\right) \\\\ \overline{C}\left(s\right) &=& b^2 h(s) + \frac{b \rho}{\overline{\xi}}\left(\frac{\overline{\xi}^2}{2} - a\theta\right)g\left(s\right) \\\\ D\left(s\right) &=& -\frac{b \rho}{\overline{\xi}}\mathcal{g}'\left(s\right) \end{array}$$

Let (G*t*)*t*≥0 denote the filtration generated by *<sup>ν</sup>*(*t*), *t* ≥ 0. Using iterated expectation and independence, we can write the conditional moment generating function of *Z*(*T*) as:

$$\begin{split} \mathbb{E}\left[\epsilon^{\mu Z(T)}|\mathcal{F}\_{t}\right] &= \mathbb{E}\left[\mathbb{E}\left[\epsilon^{\mu Z(T)}\mid\mathcal{F}\_{t}\bigcup\mathcal{G}\_{t}\right]\mid\mathcal{F}\_{t}\right] = \exp\left\{u\left(Z(t) + \int\_{t}^{T}A(s)ds\right)\right\}\nu(t)^{\frac{b\rho}{\xi}\mathfrak{g}(t)}\exp\left\{-u\rho\frac{\mathcal{G}\left(t\right)\nu(t)}{\overline{\xi}}\right\} \\ &\times \mathbb{E}\left[\exp\left\{u\left(\int\_{t}^{T}B(s)\nu(s)ds + \int\_{t}^{T}\mathcal{C}(s)\frac{1}{\nu(s)}ds + \int\_{t}^{T}D(s)\ln(\nu(s))ds + \frac{b\rho}{\xi}\ln\nu(T)\mathfrak{z}^{\xi(T)} + \rho\frac{\mathcal{G}\left(T\right)\nu(T)}{\overline{\xi}}\right\}\right\}\right] \\ &\times \mathbb{E}\left[\exp\left\{u\sqrt{1-\rho^{2}}\int\_{t}^{T}g(s)\left(\sqrt{\nu(s)} + \frac{b}{\sqrt{\nu(s)}}\right)dB(s)^{\perp}\right\} \mid\mathcal{F}\_{t}\bigcup\mathcal{G}\_{t}\right]. \end{split}$$

The inner expectation, conditioned on F*t* 5 G*<sup>t</sup>*, leads to a normal random variable with mean 0 and variance (Ito's Isometry) *u*<sup>2</sup>(<sup>1</sup> − *ρ*<sup>2</sup>) 1 *Tt g*<sup>2</sup>(*s*)(*ν*(*s*) + *b*2 *ν*(*s*) + <sup>2</sup>*b*)*ds*. Putting all together:

$$\begin{split} & \mathbb{E}\left[e^{\mu Z(T)}|\mathcal{F}\_{l}\right] = \exp\left\{u\left(Z(t) + \int\_{t}^{T} A\left(s\right)ds\right)\right\} \nu(t)^{\frac{\mathrm{d}\varphi}{\xi}\mathcal{E}\left(t\right)} \exp\left\{-u\rho\frac{\mathcal{G}\left(t\right)\nu(t)}{\tilde{\xi}}\right\} \\ & \times \mathbb{E}\left[\nu(T)^{\frac{\mathrm{d}\varphi}{\xi}\mathcal{S}\left(T\right)}\exp\left\{u\left(\int\_{t}^{T} B(s)\nu(s)ds + \int\_{t}^{T} \mathcal{C}(s)\frac{1}{\nu(s)}ds + \int\_{t}^{T} D(s)\ln(\nu(s))ds + \rho\frac{\mathcal{G}\left(T\right)\nu(T)}{\tilde{\xi}}\right)\right\} \mid \mathcal{F}\_{l}\right] . \end{split}$$

where

$$\begin{aligned} A(s) &=& \overline{A}(s) + \mu (1 - \rho^2) b \varrho^2(s) \\ B(s) &=& \overline{B}(s) + \frac{1}{2} \mu^2 (1 - \rho^2) \varrho^2(s) \\ C(s) &=& \overline{C}(s) + \frac{1}{2} \mu^2 (1 - \rho^2) b^2 \varrho^2(s) \end{aligned}$$

**Proof.** Proof of Corollary 1.

The proof starts similarly to Proposition 2. We start by defining new processes *dSij*,*<sup>t</sup>* and *dSit* with *i*, *j* = 1, 2, . . . , *n* and *t* ≥ 0.

$$\begin{split} dZ\_{i}(t) &= \left[ \left( \widetilde{c}\_{i} - \frac{1}{2} \right) \left( \sqrt{\widetilde{v}\_{i}(t)} + \frac{\widetilde{b}\_{i}}{\sqrt{\widetilde{v}\_{i}(t)}} \right) dt + \left( \sqrt{\widetilde{v}\_{i}(t)} + \frac{\widetilde{b}\_{i}}{\sqrt{\widetilde{v}\_{i}(t)}} \right) d\widetilde{W}\_{i} \right] \\ &+ \sum\_{j=1}^{n} \left[ \frac{L\_{i}}{n} dt + \left( c\_{j} - \frac{1}{2} \right) a\_{ij}^{2} \left( \sqrt{v\_{j}(t)} + \frac{b\_{j}}{\sqrt{v\_{j}(t)}} \right)^{2} dt + a\_{ij} \left( \sqrt{v\_{j}(t)} + \frac{b\_{j}}{\sqrt{v\_{j}(t)}} \right) d\mathcal{W}\_{j} \right] \\ &= dS\_{i}^{i} + \sum\_{j=1}^{n} dS\_{ij,t} \end{split}$$

By the dependence structure implied by the model, it follows that all *S* are independent for a fix *i*, hence we can transform the characteristic function using the processes *S* as follows:

$$\Phi\_{\mathbf{Z}(t),\mathbf{z}(t)}(T,\omega) = \prod\_{j=1}^{n} E\left[\exp\{i\omega^{\prime}\left(\mathcal{S}\_{j,T} - \mathcal{S}\_{j,t}\right)\} \mid \mathcal{S}\_{t},\mathbf{v}(t)\right] \prod\_{i=1}^{n} E\left[\exp\{i\omega\_{i}\left(\mathcal{S}\_{T}^{i} - \mathcal{S}\_{t}^{i}\right)\} \mid \mathcal{S}\_{t},\mathbf{v}(t)\right],$$

For each factor *j* = 1, 2, ... , *n* we define *S*∗*j*,*t* = *<sup>ω</sup> <sup>S</sup>*·*j*,*<sup>t</sup>* = ∑*ni*=<sup>1</sup> *<sup>ω</sup>iSij*,*t*, the dynamics of *S*∗*j*,*t* can be expressed as

$$\begin{aligned} dS\_{j,t}^\* &= \omega^j dS\_{j,t} \\ &= \left[ L(\omega) + h\_j(\omega) \left( \sqrt{v\_j(t)} + \frac{b\_j}{\sqrt{v\_j(t)}} \right)^2 \right] dt + g\_j \left( \omega \right) \left( \sqrt{v\_j(t)} + \frac{b\_k}{\sqrt{v\_j(t)}} \right) d\mathcal{W}\_{j,t}^\* \end{aligned}$$

where *hj*(*ω*) = ∑*ni*=<sup>1</sup> *ωi ci* − 12 *a*2*ij*, *<sup>L</sup>*(*ω*) = ∑*ni*=<sup>1</sup> *ωi Li n* and *gj* (*ω*) = ∑*ni*=<sup>1</sup> *<sup>ω</sup>iaij*. Next, we find the characteristic function for the increments of *<sup>S</sup>*<sup>∗</sup>*j*,*t*:

$$E\left[\exp\left\{i\phi\left(S^{\*}\_{\mathbf{j},T}-S^{\*}\_{\mathbf{j},\mathbf{t}}\right)\right\}\mid S^{\*}\_{\mathbf{j},t'}v\_{\mathbf{j}}(t)=v\_{\mathbf{j},\mathbf{t}}\right]=\Phi\_{\mathbf{G}}\left(T,\Phi;L,h\_{\mathbf{j}},\mathbf{g}\_{\mathbf{j}},\mathbf{x}\_{\mathbf{j}},\theta\_{\mathbf{j}},\mathbf{\xi}\_{\mathbf{j}},\mathbf{\rho}\_{\mathbf{j}},b\_{\mathbf{j}},c\_{\mathbf{j}},v\_{\mathbf{j},t},S^{\*}\_{\mathbf{j},\mathbf{t}}\right).$$

The generic function Φ*G* is provided in the Appendix B.

> Similarly for the idiosyncratic factors *dSit* with *i* = 1, 2, . . . , *n*:

$$dS\_t^{\*i} = \omega\_i dS\_t^i = h\_i(\omega) \left(\sqrt{\vec{v}\_i(t)} + \frac{\mathfrak{f}\_i}{\sqrt{\vec{v}\_i(t)}}\right)^2 dt + \mathcal{g}\_i(\omega) \left(\sqrt{\vec{v}\_i(t)} + \frac{\mathfrak{f}\_i}{\sqrt{\vec{v}\_i(t)}}\right) d\tilde{W}\_{i,t}$$

where *hi*(*ω*) = *ωi c*#*i* − 12and *gi* (*ω*) = *ωi*.

 Combining all pieces together, we obtain:

2.

$$\begin{split} \Phi\_{\mathcal{Z}(t),x(t)}(T,\omega) &= \prod\_{j=1}^{n} E\left[\exp\{i\omega'\left(S\_{j,T} - S\_{j,t}\right)\} \mid S\_{t\prime}v(t)\right] \prod\_{i=1}^{n} E\left[\exp\{i\omega\_{i}\left(S\_{T}^{i} - S\_{t}^{i}\right)\} \mid S\_{t\prime}v(t)\right] \\ &= \prod\_{j=1}^{n} \Phi\_{\mathcal{G}}\left(T, \phi\_{i}^{\prime}L, h\_{j\prime}, \mathfrak{g}\_{j\prime}, \kappa\_{j\prime}, \theta\_{j\prime}, \mathfrak{z}\_{j\prime}, \rho\_{j\prime}, c\_{j\prime}, v\_{j,t}, \mathcal{S}\_{j,t}^{\*}\right) \times \prod\_{i=1}^{n} \Phi\_{\mathcal{G}}\left(T, 1; 0, h\_{i\prime}, \mathfrak{g}\_{i\prime}, \tilde{\kappa}\_{i\prime}, \tilde{\theta}\_{i\prime}, \tilde{\rho}\_{i\prime}, \tilde{\rho}\_{i\prime}, \tilde{\upsilon}\_{i\prime}, \mathcal{S}\_{i\prime}^{\*}\right) \\ &= \prod\_{j=1}^{n} \end{split}$$

**Proof.**Proof ofCorollary

The proof uses Corollary 1, where we express the joint c.f. as the product of one dimensional c.f.s of 4/2 type.

$$\begin{split} \Phi\_{\mathbf{Z}(t),r(t)}(T,\omega) &= \prod\_{j=1}^{n} E\left[\exp\{i\omega'\left(S\_{j,T} - S\_{j,t}\right)\} \mid S\_{t\prime}v(t)\right] \prod\_{i=1}^{n} E\left[\exp\{i\omega\_{i}\left(S\_{T}^{i} - S\_{t}^{i}\right)\} \mid S\_{t\prime}v(t)\right] \\ &= \prod\_{j=1}^{n} \Phi\_{\mathbf{G}}\left(T, \varPhi\_{\mathbf{z}}^{\star}L, h\_{\mathbf{j}\cdot}\mathbf{g}\_{j\cdot}, \mathbf{\tilde{r}}\_{\mathbf{j}\cdot}\mathbf{\tilde{g}}\_{j\cdot}\mathbf{\tilde{r}}\_{\mathbf{j}\cdot}\mathbf{\tilde{r}}\_{\mathbf{j}\cdot}\mathbf{\tilde{r}}\_{\mathbf{j}\cdot}\mathbf{\tilde{r}}\_{\mathbf{j}\cdot}S\_{j,t}^{\*}\right) \times \prod\_{i=1}^{n} \Phi\_{\mathbf{G}}\left(T, \mathbf{1}\left(0, h\_{\mathbf{i}\cdot}\mathbf{\tilde{g}}\_{i\cdot}, \mathbf{\tilde{x}}\_{\mathbf{i}\cdot}\mathbf{\tilde{g}}\_{i\cdot}\mathbf{\tilde{r}}\_{\mathbf{i}\cdot}\mathbf{\tilde{r}}\_{\mathbf{i}\cdot}\mathbf{\tilde{z}}\_{\mathbf{i}\cdot}S\_{i\prime}^{\*}\right) \right) \end{split}$$

Hence, every one of these functions (Φ*G* (*<sup>T</sup>*, *φ*; *L*, *h*, *g*, *κ*, *θ*, *ξ*, *ρ*, *b*, *c*, *vt*, *Zt*) = E *euZ*(*T*) | <sup>F</sup>*t*) capture the c.f. of a process of the type:

$$\begin{aligned} dZ(t) &= \left[ L + h\left(\sqrt{v(t)} + \frac{b}{\sqrt{v(t)}}\right)^2 \right] dt + \mathfrak{g}\left(\sqrt{v(t)} + \frac{b}{\sqrt{v(t)}}\right) dW\_t \\ d\upsilon(t) &= \mathfrak{a}(\theta - v(t)) dt + \mathfrak{f}\sqrt{v(t)} dB(t) \\ \langle dB(t), dW(t) \rangle &= \rho dt \end{aligned}$$

It is not difficult to realize therefore that the c.f. given *v*(*T*) can be similarly computed for every one of those processes, hence one can infer:

$$\begin{split} \Phi\_{\mathbf{Z}(t),\mathbf{r}(T)}(T,\omega) &= \prod\_{j=1}^{n} \Phi\_{\mathbf{G},T} \left( T, \Phi; L, h\_{\mathbf{i}}, \operatorname{\mathbf{g}}\_{\mathbf{j}}, \operatorname{\mathbf{x}}\_{\mathbf{j}}, \theta\_{\mathbf{j}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{j}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{j}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{j}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{j}}} \right) \\ & \times \prod\_{i=1}^{n} \Phi\_{\mathbf{G},T} \left( T, \operatorname{\mathbf{1}}\_{i} \operatorname{\mathbf{0}}, h\_{\mathbf{i}}, \operatorname{\mathbf{\tilde{e}}\_{i}}, \operatorname{\mathbf{\tilde{e}}\_{i}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{\tilde{e}}\_{\mathbf{i}}}, \operatorname{\mathbf{S}\_{t}}^{\mathbf{i}} \right) \end{split}$$

where <sup>Φ</sup>*G*,*<sup>T</sup>* (*<sup>T</sup>*, *φ*; *L*, *h*, *g*, *κ*, *θ*, *ξ*, *ρ*, *b*, *c*, *vT*, *Zt*) = E*euZ*(*T*) | F*t* ∪ *v*(*T*) is provided next in Appendix B.
