2.2.1. Asset price

For $N=2$ , the model (3) corresponds to a one-dimensional stochastic volatility model under the identification $A=M=S$ , $\phi^1= \rho_1 \sqrt{v}$ , and $\phi^2 = \rho_2 \sqrt{v}$ , and v is a process driven by $W^1$ . Our analysis generalises [Reference Alòs, León and Vives4, Equation (2.1)] to the multi-factor case (in the continuous-path case). We refer to Section 5 for the details in the multi-factor setting and the analysis of the implied volatility.

2.2.2. VIX

The VIX is defined as $\mathrm{VIX}_{T}=\sqrt{\frac{1}{\Delta}\int_T^{T+\Delta}\mathbb{E}_T[v_t ] \mathrm{d}t}$ , where $\Delta$ is one month. The representation (2) yields that the underlying is the VIX future

$$M_{t}^{\mathrm{VIX}}\,:\!=\,\mathbb{E}_t[\mathrm{VIX}_T]=\mathbb{E}[\mathrm{VIX}_T]+(\mathbf{m}\bullet\mathbf{W})_t,\quad\text{with}\quad m^i_s=\frac{1}{2\Delta}\mathbb{E}_s\left[ \frac{1}{\mathrm{VIX}_T}\int_T^{T+\Delta}\mathrm{D}^{i}_s v_{r} \mathrm{d}r\right].$$

2.2.3. Asian options

For Asian options, the process of interest is $\mathcal{A}_T\,:\!=\,\frac{1}{T}\int_0^T S_t \mathrm{d}t$ . Using (2) we find

$$M^{\mathcal{A}}_{t}\,:\!=\,\mathbb{E}_t[\mathcal{A}_T]=\mathbb{E}[\mathcal{A}_T] + (\mathbf{m}\bullet\mathbf{W})_t,\qquad\text{with}\qquad m^i_s=\frac{1}{T} \int_s^T \mathbb{E}_s[\mathrm{D}^{i}_s S_{r}] \mathrm{d}r.$$

2.2.4. Multi-factor rough Bergomi

Rough volatility models can be written as $v_t=f(\mathbf{W}^H_t)$ , where $\mathbf{W}^H$ is an N-dimensional fractional Brownian motion with correlated components and $f\,:\,\mathbb{R}^N\to\mathbb{R}$ . For instance, in the two-factor rough Bergomi model,

\[v_t = v_0 \left( \chi \exp\left\{\nu W^{1,H}_t - \frac{\nu^2}{2} \mathbb{E}\left[\left(W^{1,H}_t\right)^2\right]\right\}+(1- \chi) \exp\left\{\eta W^{2,H}_t - \frac{\eta^2}{2} \mathbb{E}\left[\left(W^{2,H}_t\right)^2\right]\right\}\right),\]

with $\chi \in (0,1)$ , $\nu, \eta, v_0 \gt 0$ . In Example 2.2.2 we set $A=\mathrm{VIX}$ and hence $N=2$ , but in the asset price case we set $A=S$ and therefore $N=3$ even though the variance depends on only two factors.

6.2.1. Proof of Theorem 1: level

To prove this result, we draw insights from the proofs of [Reference Alòs, García-Lorite and Muguruza2, Theorem 8] and [Reference Alòs and Shiraya5, Proposition 3.1]. By definition

$$\mathcal{I}_{T}= \mathrm{BS}^{\leftarrow}(0,\mathfrak{M}_0,\mathfrak{M}_0,\Pi_0) =: \overleftarrow{\mathrm{BS}}(\Pi_0),$$

and we write $\widetilde{\mathrm{BS}}(x) \,:\!=\, \mathrm{BS}(0,x,x,u_0)$ . Using Proposition 5 at time 0, we see that $\Pi_0=\Gamma_T$ , where

$$\Gamma_t \,:\!=\, \mathbb{E}\left[\widetilde{\mathrm{BS}}(\mathfrak{M}_0) + \frac{1}{2} \int_0^t \partial_{x} G(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)|\boldsymbol\Theta_{s}|\mathrm{d}s\right],\qquad\text{for }t\in\mathbb{T},$$

which is a deterministic path. The fundamental theorem of integration reads

\begin{align*}\mathcal{I}_{T}=\overleftarrow{\mathrm{BS}}(\Gamma_T)= \overleftarrow{\mathrm{BS}}(\Gamma_0) + \int_0^T \partial_{t} \overleftarrow{\mathrm{BS}}(\Gamma_t) \mathrm{d}t&= \overleftarrow{\mathrm{BS}}(\Gamma_0) + \int_0^T \overleftarrow{\mathrm{BS}}^{\prime}(\Gamma_t) \partial_{t} \Gamma_t \mathrm{d}t\nonumber \\[3pt] & = \overleftarrow{\mathrm{BS}}(\Gamma_0) + \frac{1}{2} \int_0^T \overleftarrow{\mathrm{BS}}^{\prime}(\Gamma_t) \, \mathbb{E}\big[|\boldsymbol\Theta_{t}|\partial_{x} G_t \big] \mathrm{d}t,\end{align*}

where $G_t\,:\!=\, G(t,\mathfrak{M}_t,\mathfrak{M}_0,u_t)$ . We can deal with the integral by computing $\overleftarrow{\mathrm{BS}}^{\prime}$ and $\partial_x G$ explicitly:

\begin{align*}& \overleftarrow{\mathrm{BS}}^{\prime}(\Gamma_t) = \left(\mathrm{e}^{\mathfrak{M}_0} \mathcal{N}^{\prime}\left(d_+\left(\mathfrak{M}_t,\mathfrak{M}_0,\overleftarrow{\mathrm{BS}}(\Gamma_t)\right)\right) \sqrt{T-t} \right)^{-1}, \\[3pt] & \partial_x G (s,x,k,\sigma) = \frac{\mathrm{e}^{x} \mathcal{N}^{\prime}\big(d_+(x,k,\sigma)\big)}{\sigma\sqrt{T-s}} \left( 1 -\frac{d_+(x,k,\sigma)}{\sigma \sqrt{T-s}}\right).\end{align*}

Since $\Gamma:\mathbb{R}_+\to\mathbb{R}$ and $\overleftarrow{\mathrm{BS}}:\mathbb{R}\to\mathbb{R}$ are continuous, the following is uniformly bounded for all $T\le1$ :

\[\frac{\mathcal{N}^{\prime} \big(d_+(\mathfrak{M}_s,\mathfrak{M}_0,u_0)\big)}{\mathcal{N}^{\prime}\left(d_+\left(\mathfrak{M}_s,\mathfrak{M}_0,\overleftarrow{\mathrm{BS}}(\Gamma_s) \right)\right)} = \exp\left\{\frac{1}{8}\left( (T-s) \,\overleftarrow{\mathrm{BS}}(\Gamma_s)^2 - \mathfrak{u}_0^2\right) \right\}.\]

Therefore, by $\boldsymbol{(\mathrm{H}_{4})}$ we obtain

\[\lim_{T\downarrow0} \mathbb{E}\left[\int_0^T \overleftarrow{\mathrm{BS}}^{\prime}(\Gamma_t) |\boldsymbol\Theta_{t}|\partial_{x} G_t \mathrm{d}t \right] = \lim_{T\downarrow0} \mathbb{E}\left[ \int_0^T \frac{\mathcal{N}^{\prime}\left(d_+\left(\mathfrak{M}_t,\mathfrak{M}_0,u_t\right) \right)}{\mathcal{N}^{\prime} \left(d_+\left(\mathfrak{M}_t,\mathfrak{M}_0,\overleftarrow{\mathrm{BS}}(\Gamma_t)\right) \right)} \frac{\lvert \boldsymbol\Theta_t\lvert}{2 \mathfrak{u}_t^2 \sqrt{T}} \mathrm{d}t\right] =0.\]

Since $\Gamma_0=\mathbb{E}\left[\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right]$ and $u_0=\overleftarrow{\mathrm{BS}}\left(\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right)$ , we have

(17) \begin{align}\overleftarrow{\mathrm{BS}}(\Gamma_0)= \overleftarrow{\mathrm{BS}}\left(\mathbb{E}\left[\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right]\right) - \mathbb{E}[u_0 - u_0]= \mathbb{E}\left[ \overleftarrow{\mathrm{BS}}\Big(\mathbb{E}\left[\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right] \Big) - \overleftarrow{\mathrm{BS}}\Big(\widetilde{\mathrm{BS}}(\mathfrak{M}_0) \Big)\right] + \mathbb{E}[u_0].\end{align}

The Clark–Ocone formula yields

$$\widetilde{\mathrm{BS}}(\mathfrak{M}_0) =\mathbb{E}\left[\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right]+\sum_{i=1}^N\int_0^T \mathbb{E}_s\left[\mathrm{D}^{i}_s \widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right] \mathrm{d} W^i_s,$$

and by the gamma–vega–delta relation (12) we have

(18) \begin{align}\partial_\sigma \mathrm{BS} (0,x,x,\sigma) =\exp\left\{x-\frac{\sigma^2}{8}T\right\}\sqrt{\frac{T}{2\pi}},\end{align}

which in turn implies

(19) \begin{align}U^i_s \,:\!=\, \mathbb{E}_s\left[\mathrm{D}^{i}_s \widetilde{\mathrm{BS}}(\mathfrak{M}_0)\right]= \mathbb{E}_s \left[ \frac{\partial_{\sigma} \widetilde{\mathrm{BS}}(\mathfrak{M}_0)}{2 u_0 \sqrt T} \int_0^T \mathrm{D}^i_s {\left\|{\boldsymbol\phi_r}\right\|}^2 \mathrm{d}r \right] = \mathbb{E}_s \left[ \frac{\mathrm{e}^{\mathfrak{M}_0 - \frac{1}{8}\mathfrak{u}_0^2}}{2\sqrt{2\pi} u_0} \int_0^T \mathrm{D}^i_s {\left\|{\boldsymbol\phi_r}\right\|}^2 \mathrm{d}r \right].\end{align}

Define $\Lambda_r\,:\!=\,\mathbb{E}_r \Big[\widetilde{\mathrm{BS}}(\mathfrak{M}_0)\Big]$ , so that the difference we are interested in from (17) reads, after we apply the standard Itô’s formula,

(20) \begin{align}\overleftarrow{\mathrm{BS}}(\Gamma_0)-\mathbb{E}[u_0] & = \mathbb{E}\Big[ \overleftarrow{\mathrm{BS}} ( \Lambda_0 ) - \overleftarrow{\mathrm{BS}} (\Lambda_T ) \Big] \nonumber\\[3pt] & = -\sum_{i=1}^N \mathbb{E}\left[\int_0^T \overleftarrow{\mathrm{BS}}^{\prime}(\Lambda_s) U^i_s \mathrm{d} W^i_s+\frac{1}{2}\int_0^T\overleftarrow{\mathrm{BS}}^{\prime\prime}(\Lambda_s) (U^i_s)^2 \mathrm{d}s\right].\end{align}

The stochastic integral above has zero expectation by the same argument as used for [Reference Alòs and Shiraya5, Proposition 3.1]. Moreover, $\boldsymbol{(\mathrm{H}}_{\boldsymbol{5}}\boldsymbol{)}$ states that $\mathfrak{u}_0$ is dominated almost surely by $Z\in L^p$ , and therefore so are $\Lambda$ and

\[\overleftarrow{\mathrm{BS}}^{\prime\prime}(\Lambda_s) = \frac{\overleftarrow{\mathrm{BS}}(\Lambda_s)}{4 \mathrm{e}^{2\mathfrak{M}_s} \mathcal{N}^{\prime}\Big(d_+\big(\mathfrak{M}_s,\mathfrak{M}_0,\overleftarrow{\mathrm{BS}}(\Lambda_s)\big)\Big)^2},\]

by continuity. Plugging in the expression for $U^i$ from (19), we apply $\boldsymbol{(\mathrm{H}}_{\boldsymbol{5}}\boldsymbol{)}$ to conclude that the second integral of (20) tends to zero.

6.2.2. Proof of Theorem 2: skew

This proof follows from arguments similar to those of [Reference Alòs, León and Vives4, Proposition 5.1]. We recall that $\Pi_0(k)= \mathrm{BS}\big(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k)\big)$ . On the one hand, by the chain rule we have

(21) \begin{align}\partial_{k} \Pi_0(k) = \partial_{k}\mathrm{BS}\big(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k)\big)+\partial_{\sigma}\mathrm{BS}\big(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k)\big) \partial_{k} \mathcal{I}_{T}(k).\end{align}

On the other hand, the decomposition obtained in Proposition 5 yields

(22) \begin{align}\partial_{k} \Pi_0(k)= \mathbb{E}\big[\partial_{k}\mathrm{BS}(0,\mathfrak{M}_T,k,u_{0})\big]+\mathbb{E}\left[ \int_{0}^{T} \frac{1}{2} \partial_{xk} G(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)|\boldsymbol\Theta_{s}|\mathrm{d}s\right].\end{align}

Equating (21) and (22) gives

(23) \begin{align}\partial_{k} \mathcal{I}_{T}(k) & = \frac{\mathbb{E}\left[\partial_{k} \mathrm{BS}(0,\mathfrak{M}_T,k,u_{0})\right] - \partial_{k} \mathrm{BS}(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k))}{\partial_{\sigma} \mathrm{BS}(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k))} \nonumber \\[3pt] & \qquad +\frac{\mathbb{E}\left[ \int_{0}^{T}\partial_{xk} G(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)|\boldsymbol\Theta_{s}|\mathrm{d}s\right]}{2\partial_{\sigma}\mathrm{BS}(0,\mathfrak{M}_0,k,\mathcal{I}_{T}(k))},\end{align}

which in particular also holds for $k=\mathfrak{M}_0$ . Performing simple algebraic manipulations and using the derivatives of the Black–Scholes function ATM as in [Reference Alòs, León and Vives4, Proposition 5.1], we find the following (remember we drop the k-dependence in $\mathcal{I}_{T}$ when ATM):

$$\mathbb{E}\big[\partial_{k}\mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,u_{0}) \big]- \partial_{k}\mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})= \frac{1}{2} \mathbb{E}\left[\int_0^T \frac{1}{2}\partial_{x} G(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)|\boldsymbol\Theta_{s}|\mathrm{d}s\right].$$

By (23), this in turn yields

(24) \begin{align}\partial_{k} \mathcal{I}_{T}= \frac{\mathbb{E}\left[ \int_{0}^{T} L(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)|\boldsymbol\Theta_{s}|\mathrm{d}s\right]}{\partial_{\sigma}\mathrm{BS}\Big(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T} \Big) },\end{align}

where $L\,:\!=\,(\frac{1}{2}+\partial_{k})\frac{1}{2}\partial_{x} G$ . We write $L_s\,:\!=\,L(s,\mathfrak{M}_s,\mathfrak{M}_0,u_s)$ for simplicity and apply Lemma 2 to $L_s\, \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r$ , which yields

\begin{align*}\mathbb{E} \left[ \int_0^T L_s |\boldsymbol\Theta_s|\mathrm{d}s \right]& = \mathbb{E} \left[L_0 \int_0^T|\boldsymbol\Theta_s| \mathrm{d}s \right]+ \mathbb{E} \left[ \int_0^T (\partial_{x}^3 - \partial_{x}^2) L_s |\boldsymbol\Theta_s| \left( \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r \right) \mathrm{d}s \right] \\[3pt] & + \mathbb{E} \left[ \int_0^T \partial_{x} L_s \sum_{j=1}^N \left\{ \phi^j_s \, \mathrm{D}^j_s \left( \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r \right) \right\} \mathrm{d}s \right]=: R_1 + R_2 + R_3.\end{align*}

We combine (18) with the bound $\partial_k \partial_x^n G(t,x,k,\sigma) \le C \big(\sigma\sqrt{T-t}\big)^{-n-2}$ from Corollary 2 to obtain

\begin{align*}\frac{R_2}{\partial_\sigma \mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})}&\le \frac{C}{\sqrt{T}}\mathbb{E}\left[\int_0^T \frac{|\boldsymbol\Theta_s|}{\mathfrak{u}_s^{6}} \left( \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r \right) \mathrm{d}s\right], \\[3pt] \frac{R_3}{\partial_\sigma \mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})}&\le\frac{C}{\sqrt{T}} \mathbb{E}\left[\int_0^T \frac{1}{\mathfrak{u}_s^{4}} \sum_{j=1}^N \left\{ \phi^j_s \mathrm{D}^j_s \left( \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r \right) \right\} \mathrm{d}s\right],\end{align*}

and both converge to zero by $\boldsymbol{(\mathrm{H}}_{\boldsymbol{6}}^{\lambda}\boldsymbol{)}$ . We are left with $R_1$ . From Section 6.6, we have

$$L(0,x,x,u) = \frac{\exp\{x-\frac{u^2}{8}T\}}{u\sqrt{2\pi T}}\left( \frac14 + \frac{1}{2u^2 T} \right),$$

and therefore by (18),

$$\frac{L_0}{\partial_\sigma \mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})}= \left( \frac14 + \frac{1}{2u_0^2 T} \right) \frac{1}{u_0 T} \exp\left\{ -\frac{T}{8} \Big(u_0^2 - \mathcal{I}_{T}^2 \Big)\right\}.$$

This yields

\[\displaystyle \frac{R_1}{T^{\lambda}\partial_\sigma \mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})}= \mathbb{E}\left[ \left(\frac{\mathfrak{u}_0^2}{2}+1\right) \exp\left\{ -\frac{T}{8} \Big(u_0^2 - \mathcal{I}_{T}^2 \Big)\right\} \mathfrak{K}_T\right],\]

where $\displaystyle\mathfrak{K}_T\,:\!=\,\frac{\int_0^T |\boldsymbol\Theta_s|\mathrm{d}s}{2 T^{\frac{1}{2}+\lambda} \mathfrak{u}_0^3}$ . Furthermore,

\[\sup_{\omega\in\Omega} \left|\exp\left\{ -\frac{T}{8} \Big(u_0^2 - \mathcal{I}_{T}^2 \Big)\right\} -1\right|= \sup_{\omega\in\Omega} \left| \exp\left\{ \frac18 \left(T \mathcal{I}_{T}^2 - \mathfrak{u}_0^2 \right)\right\} -1\right|\]

not only is finite but converges to zero as T goes to zero. Hence,

\[\lim_{T\downarrow0} \mathbb{E}\left[ \left(\frac{\mathfrak{u}_0^2}{2}+1\right) \exp\left\{ -\frac{T}{8} \Big(u_0^2 - \mathcal{I}_{T}^2 \Big)\right\} \mathfrak{K}_T\right] = \lim_{T\downarrow0} \mathbb{E}\left[ \left(\frac{\mathfrak{u}_0^2}{2}+1\right) \mathfrak{K}_T\right].\]

We can finally conclude by $\boldsymbol{(\mathrm{H}}_{\boldsymbol{7}}^{\lambda}\boldsymbol{)}$ that

\[\displaystyle \lim_{T\downarrow 0} \frac{R_1}{T^{\lambda}\partial_\sigma \mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T})}= \lim_{T\downarrow 0} \mathbb{E}[\mathfrak{K}_T],\]

which has a finite limit.

6.2.3. Proof of Theorem 3: curvature

Step 1. Let us start by simply taking a second derivative with respect to k (we write $\mathrm{BS}(\mathfrak{M}_0,\mathcal{I}_{T}(k))$ as short for $\mathrm{BS}(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}(k))$ ):

\begin{align*}\partial_{k} \Big( \partial_{\sigma}\mathrm{BS}\big(\mathfrak{M}_0,\mathcal{I}_{T}(k)\big) \,\partial_{k} \mathcal{I}_{T}(k) \Big) & = \partial_{\sigma}\mathrm{BS}\big(\mathfrak{M}_0,\mathcal{I}_{T}(k)\big) \partial_{k}^2 \mathcal{I}_{T}(k) \\[3pt] &\quad+\Big[ \partial_{k\sigma}\mathrm{BS}\big(\mathfrak{M}_0, \mathcal{I}_{T}(k)\big)+ \partial_{\sigma}^2\mathrm{BS}\big(\mathfrak{M}_0,\mathcal{I}_{T}(k)\big) \partial_{k} \mathcal{I}_{T}(k) \Big] \partial_{k} \mathcal{I}_{T}(k).\end{align*}

Taking the derivative with respect to k in (24) and equating with the above formula yields

\begin{align*}\partial_{k}^2 \mathcal{I}_{T}(k) \partial_{\sigma}\mathrm{BS}(\mathfrak{M}_0,\mathcal{I}_{T}) & =- \partial_{\sigma}^2\mathrm{BS}\big(\mathfrak{M}_0, \mathcal{I}_{T}(k)\big) \partial_{k} \mathcal{I}_{T}(k)^2 - \partial_{k\sigma}\mathrm{BS}\big(\mathfrak{M}_0, \mathcal{I}_{T}(k)\big) \partial_{k} \mathcal{I}_{T}(k)\\[3pt] &\quad + \mathbb{E} \left[ \int_0^T \partial_{k} L(s,\mathfrak{M}_s,\mathfrak{M}_0,u_{s} ) |\boldsymbol\Theta_s| \mathrm{d}s \right]=: T_1 + T_2 + T_3.\end{align*}

A similar expression is presented in [Reference Alòs and León3], and we notice that $T_1$ and $T_2$ in the expression above, after being multiplied by $T^{-\lambda}$ , are identical to those from [Reference Alòs and León3, Equation (25)] and can therefore be dealt with in the same way. Step 1 shows that $T^{-\lambda} T_1$ tends to zero as $T\downarrow 0$ , and Step 2 yields $T_2= - \frac{1}{2} \partial_{k} \mathcal{I}_{T}(k)$ .

Step 2. Recall that $L = \frac{1}{2} \left(\frac{1}{2} + \partial_{k} \right) \partial_{x} G$ . We need the anticipative Itô’s formula (Lemma 2) twice on $T_3$ . Indeed, even though the bound on $\partial_{x}^n G$ worsens as n increases, it is more than compensated for by the additional integrations. The terms with more integrals (i.e. more regularity) tend to zero as T goes to zero, by $\boldsymbol{(\mathrm{H}}_{\boldsymbol{8}}^{\gamma}\boldsymbol{)}$ , and we compute the others in closed form. For clarity we write $L_s = L (s,\mathfrak{M}_s,\mathfrak{M}_0,u_s )$ for all $s\ge0$ . By a first application of Lemma 2 on $\partial_{k} L_s \int_s^T |\boldsymbol\Theta_s| \mathrm{d}s$ we obtain

\begin{align*}T_3= & \, \mathbb{E} \left[\partial_{k} L_0 \int_0^T |\boldsymbol\Theta_s| \mathrm{d}s\right]+ \mathbb{E} \left[ \int_0^T (\partial_{x}^3 - \partial_{x}^2) \partial_{k} L_s |\boldsymbol\Theta_s| \left( \int_s^T |\boldsymbol\Theta_r|\mathrm{d}r \right) \mathrm{d}s \right] \\[3pt] & + \mathbb{E} \left[ \int_0^T \partial_{xk} L_s \sum_{j=1}^N \left\{ \phi^j_s \, \mathrm{D}^j_s \left( \int_s^T |\boldsymbol\Theta_r| \mathrm{d}r \right) \right\} \mathrm{d}s \right]=: S_1 + S_2 + S_3.\end{align*}

To deal with $S_2$ , we apply Lemma 2 again on $(\partial_{x}^3 - \partial_{x}^2) \partial_{k} L_s \int_s^T |\boldsymbol\Theta_r|\left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \mathrm{d}r =: H_s Z_s$ , which yields

\[S_2= \mathbb{E} \left[ H_0 Z_0+ \int_0^T (\partial_{x}^3 - \partial_{x}^2) H_s |\boldsymbol\Theta_s| Z_s \mathrm{d}s+ \int_0^T \partial_{x} H_s \sum_{j=1}^N \left( \phi^j_s \mathrm{D}^j_s Z_s \right) \mathrm{d}s\right]=: S_2^a + S_2^b + S_2^c.\]

We will deal with these terms in the last step. For $S_3$ , we apply Lemma 2 once more to

$$\partial_{xk} L_s \int_s^T \sum_{j=1}^N \left\{ \phi^j_r \, \mathrm{D}^j_r \left( \int_r^T |\boldsymbol\Theta_{y}| \mathrm{d}y \right) \right\} \mathrm{d}r=: \widetilde{H}_s \widetilde{Z}_s,$$

and obtain

$$S_3= \mathbb{E} \left[ \widetilde{H}_0 \widetilde{Z}_0+ \int_0^T (\partial_{x}^3 - \partial_{x}^2) \widetilde{H}_s |\boldsymbol\Theta_s| \widetilde{Z}_s \mathrm{d}s+ \int_0^T \partial_{x} \widetilde{H}_s \sum_{j=1}^N \left( \phi^j_s \mathrm{D}^j_s \widetilde{Z}_s \right) \mathrm{d}s \right]=: S_3^a + S_3^b + S_3^c.$$

Step 3. We now evaluate the derivative at $k=\mathfrak{M}_0$ and drop the k-dependence. To summarise,

$$\partial_{k}^2 \mathcal{I}_{T} = \frac{T_1 + T_2 + S_1 + S_2^a + S_2^b + S_2^c + S_3^a + S_3^b + S_3^c }{\partial_\sigma \mathrm{BS} \big(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}\big)},$$

where

\begin{align*}S_1 & = \mathbb{E} \left[\partial_{k} L_0 \int_0^T |\boldsymbol\Theta_s| \mathrm{d}s \right],\\[3pt] S_2^a & = \mathbb{E} \left[H_0 \int_0^T |\boldsymbol\Theta_r| \left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \mathrm{d}r\right], \\[3pt] S_3^a &= \mathbb{E} \left[\widetilde{H}_0 \int_0^T \sum_{j=1}^N \left\{ \phi^j_r \, \mathrm{D}^j_r \left( \int_r^T |\boldsymbol\Theta_{y}| \mathrm{d}y \right) \right\} \mathrm{d}r\right], \\[3pt] S_2^b &= \mathbb{E}\left[\int_0^T (\partial_{x}^3 - \partial_{x}^2) H_s |\boldsymbol\Theta_s|\left( \int_s^T \lvert \boldsymbol\Theta_r\lvert \left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \mathrm{d}r \right) \mathrm{d}s\right], \\[3pt] S_2^c &= \mathbb{E} \left[\int_0^T \partial_{x} H_s \sum_{j=1}^N \left( \phi^j_s \mathrm{D}^j_s \left( \int_s^T |\boldsymbol\Theta_r| \left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \mathrm{d}r \right) \mathrm{d}s \right) \mathrm{d}s\right],\\[3pt] S_3^b &= \mathbb{E} \left[\int_0^T (\partial_{x}^3 - \partial_{x}^2) \widetilde{H}_s |\boldsymbol\Theta_s|\int_s^T \sum_{j=1}^N \left\{ \phi^j_r \, \mathrm{D}^j_r \left( \int_r^T |\boldsymbol\Theta_{y}| \mathrm{d}y \right) \right\} \mathrm{d}r \mathrm{d}s\right], \\[3pt] S_3^c &= \mathbb{E} \left[\int_0^T \partial_{x} \widetilde{H}_s \sum_{k=1}^N \left\{ \phi^k_s \mathrm{D}^k_s \left( \int_s^T \sum_{j=1}^N \left\{ \phi^j_r \, \mathrm{D}^j_r \left( \int_r^T |\boldsymbol\Theta_{y}| \mathrm{d}y \right) \right\} \mathrm{d}r \right) \right\} \mathrm{d}s\right].\end{align*}

We recall once again the bound $\partial_x^n G(t,x,k,\sigma) \le C \big(\sigma\sqrt{T-t}\big)^{-n-1}$ as $T-t$ goes to zero. We observe that H and $\widetilde H$ consist of derivatives of G up to the sixth and the fourth order, respectively; therefore $S_2^b, S_2^c,S_3^b,S_3^c$ tend to zero by $\boldsymbol{(\mathrm{H}}_{\boldsymbol{8}}^{\gamma}\boldsymbol{)}$ . In order to deal with $S_1$ , $S_2^a$ , and $S_3^a$ , we use the explicit partial derivatives from Section 6.6 and (18); as in the proof of Theorem 2, $\boldsymbol{(\mathrm{H}}_{\boldsymbol{9}}^{\gamma}\boldsymbol{)}$ implies that only the higher derivatives of $\mathfrak{u}_0$ remain in the limit:

\begin{align*}\lim_{T\downarrow 0} \frac{S_1}{T^{\lambda}\partial_\sigma \mathrm{BS}\big(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}\big)}& = \lim_{T\downarrow 0} \frac{1}{T^{\lambda}}\frac{\mathbb{E}\left[ \partial_{k} L_0 \int_0^T |\boldsymbol\Theta_s| \mathrm{d}s \right]}{\partial_\sigma \mathrm{BS}\big(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}\big)}\\[3pt] & = \lim_{T\downarrow 0} \frac{1}{T^{\lambda}}\mathbb{E} \left[ \frac{1}{u_0 T} \left( \frac18 + \frac{1}{2 u_0^2 T}\right) \int_0^T |\boldsymbol\Theta_s| \mathrm{d}s \right] \\[3pt] & = \lim_{T\downarrow 0} \mathbb{E} \left[ \frac{1}{2 \mathfrak{u}_0^3 \,T^{\frac{1}{2}+\lambda}} \int_0^T |\boldsymbol\Theta_s| \mathrm{d}s \right]= \lim_{T\downarrow 0} \frac{\mathcal{S}_T}{T^{\lambda}},\\[3pt] \lim_{T\downarrow 0} \frac{S^a_2}{T^{\lambda}\partial_\sigma \mathrm{BS}\big((0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}\big))}& = \lim_{T\downarrow 0} \frac{1}{T^{\lambda}} \mathbb{E} \left[ \frac{Z_0}{\mathfrak{u}_{0} \sqrt{T}} \left({-} \frac{15}{2\mathfrak{u}_0^6} - \frac{3}{2\mathfrak{u}_0^4} - \frac{5}{32\mathfrak{u}_0^2} - \frac{1}{64} \right) \right]\\[3pt] & = \lim_{T\downarrow 0} \mathbb{E} \left[ \frac{-15}{2\mathfrak{u}_0^7 T^{\frac{1}{2}+\lambda}} \int_0^T |\boldsymbol\Theta_r|\left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \mathrm{d}r\right],\\[3pt] \lim_{T\downarrow 0} \frac{S^a_3 }{T^{\lambda}\partial_\sigma \mathrm{BS}\big(0,\mathfrak{M}_0,\mathfrak{M}_0,\mathcal{I}_{T}\big)}& = \lim_{T\downarrow 0} \frac{1}{T^{\lambda}}\mathbb{E} \left[ \frac{\widetilde{Z}_0}{\mathfrak{u}_0 \sqrt T} \left( \frac{3}{2 \mathfrak{u}_0^4} + \frac{3}{8\mathfrak{u}_0^2} + \frac{1}{16} \right) \right]\\[3pt] & = \lim_{T\downarrow 0} \frac32 \mathbb{E} \left[ \frac{1}{\mathfrak{u}_0^5 T^{\frac{1}{2}+\lambda}} \int_0^T \sum_{j=1}^N \left\{ \phi^j_r \, \mathrm{D}^j_r \left( \int_r^T |\boldsymbol\Theta_{y}|\mathrm{d}y \right) \right\} \mathrm{d}r \right].\end{align*}

Hence, to conclude, the claim follows from

\[\lim_{T\downarrow 0} \frac{\mathcal{C}_T}{T^{\lambda}}= \lim_{T\downarrow 0} \frac{T_2 + S_1 + S_2^a + S_3^a}{T^{\lambda} \partial_\sigma \mathrm{BS}(\mathfrak{M}_0,\mathcal{I}_{T})}.\]

6.3.1. Convergence lemmas

We require some preliminaries before we dive into the computations. We present three versions of integral convergence tailored to our purposes, which are essential for computing the limits in Theorems 1, 2, and 3. The conditions they require hold thanks to the continuity of $\boldsymbol{(\mathrm{C}_{4})}$ . Recall the local Taylor theorem: if a function $g({\cdot})$ is continuous on $[0, \delta]$ for some $\delta \gt 0$ , then there exists a continuous function $\varepsilon({\cdot})$ on $[0, \delta]$ with $\lim_{x\downarrow 0}\varepsilon(x)=0$ such that $g(x) = g(0) + \varepsilon(x)$ for any $x \in [0, \delta]$ .

Lemma 3. If $f\,:\,\mathbb{R}_+^2\to\mathbb{R}$ is such that $f(T,\cdot)$ is continuous on $[0,\delta_0]$ for some $\delta_0 \gt 0$ and $\lim\limits_{T\downarrow0}f(T,0)=f(0,0)$ , then

(26) \begin{align} \lim_{T\downarrow0}\frac1T\int_0^T f(T,y)\,\mathrm{d}y =f(0,0). \end{align}

Proof. For $T \lt \delta_0$ , we can write

\begin{align*} \frac1T \int_0^T f(T,y) \mathrm{d}y = \frac1T \int_0^T [f(T,0) + \varepsilon_0(y) ] \mathrm{d}y = f(T,0) + \frac1T \int_0^T\varepsilon_0(y) \mathrm{d}y, \end{align*}

where the function $\varepsilon_0$ is continuous on $[0,\delta_0]$ and converges to zero at the origin. Hence, for any $\eta_0 \gt 0$ , there exists $\widetilde \delta_0 \gt 0$ such that, for any $y\le\widetilde \delta_0$ , $|\varepsilon_0(y)| \lt \eta_0$ . For all $T \lt \widetilde \delta_0 \wedge \delta_0$ ,

$$ \bigg\lvert\frac1T \int_0^T \varepsilon_0(y) \mathrm{d}y \bigg\lvert \le \eta_0. $$

Since $\eta_0$ can be taken as small as desired, the fact that $\lim_{T\downarrow0} f(T,0)=f(0,0)$ concludes the proof.

Lemma 4. Let $f\,:\,\mathbb{R}^3_+\to\mathbb{R}$ be such that, for each $y\le T$ , $f(T,y,\cdot)$ is continuous on $[0,\delta_0]$ with $\delta_0 \gt 0$ , $f(T,\cdot,0)$ is continuous on $[0,\delta_1]$ with $\delta_1 \gt 0$ , and $\lim_{T\downarrow0}f(T,0,0)=f(0,0,0)$ . Then

(27) \begin{align} \lim_{T\downarrow0} \frac{1}{T^2} \int_0^T \int_0^y f(T,y,s) \mathrm{d}s \mathrm{d}y = \frac{f(0,0,0)}{2}. \end{align}

Proof. For $T \lt \delta_0\wedge\delta_1$ , we can write

\begin{align*} \frac{1}{T^2}\int_{0}^{T}\left\{\int_{0}^{y}f(T, y, s)\mathrm{d} s\right\}\mathrm{d} y & = \frac{1}{T^2}\int_{0}^{T}\left\{\int_{0}^{y}\left[f(T, y, 0) + \varepsilon_{0}(s)\right]\mathrm{d} s\right\}\mathrm{d} y\\[3pt] & = \frac{1}{T^2}\int_{0}^{T}\left\{f(T, y, 0)y + \int_{0}^{y}\varepsilon_{0}(s)\mathrm{d} s\right\}\mathrm{d} y\\[3pt] & = \frac{1}{T^2}\int_{0}^{T}\left\{\Big(f(T, 0, 0)+\varepsilon_{1}(y)\Big)y + \int_{0}^{y}\varepsilon_{0}(s)\mathrm{d} s\right\}\mathrm{d} y\\[3pt] & = \frac{f(T,0,0)}{2} + \frac{1}{T^2}\int_{0}^{T}\left\{\varepsilon_{1}(y)y + \int_{0}^{y}\varepsilon_{0}(s)\mathrm{d} s\right\}\mathrm{d} y, \end{align*}

where $\varepsilon_{1}({\cdot})$ is continuous on $[0,\delta_1]$ and $\varepsilon_{0}({\cdot})$ is continuous on $[0,\delta_0]$ , and both are null at the origin. For any $\eta_{1} \gt 0$ , there exists $\widetilde{\delta}_{1} \gt 0$ such that, for any $y \in [0, \widetilde{\delta}_{1}]$ , we have $|\varepsilon_{1}(y)| \lt \eta_{1}$ . Therefore, for the first integral, we have, for $T \lt \widetilde{\delta}_{1}\wedge\delta_0\wedge\delta_1$ ,

$$ \left|\frac{1}{T^2}\int_{0}^{T}\varepsilon_{1}(y)y\mathrm{d} y\right| \leq \frac{1}{T^2}\int_{0}^{T}\left|\varepsilon_{1}(y)\right|y\mathrm{d} y \leq \frac{1}{T^2}\int_{0}^{T}\eta_1 y\mathrm{d} y \leq \frac{\eta_{1}}{2}. $$

Likewise, since $\varepsilon_{0}({\cdot})$ tends to zero at the origin, for any $\eta_{0} \gt 0$ there exists $\widetilde{\delta}_{0} \gt 0$ such that, for any $y \in [0, \widetilde{\delta}_{0}]$ , we have $|\varepsilon_{0}(y)| \lt \eta_{0}$ . Therefore, for the second integral, we have, for $T \lt \widetilde{\delta}_{0}\wedge\delta_0\wedge\delta_1$ ,

$$ \left|\frac{1}{T^2}\int_{0}^{T}\int_{0}^{y}\varepsilon_{0}(s)\mathrm{d} s\mathrm{d} y\right| \leq \frac{1}{T^2}\int_{0}^{T}\int_{0}^{y}\left|\varepsilon_{0}(s)\right|\mathrm{d} s\mathrm{d} y \leq \frac{1}{T^2}\int_{0}^{T}\int_{0}^{y}\eta_{0} \mathrm{d} s \mathrm{d} y \leq \frac{\eta_{0}}{2}. $$

Since $\eta_1$ and $\eta_0$ can be taken as small as desired, taking the limit of f(T, 0, 0) as T goes to zero concludes the proof.

Lemma 5. Let $f\,:\,\mathbb{R}^4_+\to\mathbb{R}$ be such that, for all $0\le s\le y\le T$ , the functions $f(T,y,s,\cdot)$ , $f(T,y,\cdot,0)$ , $f(T,\cdot,0,0)$ are continuous on $[0,\delta_0]$ , $[0,\delta_1]$ , $[0,\delta_2]$ , respectively, for some $\delta_0,\delta_1,\delta_2 \gt 0$ , and $\lim_{T\downarrow0}f(T,0,0,0)=f(0,0,0,0)$ . Then the following limit holds:

(28) \begin{align} \lim_{T\downarrow0} \frac{1}{T^3} \int_0^T \int_0^y \int_0^s f(T,y,s,t) \mathrm{d}t \mathrm{d}s \mathrm{d}y = \frac{f(0,0,0,0)}{6}. \end{align}

Proof. For $T \lt \delta_0\wedge\delta_1\wedge\delta_2$ , we can write

\begin{align*} \frac{1}{T^3}&\int_{0}^{T}\left\{\int_{0}^{y} \left( \int_0^s f(T, y, s,t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y\\[3pt] &=\frac{1}{T^3}\int_{0}^{T}\left\{\int_{0}^{y} \left( \int_0^s [f(T, y, s,0)+\varepsilon_0(t)]\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y\end{align*}

\begin{align*} & = \frac{1}{T^3}\int_{0}^{T}\left\{\int_{0}^{y} \left( f(T, y, s,0)s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y\\[3pt] & = \frac{1}{T^3}\int_{0}^{T}\left\{\int_{0}^{y} \left( \big[f(T, y, 0,0)+\varepsilon_1(s)\big]s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y\\[3pt] & = \frac{1}{T^3}\int_{0}^{T} \left\{f(T, y, 0,0) \frac{y^2}{2}+ \int_{0}^{y}\left( \varepsilon_1(s)s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y \\[3pt] & = \frac{1}{T^3}\int_{0}^{T} \left\{ \big[f(T, 0, 0,0)+\varepsilon_2(y)\big] \frac{y^2}{2}+ \int_{0}^{y}\left( \varepsilon_1(s)s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y\\[3pt] & = \frac{f(T,0,0,0)}{6} + \frac{1}{T^3}\int_{0}^{T} \left\{ \varepsilon_2(y) \frac{y^2}{2} +\int_{0}^{y}\left( \varepsilon_1(s)s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y, \end{align*}

where the function $\varepsilon_2$ is continuous on $[0,\delta_2]$ , the function $\varepsilon_{1}$ is continuous on $[0,\delta_1]$ , and the function $\varepsilon_{0}$ is continuous on $[0,\delta_0]$ , all converging to zero at the origin. By the same argument as in the previous proof, for any $\eta_0,\eta_1,\eta_2 \gt 0$ , there exists $\widetilde\delta \gt 0$ such that for all $T\le\widetilde \delta$ , we have $|\varepsilon_0(T)|\le \eta_0$ , $|\varepsilon_1(T)|\le \eta_1$ , and $|\varepsilon_2(T)|\le \eta_2$ . This implies

$$ \bigg\lvert\frac{1}{T^3}\int_{0}^{T} \left\{ \varepsilon_2(y) \frac{y^2}{2} +\int_{0}^{y}\left( \varepsilon_1(s)s + \int_0^s \varepsilon_0(t)\mathrm{d}t \right) \mathrm{d}s\right\}\mathrm{d}y \bigg\lvert \le \frac{\eta_2+\eta_1+\eta_0}{6}. $$

Since $\eta_2$ , $\eta_1$ and $\eta_0$ can be taken as small as desired, taking the limit of f(T,0,0,0) as T goes to zero concludes the proof.

To apply these lemmas, we will use a modified version of the martingale convergence theorem, which holds in our setting thanks to domination provided by $\boldsymbol{(\mathrm{C}_{1})}$ and $\boldsymbol{(\mathrm{C}_{2})}$ and the continuity of $\boldsymbol{(\mathrm{C}_{4})}$ .

Lemma 6. Let $(X_t)_{t\ge0}$ be almost surely continuous in a neighbourhood of zero, with $\sup_{t\le 1} |X_t|\le Z\in L^1$ . Then the conditional expectation process $(\mathbb{E}_t[X_t])_{t\ge0}$ is also almost surely continuous in a neighbourhood of zero. In particular,

\[ \lim_{t\downarrow0} \mathbb{E}_t[X_t] =\mathbb{E}[X_0]. \]

Proof. All the limits are taken in the almost sure sense. Let $\delta \gt 0$ be such that X is continuous on $[0,\delta]$ , and fix $t \lt \delta$ . We set a sequence $\{t_n\}_{n\in\mathbb{N}}$ on $[0,\delta]$ which converges to t as n goes infinity. Assume first that $\{t_n\}_{n\in\mathbb{N}}$ is a monotone sequence. Since $\mathcal{F}_{t_n}$ tends monotonically to $\mathcal{F}_t$ and X is dominated, the classical martingale convergence theorem (MCT) asserts that $\lim_{n\uparrow\infty} \mathbb{E}_{t_n}[X_t] = \mathbb{E}_t[X_t]$ . For fixed $n\in\mathbb{N}$ and any $\mathfrak{q}\ge |t_n-t|$ ,

(29) \begin{align} |X_{t_n}-X_t| \le \sup_{|\mathfrak{p}-t|\le \mathfrak{q}} |X_{\mathfrak{p}} -X_t|. \end{align}

Let us fix $\varepsilon \gt 0$ . By the MCT, there exists $n_0\in\mathbb{N}$ such that, if $n\ge n_0$ , then

$$ \bigg\lvert\mathbb{E}_{t_n} \bigg[\sup_{|\mathfrak{p}-t|\le \mathfrak{q}} |X_{\mathfrak{p}} -X_t| \bigg]- \mathbb{E}_t \bigg[\sup_{|\mathfrak{p}-t|\le \mathfrak{q}} |X_{\mathfrak{p}} -X_t| \bigg] \bigg\lvert \lt \varepsilon, $$

and by dominated convergence there exists $\delta^{\prime} \gt 0$ with $\mathbb{E}_t \Big[\sup_{|\mathfrak{p}-t|\le \delta^{\prime}} |X_{\mathfrak{p}} -X_t| \Big] \lt \varepsilon$ . There exists $n_1\in\mathbb{N}$ such that $|t_n-t|\le\delta^{\prime}$ for all $n\ge n_1$ ; thus if $n\ge n_0\vee n_1$ , then (29) yields $\mathbb{E}_{t_n}[|X_{t_n}-X_t|] \lt 2\varepsilon$ and

(30) \begin{align} \lim_{n\uparrow\infty} \mathbb{E}_{t_n}[X_{t_n}] = \mathbb{E}_t[X_t]. \end{align}

Now we consider the general case where $\{t_n\}_{n\in\mathbb{N}}$ is not monotone. From every subsequence of $\{t_n\}_{n\in\mathbb{N}}$ , one can extract a further subsequence which is monotone. Let us call this sub-subsequence $\{t_{n_k}\}_{k\in\mathbb{N}}$ . Therefore, (30) holds with $t_{n_k}$ instead of $t_{n}$ . Since every subsequence of $(\mathbb{E}_{t_n}[X_{t_n}])_{n\in\mathbb{N}}$ has a further subsequence that converges to the same limit, the original sequence also converges to this limit.

For convenience, we use the following definition.

Definition 2. Let $k,n\in\mathbb{N}$ with $k\le n$ . For a function $f\,:\,\mathbb{R}_+^n\to\mathbb{R}$ , we define

$$ \lim_{0\le x_1\le x_2\le\cdots\le x_k\downarrow0} f(x_1,\cdots,x_n)\,:\!=\, \lim_{x_k\downarrow0} \cdots \lim_{x_2\downarrow0} \lim_{x_1\downarrow0} f(x_1,\cdots,x_n). $$

Notice that the right-hand sides of (26), (27), and (28) correspond to

$$\lim\limits_{y\le T\downarrow0} f(T,y), \qquad \displaystyle\frac{1}{2}\lim\limits_{s\le y\le T\downarrow0} f(T,y,s), \qquad\text{and} \quad \displaystyle\frac{1}{6}\lim\limits_{t\le s\le y\le T\downarrow0} f(T,y,s,t),$$

respectively.

6.3.2. Proof of Proposition 1

Let us recall some important quantities:

(31) \begin{align}M_y &= \mathbb{E}_y\left[\mathrm{VIX}_T\right] = \mathbb{E}_y\left[ \sqrt{\frac1\Delta \int_T^{T+\Delta} \mathbb{E}_T v_r \mathrm{d}r}\right],\nonumber\\[3pt] m^i_y &= \mathbb{E}_y[\mathrm{D}^i_y M_y]= \mathbb{E}_y \left[\frac{\int_T^{T+\Delta} \mathrm{D}^i_y \mathbb{E}_T v_r \mathrm{d}r }{2\Delta \mathrm{VIX}_T}\right]= \mathbb{E}_y \left[\frac{\int_T^{T+\Delta} \mathrm{D}^i_y v_r \mathrm{d}r }{2\Delta \mathrm{VIX}_T}\right], \nonumber\\[3pt] \phi_y^i&= \frac{m_y^i}{M_y} = \frac{\mathbb{E}_y\left[\left(\int_T^{T+\Delta}\mathrm{D}_y^i v_r \mathrm{d}r\right)/(2\Delta \mathrm{VIX}_T)\right]}{\mathbb{E}_y[\mathrm{VIX}_T]}.\end{align}

We also recall that $J_i$ and $G_{ij}$ , $i,j\in [\![ 1,N ]\!]$ , were defined in (9). In this proof we will define $f(0)\,:\!=\,\lim_{x\downarrow0}f(x)$ , for every $f\,:\,\mathbb{R}_+\to\mathbb{R}$ , as soon as the limit exists and even if f is not actually continuous around zero. In this way we make it continuous, which allows us to apply the convergence lemmas.

Level. By $\boldsymbol{(\mathrm{C}_{1})}$ and the MCT, $\lim_{y\downarrow0}\mathbb{E}_y[\mathrm{VIX}_T]=\mathbb{E}[\mathrm{VIX}_T]$ and $(M_y)_{y\ge0}$ is continuous around zero, almost surely. By $\boldsymbol{(\mathrm{C}_{4})}$ and the dominated convergence theorem (DCT), we have $\lim_{y\downarrow0} \int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r=\int_T^{T+\Delta} \mathrm{D}^i_0 v_r\mathrm{d}r$ and $\big(\int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r)_{y\ge0}$ is continuous around zero, almost surely. Let $i\in [\![ 1,N ]\!]$ ; from $\boldsymbol{(\mathrm{C}_{1})}$ and $\boldsymbol{(\mathrm{C}_{2})}$ we also obtain that almost surely

$$\frac{1}{\mathrm{VIX}_T}\int_T^{T+\Delta}\mathrm{D}_y^i v_r \mathrm{d}r\le X^2 \Big\{(T+\Delta-y)^{H_{+}} - (T-y)^{H_{+}}\Big\},$$

for some $X\in L^2$ . Therefore it is dominated, and by Lemma 6, almost surely $m_y^i$ is continuous at zero and

$$\lim_{y\downarrow0} m^i_y = \mathbb{E}\left[\frac{\int_T^{T+\Delta} \mathrm{D}^i_0 v_r\mathrm{d}r}{2\Delta \mathrm{VIX}_T}\right].$$

Since $M_y \gt 0$ for all $y\le T$ , $\phi^i$ is also continuous at zero and $\lim_{y \le T\downarrow0} \phi^i_y = J_i / (2\Delta\mathrm{VIX}_0^2)$ . By virtue of Theorem 1 and Lemma 3, we obtain

\[\lim_{T\downarrow0} \mathcal{I}_{T} = \lim_{T\downarrow0} \mathbb{E}[u_0] = \lim_{y\le T\downarrow0} {\left\|{\boldsymbol\phi_y}\right\|}= \frac{{\left\|{\boldsymbol J}\right\|}}{2\Delta\mathrm{VIX}_0^2}.\]

Skew. To obtain the skew limit we need to compute a few Malliavin derivatives. For all $i,j\in [\![ 1,N ]\!]$ ,

\begin{align*}\mathrm{D}^j_s m^i_y &= \mathbb{E}_y \left[ \frac{\int_T^{T+\Delta}\mathrm{D}^j_s \mathrm{D}^i_y v_r \mathrm{d}r\, \mathrm{VIX}_T - \int_T^{T+\Delta} \mathrm{D}^i_y v_r \mathrm{d}r \, \mathrm{D}^j_s \mathrm{VIX}_T}{2\Delta \mathrm{VIX}_T^2}\right]\\[3pt] &= \mathbb{E}_y\left[\frac{\int_T^{T+\Delta}\mathrm{D}^j_s \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta\mathrm{VIX}_T} - \frac{\int_T^{T+\Delta}\mathrm{D}^i_y v_r \mathrm{d}r \int_T^{T+\Delta}\mathrm{D}^j_s v_r \mathrm{d}r}{4\Delta^2\mathrm{VIX}_T^3} \right],\end{align*}

which yields

\begin{align*}\mathrm{D}_s^j \phi_y^i & = \frac{\mathrm{D}^j_s m^i_y}{M_y} - \frac{ m^i_y \mathrm{D}^j_s M_y}{M_y^2}\\[3pt] & = \mathbb{E}_y\left[\frac{\int_T^{T+\Delta}\mathrm{D}^j_s \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta \mathrm{VIX}_T M_y} - \frac{ \int_T^{T+\Delta}\mathrm{D}_y^i v_r \mathrm{d}r \int_T^{T+\Delta}\mathrm{D}_s^j v_r \mathrm{d}r}{4\Delta^2 \mathrm{VIX}^3_T M_y} - \frac{ m_y^i \int_T^{T+\Delta} \mathrm{D}^j_s v_r \mathrm{d}r}{2\Delta\mathrm{VIX}_T M_y^2}\right]\\[3pt] & =: \mathbb{E}_y\left[A_T^{ij}(y,s) + B_T^{ij}(y,s) + C_T^{ij}(y,s)\right].\end{align*}

Based on $\boldsymbol{(\mathrm{C}_{1})}$ , $\boldsymbol{(\mathrm{C}_{2})}$ , and $\boldsymbol{(\mathrm{C}_{4})}$ , for each $T\ge0$ , $A^{ij}_T$ , $B^{ij}_T$ , and $C^{ij}_T$ are dominated and almost surely continuous in both arguments. For each $s\ge0$ , Lemma 6 and the DCT yield, almost surely, that $(\mathrm{D}^j_s \phi^i_y)_{y\ge0}$ and $(\mathrm{D}^j_s \phi^i_0)_{s\ge0}$ are continuous around zero. In particular,

\begin{align*}\lim_{s\downarrow0} \mathbb{E}_y\big[ A_T^{ij}(y,s)+B_T^{ij}(y,s)+C_T^{ij}(y,s)\big] &= \mathbb{E}\big[ A_T^{ij}(y,0)+B_T^{ij}(y,0)+C_T^{ij}(y,0)\big],\\[3pt] \lim_{y\downarrow0} \mathbb{E}_y\big[ A_T^{ij}(y,0)+B_T^{ij}(y,0)+C_T^{ij}(y,0)\big] &= \mathbb{E}\big[ A_T^{ij}(0,0)+B_T^{ij}(0,0)+C_T^{ij}(0,0)\big].\end{align*}

By the DCT again this yields

\begin{align*}\lim_{T\downarrow0}\mathbb{E}[ A_T^{ij}(0,0)] = \frac{G_{ij}}{2\Delta\mathrm{VIX}_0^2}\qquad \text{and}\qquad\lim_{T\downarrow0}\mathbb{E}[ B_T^{ij}(0,0)] =\lim_{T\downarrow0} \mathbb{E}[C_T^{ij}(0,0)]= -\frac{J_i J_j}{4\Delta^2 \mathrm{VIX}_0^4}.\end{align*}

Therefore $\phi_s^j \mathrm{D}_s^j (\phi_y^i)^2$ satisfies the continuity requirements of f(T, y, s) in Lemma 4. We combine this lemma with the limits above to see that, almost surely,

\begin{align*}\lim_{T\downarrow0} \frac{1}{T^2} \int_0^T \phi_s^j \int_s^T \mathrm{D}_s^j (\phi_y^i)^2 \mathrm{d}y \mathrm{d}s&= \lim_{T\downarrow0} \frac{1}{T^2} \int_0^T \int_0^y \phi_s^j \mathrm{D}_s^j (\phi_y^i)^2 \mathrm{d}s \mathrm{d}y= \frac{1}{2} \lim_{s\le y \le T\downarrow0} \phi_s^j \mathrm{D}_s^j (\phi_y^i)^2\\[3pt] &= \frac{J_j}{4\Delta \mathrm{VIX}_0^2}\left[ \frac{J_i G_{ij}}{2\Delta^2\mathrm{VIX}_0^4} - \frac{J_i^2 J_j}{2\Delta^3 \mathrm{VIX}_0^6} \right].\end{align*}

We also recall that $\lim_{T\downarrow0} u_0 =\frac{{\left\|{\boldsymbol J}\right\|}}{2\Delta\mathrm{VIX}_0^2}$ almost surely; hence, with $\boldsymbol{(\mathrm{C}_{2})}$ and $\boldsymbol{(\mathrm{C}_{3})}$ , the DCT implies

(32) \begin{align}\lim_{T\downarrow0} \mathcal{S}_T=\sum_{i,j=1}^N\lim_{T\downarrow0} \frac{1}{2} \mathbb{E}\left[\frac{\int_0^T \phi_s^j \int_s^T \mathrm{D}_s^j (\phi_y^i)^2 \mathrm{d}y \mathrm{d}s}{u_0^3 T^{2}}\right]= \frac{1}{2{\left\|{\boldsymbol J}\right\|}^3}\sum_{i,j=1}^N J_i J_j \left(G_{ij} - \frac{J_i J_j}{\Delta\mathrm{VIX}_0^2}\right).\end{align}

Curvature. We now turn our attention to the curvature. By the same arguments as above we have

\begin{align*}\lim_{T\downarrow0} \mathbb{E}\left[ \frac{\left(\sum_{i,j=1}^N \int_0^T \phi_s^j \int_s^T \mathrm{D}^j_s (\phi_y^i)^2 \mathrm{d}y \mathrm{d}s\right)^2}{u_0^7 T^4}\right]&= \frac{2\Delta\mathrm{VIX}^2_0}{{\left\|{\boldsymbol J}\right\|}^7}\left(\sum_{i,j=1}^N J_i J_j \left(G_{ij} - \frac{J_i J_j}{\Delta\mathrm{VIX}_0^2}\right)\right)^2.\end{align*}

For the last term of (8) we need to go one step further and compute more Malliavin derivatives, since

\begin{align*}\mathrm{D}^k_t \Theta^j_s &= \sum_{i=1}^N\left(\mathrm{D}^k_t \phi_s^j \int_s^T \mathrm{D}^j_s (\phi_y^i)^2\mathrm{d}y+2 \phi^j_s \int_s^T \mathrm{D}^k_t \phi^i_y \mathrm{D}^j_s \phi^i_y \mathrm{d}y + 2 \phi_s^j \int_s^T \phi^i_y \mathrm{D}^k_t \mathrm{D}^j_s \phi^i_y \mathrm{d}y\right) \\[3pt] & =: \sum_{i=1}^N \int_s^T \Upsilon^{ijk}(t,s,y,T)\mathrm{d}y. \nonumber\end{align*}

Thus we zoom in on the last term of the display above:

\begin{align*}\mathrm{D}^k_t \mathrm{D}^j_s \phi_y^i &= \frac{\mathrm{D}^k_t \mathrm{D}^j_s m^i_y \, M_y -\mathrm{D}^j_s m^i_y \mathrm{D}^k_t M_y}{M_y^2}- \frac{\mathrm{D}^k_t m^i_y \mathrm{D}^j_s M_y + m_y^i \mathrm{D}^k_t \mathrm{D}^j_s M_y}{M_y^2} + \frac{2 m^i_y \mathrm{D}^j_s M_y \mathrm{D}^k_t M_y}{M_y^3}\\[3pt] &=:\sum_{n=1}^5 Q_n^{ijk}(t,s,y,T),\end{align*}

We zoom in again on $Q_1^{ijk}(t,s,y,T)$ :

\begin{align*}\mathrm{D}^k_t \mathrm{D}^j_s m_y^i = \mathrm{D}^k_t \mathrm{D}^j_s \mathrm{D}^i_y M_y & = \mathrm{D}^k_t \mathbb{E}_y\left[ \frac{\int_T^{T+\Delta} \mathrm{D}^j_s \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta\mathrm{VIX}_T} - \frac{\int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r \int_T^{T+\Delta} \mathrm{D}^j_s v_r\mathrm{d}r}{4\Delta^2\mathrm{VIX}_T^3}\right] \\[3pt] & =: \mathbb{E}_y\left[ \alpha_T^{ijk}+\beta_T^{ijk}\right].\end{align*}

Some additional computations lead to

\begin{align*}\alpha_T^{ijk}&= \frac{\mathrm{VIX}_T \int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^j_s\mathrm{D}^i_y v_r\mathrm{d}r - \mathrm{D}^k_t \mathrm{VIX}_T \, \int_T^{T+\Delta} \mathrm{D}^j_s \mathrm{D}^i_y v_r\mathrm{d}r}{2\Delta \mathrm{VIX}_T^2}\\[3pt] &=\frac{\int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^j_s\mathrm{D}^i_y v_r\mathrm{d}r}{2\Delta \mathrm{VIX}_T} - \frac{\int_T^{T+\Delta} \mathrm{D}^k_t v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^j_s\mathrm{D}^i_y v_r\mathrm{d}r}{4\Delta^2\mathrm{VIX}_T^3},\end{align*}

\begin{align*} \beta_T^{ijk} &=- \frac{\int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^j_s v_r\mathrm{d}r + \int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^j_s v_r\mathrm{d}r}{4\Delta^2 \mathrm{VIX}_T^3} \\[3pt] & \qquad\qquad + \frac{\mathrm{D}^k_t \mathrm{VIX}_T^3 \int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^j_s v_r\mathrm{d}r}{4\Delta^2 \mathrm{VIX}_T^6} \\[3pt] &=- \frac{\int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^j_s v_r\mathrm{d}r + \int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^k_t \mathrm{D}^j_s v_r\mathrm{d}r}{4\Delta^2 \mathrm{VIX}_T^3}\\[3pt] &\qquad\qquad + \frac{3 \int_T^{T+\Delta} \mathrm{D}^k_t v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^i_y v_r\mathrm{d}r \, \int_T^{T+\Delta} \mathrm{D}^j_s v_r\mathrm{d}r}{8\Delta^3 \mathrm{VIX}_T^5} .\end{align*}

We notice, crucially, that we have already justified the continuity of $\phi$ and $\mathrm{D}\phi$ around zero in the proofs of level and skew, respectively. Furthermore, by Lemma 6, the first two terms in $\Upsilon^{ijk}$ as well as $Q_2,Q_3,Q_4,Q_5$ all converge to some finite limit as $t\le s\le y\downarrow0$ and are continuous around zero, almost surely. Similarly, $\beta_T$ and the second term in $\alpha_T$ are almost surely continuous around zero, and their conditional expectation converges almost surely to some finite limit as $t\le s\le y\downarrow0$ by the DCT and Lemma 6. Taking the limit as T goes to zero afterwards, we see that all of the aforementioned terms tend to a finite limit. On the other hand, by $\boldsymbol{(\mathrm{C}_{4})}$ , the DCT, and Lemma 6 we know that the conditional expectation of the first term in $\alpha_T$ is almost surely continuous around zero, and its limit is

\[\lim_{t\le s\le y\downarrow0} \mathbb{E}_y\left[\frac{\int_T^{T+\Delta} \mathrm{D}_t^k \mathrm{D}_s^j \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta\mathrm{VIX}_T}\right] = \mathbb{E}\left[\frac{\int_T^{T+\Delta} \mathrm{D}_0^k \mathrm{D}_0^j \mathrm{D}^i_0 v_r \mathrm{d}r}{2\Delta\mathrm{VIX}_T}\right].\]

Since $\gamma \lt 0$ , only this term contributes in the limit:

\begin{align*}\lim_{t\le s\le y\le T\downarrow0} \frac{ \phi_t^k\,\Upsilon^{ijk}(t,s,y,T)}{T^{\gamma}}&= \lim_{t\le s\le y\le T\downarrow0} 2 \phi_t^k \phi^j_s \phi^i_y \mathbb{E}_y\left[\frac{\int_T^{T+\Delta} \mathrm{D}_t^k \mathrm{D}_s^j \mathrm{D}^i_y v_r \mathrm{d}r}{2 T^\gamma\Delta\mathrm{VIX}_T M_y}\right] \\[3pt] &= \frac{J_i J_j J_k}{8\Delta^4\mathrm{VIX}_0^8} \, \lim_{T\downarrow0} \frac{\mathbb{E}\left[\int_T^{T+\Delta} \mathrm{D}_0^k \mathrm{D}_0^j \mathrm{D}^i_0 v_r \mathrm{d}r\right]}{T^\gamma},\end{align*}

where we applied the DCT at the end. Moreover, we know by $\boldsymbol{(\mathrm{C}_{2})}$ that this limit is finite for $\gamma=3H-\frac{1}{2}$ ; hence the conditions of Lemma 5 are satisfied. We also recall that $\lim_{T\downarrow0} u_0 =\frac{{\left\|{\boldsymbol J}\right\|}}{2\Delta\mathrm{VIX}_0^2}$ almost surely; hence Lemma 5 yields the almost sure limit

\begin{align*}& \lim_{T\downarrow0}\frac{1}{u_0^5 T^{3+\gamma}} \int_0^T \sum_{k=1}^N \left\{ \phi^k_t \mathrm{D}^k_t \left( \int_t^T |\boldsymbol\Theta_{s}|\mathrm{d}s \right) \right\} \mathrm{d}t \\[3pt] & \qquad = \sum_{i,j,k=1}^N \lim_{T\downarrow0}\frac{1}{u_0^5 T^{3}} \int_0^T \int_0^y\int_0^s \frac{\phi^k_t\,\Upsilon^{ijk}(t,s,y,T)}{T^{3H-\frac{1}{2}}} \mathrm{d}y\mathrm{d}s\mathrm{d}t \\[3pt] & \qquad = \frac{2 \Delta \mathrm{VIX}_0^{2}}{3{\left\|{\boldsymbol J}\right\|}^5} \sum_{i,j,k=1}^N J_i J_j J_k \, \lim_{T\downarrow0} \frac{\mathbb{E}\left[\int_T^{T+\Delta} \mathrm{D}_0^k \mathrm{D}_0^j \mathrm{D}^i_0 v_r \mathrm{d}r\right]}{T^{3H-\frac{1}{2}}} .\end{align*}

The first two terms in (8) tend to zero since $\gamma \lt 0$ ; hence Theorem 3 and the DCT yield the final result:

\begin{align*}\lim_{T\downarrow0} \frac{\mathcal{C}_T}{T^{3H-\frac{1}{2}}}&=\frac{2\Delta \mathrm{VIX}_0^{2}}{3{\left\|{\boldsymbol J}\right\|}^5} \sum_{i,j,k=1}^N J_i J_j J_k \lim_{T\downarrow0} \,\frac{\int_T^{T+\Delta} \mathbb{E}\left[\mathrm{D}^k_0 \mathrm{D}^j_0\mathrm{D}^i_0 v_r\right]\mathrm{d}r}{T^{3H-\frac{1}{2}}}. \end{align*}

6.4.1. Proof of Proposition 2

We start with a useful lemma for Gaussian processes.

Proof. The Borell–TIS inequality asserts that $\mathbb{E}[{\left\|{B}\right\|}_T] \lt \infty$ and

$$\mathbb{P}({\left\|{B}\right\|}_T-\mathbb{E}{\left\|{B}\right\|}_T \gt x)\le \exp\left\{-\frac{x^2}{2\sigma^2_T}\right\},$$

where $\sigma^2_T\,:\!=\,\sup_{t\le T} \mathbb{E}[B_t^2]$ ; see [Reference Adler and Taylor1, Theorem 2.1.1]. We then follow the proof of [Reference Adler and Taylor1, Theorem 2.1.2]:

\begin{align*} \mathbb{E} \left[\mathrm{e}^{p {\left\|{B}\right\|}_T}\right] = \int_0^\infty \mathbb{P}\left( \mathrm{e}^{p{\left\|{B}\right\|}_T} \gt x\right) \mathrm{d}x \le \mathrm{e}^p + \mathbb{E}[{\left\|{B}\right\|}_T] + \int_{\mathrm{e}^p \vee \mathbb{E}[{\left\|{B}\right\|}_T]}^\infty \mathbb{P} \left({\left\|{B}\right\|}_T \gt \frac{\log(x)}{p} \right) \mathrm{d}x. \end{align*}

The Borell–TIS inequality in particular reads as follows:

\[ \mathbb{P} \left({\left\|{B}\right\|}_T \gt \log(x^{1/p}) \right) \le \exp\left\{-\frac{\left(\log(x^{1/p})-\mathbb{E}[{\left\|{B}\right\|}_T]\right)^2}{2\sigma_T^2}\right\}, \qquad \text{for all } u \gt \mathbb{E}[{\left\|{B}\right\|}_T]. \]

After a change of variable this yields

\[ \int_{\mathrm{e}^p \vee \mathbb{E}[{\left\|{B}\right\|}_T]}^\infty \mathbb{P} \left({\left\|{B}\right\|}_T \gt \frac{\log(x)}{p} \right) \mathrm{d}x \le \int_{\frac{\log\left(\mathrm{e}^p \vee \mathbb{E}[{\left\|{B}\right\|}_T]\right)}{p}}^\infty \exp\left\{-\frac{\left(x-\mathbb{E}[{\left\|{B}\right\|}_T]\right)^2}{2\sigma_T^2}\right\} p \mathrm{e}^{px} \mathrm{d}x, \]

which is finite as desired.

By the above lemma, ${\left\|{v}\right\|}_T\in L^p$ , so that we can compute its Malliavin derivatives

(33) \begin{align}\mathrm{D}^1_y v_r = v_0 (r-y)^{H_{-}} \Big(\chi \nu \mathcal{E}^1_r + \overline{\chi} \eta \rho \mathcal{E}_r^2\Big)\qquad\text{and}\qquad\mathrm{D}^2_y v_r = v_0 \overline{\chi} \eta \overline{\rho} (r-y)^{H_{-}} \mathcal{E}_r^2.\end{align}

Without explicitly computing further derivatives, one notices that $\boldsymbol{(\mathrm{C}_{4})}$ holds and that there exist $C \gt 0$ and a random variable $X=C{\left\|{\mathcal{E}_r^1+\mathcal{E}_r^2}\right\|}_T\in L^p$ for all $p \gt 1$ such that $\mathrm{D}^i_y v_r \le X (r-y)^{H_{-}}$ , $\mathrm{D}^j_s \mathrm{D}^i_y v_r \le X (r-s)^{H_{-}} (r-y)^{H_{-}}$ , and $\mathrm{D}^k_t \mathrm{D}^j_s \mathrm{D}^i_y v_r \le X (r-t)^{H_{-}} (r-s)^{H_{-}} (r-y)^{H_{-}}$ , implying $\boldsymbol{(\mathrm{C}_{2})}$ . The following lemma yields $\boldsymbol{(\mathrm{C}_{1})}$ .

Lemma 8. In the two-factor rough Bergomi model (10) with $0\leq T_1 \lt T_2$ ,

$$ \mathbb{E}\left[ \sup_{y\le T_1}\left(\mathbb{E}_{y}\left[\frac{1}{T_2-T_1}\int_{T_1}^{T_2} v_{r} \mathrm{d}r\right]\right)^{-p} \right] $$

is finite for all $p \gt 1$ . In particular, $1/M$ is dominated in $L^p$ .

Proof. We first apply an $\exp$ – $\log$ identity, then Jensen’s inequality (using the concavity of the logarithm function), to obtain

\begin{align*}\mathbb{E}_{y}\left[\frac{1}{T_2-T_1}\int_{T_1}^{T_2} v_{r} \mathrm{d}r\right]^{-p} & = \exp\left\{-p \log \left(\frac{1}{T_2-T_1}\int_{T_1}^{T_2} v_{r}\mathrm{d}r \right)\right\}\\[3pt] & \le \exp\left\{-\frac{p}{T_2-T_1}\int_{T_1}^{T_2} \log(\mathbb{E}_y [v_r]) \mathrm{d}r\right\}.\end{align*}

We further bound $\log\mathbb{E}_y [v_r]$ as follows, using the concavity of the logarithm and (10):

(34) \begin{align}-\log\mathbb{E}_y [v_r] \le -\frac{1}{2} \Big\{\log\left(2\chi v_0 \mathbb{E}_y[\mathcal{E}^1_r]\right) + \log\left(2\overline{\chi} v_0 \mathbb{E}_y[\mathcal{E}^2_r]\right) \Big\},\end{align}

which we now compute as

(35) \begin{align} \mathbb{E}_y[\mathcal{E}^1_r]& = \exp\left\{-\frac{\nu^2 r^{2H}}{4H} + \nu \int_0^y (r-s)^{H_{-}}\mathrm{d} W^1_s\right\} \mathbb{E}_y\left[\exp\left\{\nu \int_y^r (r-s)^{H_{-}}\mathrm{d} W^1_s\right\}\right] \nonumber\\[3pt] & = \exp\left\{\frac{\nu^2}{4H} \left[(r-y)^{2H}-r^{2H}\right] + \nu \int_{0}^{y} (r-s)^{H_{-}} \mathrm{d} W^1_s\right\}, \nonumber \\[3pt] \mathbb{E}_y[\mathcal{E}^2_r] &=\exp\left\{\frac{\eta^2}{4H}\left[(r-y)^{2H}-r^{2H}\right] + \eta \int_0^{y} (r-s)^{H_{-}} \mathrm{d} (\rho W^1_s+\overline{\rho} W^2_s)\right\}.\end{align}

Let us deal with the first term of (34), as the second one is analogous. We have

\[ \int_{T_1}^{T_2} \left[(r-y)^{2H}-r^{2H}\right] \mathrm{d}r = \frac{(T_2-u)^{2H_+}-(T_1-y)^{2H_+}-T_2^{2H_+}+T_1^{2H_+}}{2H_+}, \]

which is clearly bounded below for all $0\le u\le T_1$ . Moreover, by Fubini’s theorem,

$$\int_{T_1}^{T_2} \int_0^y (r-t)^{H_{-}} \mathrm{d} W^1_t \mathrm{d}r=\int_0^y \int_{T_1}^{T_2} (r-t)^{H_{-}} \mathrm{d}r\mathrm{d} W^1_t= \int_0^y \frac{(T_2-t)^{H_{+}}-(T_1-t)^{H_{+}}}{H_{+}} \mathrm{d} W^1_t=: \overline{B}_t$$

is a Gaussian process. Since $\exp\{\cdot\}$ is increasing, $\sup_{t\in[0,T]} \exp\{\overline{B}_t\} = \exp\{\sup_{t\in[0,T]} \overline{B}_t\}$ ; thus

\[ \mathbb{E}\left[\sup_{y\le T_1} \exp\left({-}\frac{p}{T_2-T_1}\int_{T_1}^{T_2} \int_0^y (r-s)^{H_{-}} \mathrm{d} W^1_s\mathrm{d}r\right)\right] \le \mathbb{E}\left[\exp\left(\frac{p}{T_2-T_1} {\left\|{\overline{B}}\right\|}_T\right)\right] \lt \infty, \]

by Lemma 7, which concludes the proof.

Combining (33) and (35), we obtain $\mathbb{E}_y[\mathrm{D}^i_y v_r]$ , $i=1,2$ . The following lemma proves that $\boldsymbol{(\mathrm{C}_{3})}$ is satisfied.

Proof. Since $\nu,\eta,\rho+\overline{\rho} \gt 0$ , we have $\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r \gt 0$ almost surely for all $y\le r$ . Moreover, VIX and $1/\mathrm{VIX}$ are dominated by some $X \in L^p$ for all $p \gt 1$ , so, almost surely and independently of the sign of the numerator, we obtain

\[ m^i_y = \mathbb{E}_y\left[\frac{\int_T^{T+\Delta} \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta \mathrm{VIX}_T}\right] \ge \mathbb{E}_y\left[\frac{\int_T^{T+\Delta} \mathrm{D}^i_y v_r \mathrm{d}r}{2\Delta X}\right]. \]

Therefore, using that $1/M$ is dominated by X and Jensen’s inequality, we get

(36) \begin{align} \frac{1}{u_s^2} = \frac{T-s}{\int_s^T \sum_{i=1}^N (\phi^i_y)^2\mathrm{d}y} & \le \frac{X^2(T-s)}{\int_s^T \sum_{i=1}^N (m^i_y)^2\mathrm{d}y} \le \frac{X^2 N(T-s)}{\int_s^T \left(\sum_{i=1}^N m^i_y \right)^2\mathrm{d}y} \le X^2 N \left( \frac{T-s}{\int_s^T \sum_{i=1}^N m^i_y\mathrm{d}y}\right)^2 \nonumber\\[3pt] &\le 4 X^2 N \left(\frac{\int_s^T\int_T^{T+\Delta}\sum_{i=1}^N \mathbb{E}_y\left[\mathrm{D}^i_y v_r /X\right] \mathrm{d}r\mathrm{d}y}{\Delta(T-s)}\right)^{-2}. \end{align}

Hence we turn our attention to

(37) \begin{align} &\mathbb{E}\left[ \left(\frac{1}{\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \mathbb{E}_y\left[\frac{\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r}{X}\right] \mathrm{d}r \mathrm{d}y\right)^{-p}\right] \nonumber\\[3pt] &= \mathbb{E}\left[ \exp\left\{-p\log \left(\frac{1}{\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \mathbb{E}_y\left[\frac{\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r}{X}\right] \mathrm{d}r \mathrm{d}y\right)\right\}\right] \nonumber \\[3pt] &\le \mathbb{E}\left[ \exp\left\{-\frac{p}{\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \mathbb{E}_y\left[\log \left(\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r\right) -\log(X)\right] \mathrm{d}r \mathrm{d}y \right\}\right] \nonumber\\[3pt] &\le \left(\mathbb{E}\left[ \exp\left\{-\frac{2p}{\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \mathbb{E}_y\left[\log \left(\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r \right)\right] \mathrm{d}r \mathrm{d}y \right\}\right]\right)^\frac{1}{2} \sqrt{\mathbb{E}\left[X^{2p}\right]}, \end{align}

using Jensen’s inequality, the Cauchy–Schwarz inequality, and the fact that $\mathrm{e}^{p\mathbb{E}_y[\log(X)]}\le \mathbb{E}_y[X^p]$ . Convexity and (33) imply

\begin{align*} -\log \left(\mathrm{D}^1_y v_r+\mathrm{D}^2_y v_r\right) \le &-\frac12 \Big\{\log \left(2 v_0\chi \nu(r-y)^{H_{-}} \mathcal{E}^1_r \right) + \log\left(2 v_0 \overline{\chi} \eta(\rho+\overline{\rho})(r-y)^{H_{-}} \mathcal{E}_r^2\right) \Big\}. \end{align*}

We focus on the first term; the other can be treated identically. From (35) we have

(38) \begin{align} \mathbb{E}_y\left[\log \left(2 v_0 \chi \nu(r-y)^{H_{-}} \mathcal{E}^1_r \right)\right] = \log\left(2 v_0 \chi \nu(r-y)^{H_{-}}\right) - \frac{\nu^2 r^{2H}}{4H} + \nu \int_0^y (r-t)^{H_{-}} \mathrm{d} W^1_t . \end{align}

Let us start with

\begin{align*} &\int_s^T \int_T^{T+\Delta} \log\left(2 v_0 \chi \nu(r-y)^{H_{-}}\right) \mathrm{d}r \mathrm{d}y \\[3pt] &= 2(T-s)\Delta v_0 \chi \nu + H_- \int_s^T \int_T^{T+\Delta} \log(r-y) \mathrm{d}r \mathrm{d}y \\[3pt] &= 2(T-s)\Delta v_0 \chi \nu + H_- \\[3pt] & \qquad \int_s^T \Big[(T+\Delta-y)\log(T+\Delta-y) - (T+\Delta-y)- (T-y)\log(T-y) + (T-y) \Big]\mathrm{d}y \\[3pt] & = 2(T-s)\Delta v_0 \chi \nu + H_- \left\{-\Delta (T-s) - \int_\Delta^{T+\Delta-s} x\log(x)\mathrm{d}x +\int_0^{T-s} x\log(x)\mathrm{d}x \right\} \end{align*}

\begin{align*} & = 2(T-s)\Delta v_0 \chi \nu + H_- \Bigg\{-\Delta (T-s)+ \left(\frac{(T-s)^2 \log(T-s)}{2} - \frac{(T-s)^2}{4}\right) \\[3pt] & \qquad\qquad\quad - \left(\frac{(T+\Delta-s)^2 \log(T+\Delta-s)}{2} - \frac{(T+\Delta-s)^2}{4}\frac{\Delta^2\log(\Delta)}{2} +\frac{\Delta^2}{4}\right) \Bigg\}\\[3pt] &= 2(T-s)\Delta v_0 \chi \nu +(T-s) H_- \Bigg\{ -\Delta + \left(\frac{(T-s) \log(T-s)}{2} - \frac{(T-s)}{4}\right) + \frac{2\Delta +(T-s)}{4}\\[3pt] & \quad \qquad\qquad\qquad - \Delta\log(T+\Delta-s) - \frac{(T-s)\log(T+\Delta-s)}{2} \Bigg\} \\[3pt] & \qquad\qquad\qquad\quad -\frac{\Delta^2}{2} \big(\log(T+\Delta-s)-\log(\Delta) \big). \end{align*}

By Taylor’s theorem, $\log(T+\Delta-s)-\log(\Delta) = \frac{T-s}{\Delta} + \varepsilon(T-s)$ , where $\varepsilon:\mathbb{R}_+\to\mathbb{R}_+$ is such that $\varepsilon(x)/x$ tends to zero at the origin. We conclude that

\[ -\frac{p}{2\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \log\left(2 v_0 \chi \nu(r-y)^{H_{-}}\right) \mathrm{d}r \mathrm{d}y \]

is uniformly bounded. Now we study the second term of (38):

\[ - \int_s^T \int_T^{T+\Delta} r^{2H} \mathrm{d}r\mathrm{d}y = (T-s) \frac{T^{2H_+}-(T+\Delta)^{2H_+}}{2H_+}. \]

Therefore the following is uniformly bounded:

\[ \frac{p}{2\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \frac{\nu^2 r^{2H}}{4H} \mathrm{d}r \mathrm{d}y. \]

For the last term, by the stochastic Fubini theorem [Reference Protter48, Theorem 65], we get

\begin{align*} \int_s^T \int_T^{T+\Delta} \int_0^y (r-t)^{H_{-}} \mathrm{d} W^1_t \mathrm{d}r\mathrm{d}y & = \int_s^T \int_0^y \int_T^{T+\Delta} (r-t)^{H_{-}} \mathrm{d}r \mathrm{d} W^1_t \mathrm{d}y\\[3pt] & =\int_0^T \int_{s\vee t}^T \frac{(T+\Delta-t)^{H_{+}} - (T-t)^{H_{+}}}{H_{+}} \mathrm{d}y \mathrm{d} W^1_t. \end{align*}

Standard Gaussian computations then yield

(39) \begin{align} &\mathbb{E}\left[\exp\left\{-\frac{p}{4\Delta(T-s)} \int_s^T \int_T^{T+\Delta} \nu \int_0^y (r-t)^{H_{-}} \mathrm{d} W^1_t \mathrm{d}r\mathrm{d}y \right\}\right] \nonumber \\[3pt] &= \exp\left\{\frac{1}{2} \left(\frac{p\nu}{4\Delta(T-s)}\right)^2 \int_0^T \left( \int_{s\vee t}^T \frac{(T+\Delta-t)^{H_{+}} - (T-t)^{H_{+}}}{H_{+}} \mathrm{d}y\right)^2\mathrm{d}t \right\}. \end{align}

The incremental function $x\mapsto (x+\Delta)^{H_{+}}-x^{H_{+}}$ is decreasing by concavity; hence $(T+\Delta-t)^{H_{+}} - (T-t)^{H_{+}} \le \Delta^{H_{+}}$ , and we obtain

\[ \int_0^T (T-s\vee t)^2 \mathrm{d}t = \int_0^s (T-s)^2\mathrm{d}t + \int_s^T (T-t)^2\mathrm{d}t = s(T-s)^2 + \frac{(T-s)^3}{3}, \]

which implies that (39) is uniformly bounded. We have thus shown that (37) is uniformly bounded in s, T.

Coming back to (36), by the Cauchy–Schwarz inequality we have

\[ \mathbb{E}[u_s^{-p}]^2 \le 2^{p} \mathbb{E}[X^{p}] \, \mathbb{E}\left[ \left(\frac{T-s}{\int_s^T (m^1_y + m^2_y) \mathrm{d}y} \right)^{2p} \right], \]

which is uniformly bounded for all $s\le T$ , and this concludes the proof.

6.4.2. Proof of Proposition 3

Level. We start with the derivatives

\begin{align*}\mathrm{D}^1_s v_t = v_0\Big[\chi \nu(t-s)^{H_{-}} \mathcal{E}^1_t + \overline{\chi} \eta \rho (t-s)^{H_{-}} \mathcal{E}_t^2\Big]\qquad\text{and}\qquad\mathrm{D}^2_s v_t = v_0 \overline{\chi} \eta \overline{\rho} (t-s)^{H_{-}} \mathcal{E}_t^2,\end{align*}

and recall, from the definitions in (9),

$$ J_1 = \int_{0}^\Delta v_{0} {\mathbb{E}}\left[\chi \nu r^{H_{-}} \mathcal{E}^1_r +\overline{\chi}\eta \rho r^{H_{-}} \mathcal{E}^2_r \right] \mathrm{d}r = v_0(\chi\nu+\overline{\chi}\eta\rho)\frac{\Delta^{H_{+}}}{H_{+}}\quad \text{and}\quad J_{2} = v_0\overline{\chi} \eta \overline{\rho} \frac{\Delta^{H_{+}}}{H_{+}}.$$

We also note that $\mathbb{E}[\mathcal{E}^i_t]=1$ . This yields the norm

\[{\left\|{\boldsymbol J}\right\|} \,:\!=\, \left(J_1^2 + J_2^2 \right)^{\frac{1}{2}}= \frac{v_0 \Delta^{H_{+}}}{H_{+}}\sqrt{ (\chi\nu+\overline{\chi}\eta\rho)^2 + \overline{\chi}^2\eta^2\overline{\rho}^2}= \frac{v_0 \Delta^{H_{+}}}{H_{+}} \psi(\rho,\nu,\eta,\chi),\]

with the function $\psi$ defined in the proposition, which grants us the first limit by Proposition 1. To simplify the notation below, we introduce $\mathfrak{w} \,:\!=\, \chi\nu + \overline{\chi}\eta\rho$ .

Skew. We compute the further derivatives

\begin{align*}\mathrm{D}^1_0 \mathrm{D}^1_0 v_t &= v_0\left(\chi\nu^2 t^{2H-1} \mathcal{E}_t^1 + \overline{\chi}\eta^2 \rho^2 t^{2H-1} \mathcal{E}_t^2 \right),\\[3pt] \mathrm{D}^1_0 \mathrm{D}^2_0 v_t & = v_0\overline{\chi}\eta^2 \rho \overline{\rho} t^{2H-1} \mathcal{E}^2_t,\\[3pt] \mathrm{D}^2_0 \mathrm{D}^2_0 v_t & = v_0\overline{\chi} \eta^2 \overline{\rho}^2 t^{2H-1} \mathcal{E}^2_t.\end{align*}

Similarly to J, we recall that $G_{ij}= \int_0^{\Delta} \mathbb{E}\big[\mathrm{D}^j_0 \mathrm{D}^i_0 v_r \big]\mathrm{d}r$ , so that

\begin{align*}G_{11}=\frac{\Delta^{2H}}{2H} v_0 (\chi\nu^2 + \overline{\chi}\eta^2 \rho^2), \qquad G_{12} = \frac{\Delta^{2H}}{2H} v_0\overline{\chi} \eta^2\rho \overline{\rho}, \qquad G_{22}= \frac{\Delta^{2H}}{2H} v_0\overline{\chi} \eta^2 \overline{\rho}^2.\end{align*}

Notice that $\mathrm{VIX}^2_0=v_0$ ; thus we have

\begin{align*}J_1^2 \left(G_{11} -\frac{ J_1^2}{\Delta\mathrm{VIX}_0^2} \right)&= \frac{v_0^3\Delta^{4H+1}}{2HH_+^2}\mathfrak{w}^2 (\chi\nu^2+\overline{\chi}\eta^2\rho^2)- \frac{v_0^3 \Delta^{4H+1}}{H_+^4} \mathfrak{w}^4\\[3pt] &= v_0^3 \frac{\Delta^{4H+1}}{H_+^2} \mathfrak{w}^2\left[ \frac{\chi\nu^2+\overline{\chi}\eta^2\rho^2}{2H}-\frac{\mathfrak{w}^2}{H_+^2}\right],\end{align*}

\begin{align*} J_1 J_2 \left( G_{12} -\frac{J_1 J_2}{\Delta\mathrm{VIX}_0^2} \right)&= \frac{v_0^3\Delta^{4H+1}}{2HH_+^2}\mathfrak{w}\overline{\chi}^2\eta^3\rho\overline{\rho}^2- \frac{v_0^3\Delta^{4H+1}}{H_+^4} \mathfrak{w}^2\overline{\chi}^2\eta^2\overline{\rho}^2 \\[3pt] &= v_0^3 \frac{\Delta^{4H+1}}{H_+^2} \mathfrak{w}\overline{\chi}^2\eta^2\overline{\rho}^2 \left[ \frac{\eta\rho}{2H} - \frac{\mathfrak{w}}{H_+^2}\right],\\[3pt] J_2^2 \left( G_{22} -\frac{J_2^2}{\Delta\mathrm{VIX}_0^2} \right)&= v_0^3 \frac{\Delta^{4H+1}}{2HH_+^2} \overline{\chi}^3 \eta^4\overline{\rho}^4 - v_0^3 \overline{\chi}^4 \frac{\Delta^{4H+1}}{H_+^4} \eta^4\overline{\rho}^4 \\[3pt] &= v_0^3 \frac{\Delta^{4H+1}}{H_+^2}\overline{\chi}^3 \eta^4\overline{\rho}^4 \left(\frac{1}{2H}-\frac{\overline{\chi}}{H_+^2}\right). \end{align*}

Finally, by Proposition 1 we obtain

\begin{align*}\lim_{T\downarrow0} \mathcal{S}_T =& \frac{H_+\Delta^{H_{-}}}{2\psi(\rho,\nu,\eta,\chi)^{3}} \Bigg\{\mathfrak{w}^2 \left[ \frac{\chi\nu^2+\overline{\chi}\eta^2\rho^2}{2H}-\left(\frac{\mathfrak{w}}{H_{+}}\right)^2\right]\\[3pt] & + 2 \mathfrak{w}\overline{\chi}^2\eta^2\overline{\rho}^2 \left[ \frac{\eta\rho}{2H} - \frac{\nu+\eta\rho}{H_+^2}\right]+ \overline{\chi}^3 \eta^4\overline{\rho}^4 \left(\frac{1}{2H}-\frac{1}{H_+^2}\right) \Bigg\}.\end{align*}

Curvature. For the last step we go one step further:

\begin{alignat*}{2}\mathrm{D}^1_0\mathrm{D}^1_0 \mathrm{D}^1_0 v_t &= v_0 \left(\chi\nu^3 \mathcal{E}_t^1 + \overline{\chi}\eta^3 \rho^3 \mathcal{E}_t^2 \right)t^{3H_-}, \qquad\qquad&&\mathrm{D}^2_0 \mathrm{D}^1_0 \mathrm{D}^1_0 v_t = v_0\overline{\chi} \eta^3 \rho^2 \overline{\rho} t^{3H_-} \mathcal{E}_t^2, \\[3pt] \mathrm{D}^2_0 \mathrm{D}^2_0 \mathrm{D}^1_0 v_t &= v_0\overline{\chi} \eta^3 \rho \overline{\rho}^2 t^{3H_-} \mathcal{E}_t^2, \qquad\qquad&&\mathrm{D}^2_0 \mathrm{D}^2_0 \mathrm{D}^2_0 v_t = v_0\overline{\chi} \eta^3 \overline{\rho}^3 t^{3H_-} \mathcal{E}_t^2.\end{alignat*}

We notice that

\[\lim_{T\downarrow0} \frac{\int_T^{T+\Delta} r^{3H_-} \mathrm{d}r}{T^{3H-\frac{1}{2}}} = \lim_{T\downarrow0} \frac{(T+\Delta)^{3H-\frac{1}{2}} - T^{3H-\frac{1}{2}}}{T^{3H-\frac{1}{2}} (3H-\frac{1}{2})} = \frac{2}{1-6H}.\]

By the curvature limit in Proposition 1, we have

\begin{alignat*}{2}& \lim_{T\downarrow0} \frac{\mathcal{C}_T}{T^{3H-\frac{1}{2}}}&&= \frac{2\Delta v_0}{3\left(\frac{v_0 \Delta^{H_{+}}}{2H_+}\right)^5 \psi(\rho,\nu,\eta,\chi)^{5}} \Bigg\{v_0^3\frac{\Delta^{3H_+}}{H_+^3} \mathfrak{w}^3 v_0 \frac{\chi\nu^3+\overline{\chi}\eta^3\rho^3}{\frac{1}{2}-3H}\\[3pt] & + 3v_0^3\frac{\Delta^{3H_+}}{H_+^3}&& \mathfrak{w}^2 \overline{\chi}\eta\overline{\rho} v_0 \frac{\overline{\chi}\eta^3\rho^2\overline{\rho}}{\frac{1}{2}-3H}\!+\! 3v_0^3\frac{\Delta^{3H_+}}{H_+^3} \mathfrak{w} \overline{\chi}^2\eta^2\overline{\rho}^2 v_0\frac{\overline{\chi}\eta^3\rho\overline{\rho}^2}{\frac{1}{2}-3H}\!+\! v_0\frac{\Delta^{3H_+}}{H_+^3} \overline{\chi}^3 \eta^3\overline{\rho}^3 v_0 \frac{\overline{\chi}\eta^3\overline{\rho}^3}{\frac{1}{2}-3H} \Bigg\} \\[3pt] & &&= \frac{128 \Delta^{-2H} H_+^2}{3\psi(\rho,\nu,\eta,\chi)^{5}(1-6H)} \\[3pt] & && \qquad \quad \Big\{ \mathfrak{w}^3 (\chi\nu^3+\overline{\chi}\eta^3\rho^3)+ 3 \mathfrak{w}^2 \overline{\chi}^2 \eta^4\overline{\rho}^2 \rho^2+ 3 \mathfrak{w}\overline{\chi}^3 \eta^5\overline{\rho}^4 \rho+ \overline{\chi}^4\eta^6\overline{\rho}^6 \Big\},\end{alignat*}

which yields the claim.

6.5.1. Proof of Proposition 4

Since $\phi$ and $u^{-p}$ are dominated by the conditions (i) and (iii) respectively, with the same notation as in the proof of Proposition 7, we obtain by (ii), as T goes to zero,

\begin{align*}\mathrm{D} \phi_s \lesssim (T-s)^{H_{-}}, \qquad\boldsymbol\Theta_s \lesssim (T-s)^{H_{+}}, \qquad\mathrm{D} \boldsymbol\Theta_s \lesssim (T-s)^{2H}, \qquad\mathrm{D} \mathrm{D} \boldsymbol\Theta_s \lesssim (T-s)^{3H-\frac{1}{2}}.\end{align*}

Under our three assumptions it is straightforward to see that $\boldsymbol{(\mathrm{H}_{12345})}$ are satisfied. Moreover, the terms in $\boldsymbol{(\mathrm{H}}_{\boldsymbol{6}}^{\lambda}\boldsymbol{)}$ behave as $T^{2H-\lambda}$ and the one in $\boldsymbol{(\mathrm{H}}_{\boldsymbol{7}}^{\lambda}\boldsymbol{)}$ as $T^{H_{-}{-\lambda}}$ , which means that if we set $\lambda=H_-$ the former vanishes and the second yields a non-trivial behaviour.

Let us have a look at the short-time implied volatility. By Lemma 3 and the continuity of v we have $\lim_{T\downarrow0} u_0 = \sqrt{\sum_{i=1}^N v_0 \rho_i^2}=\sqrt{v_0}$ almost surely; hence by Theorem 1 and the DCT,

\[\lim_{T\downarrow0}\widehat{\mathcal{I}}_{T} = \lim_{T\downarrow0} \mathbb{E}[u_0] = \sqrt{v_0}.\]

We then turn our attention to the short-time skew. With $\lambda=H_-$ , Theorem 2 and the DCT imply

\begin{align*}\lim_{T\downarrow0} \frac{\widehat{\mathcal{S}}_T}{T^{H_{-}}}=\sum_{i,j=1}^N\lim_{T\downarrow0} \frac{1}{2} \mathbb{E}\left[\frac{\int_0^T \phi_s^j \int_s^T \mathrm{D}_s^j (\phi_y^i)^2 \mathrm{d}y \mathrm{d}s}{u_0^3 T^{\frac32+H}}\right]&= \sum_{j=1}^N \frac{\rho_j}{2 v_0^{3/2}} \mathbb{E}\left[ \lim_{T\downarrow0} \frac{\int_0^T \int_s^T \sqrt{v_s}\, \mathrm{D}^j_{s} v_y\mathrm{d}y\mathrm{d}s}{T^{\frac32+H}} \right],\end{align*}

where we used $\sum_{i=1}^N \rho_i^2=1$ . For any $j\in [\![ 1,N ]\!]c$ , the Cauchy–Schwarz inequality yields

\[\mathbb{E}\left[ \left( \sqrt{\frac{v_s}{v_0}}-1\right) \mathrm{D}^j_s v_y \right] \le \mathbb{E}\left[ \left( \sqrt{\frac{v_s}{v_0}}-1\right)^2\right]^\frac{1}{2} \, \mathbb{E}\big[ (\mathrm{D}^j_s v_y)^2 \big]^\frac{1}{2},\]

where $\mathbb{E}\big[ (\mathrm{D}^j_s v_y)^2 \big]^\frac{1}{2}\le C (y-s)^{H_{-}}$ for some finite constant C by (ii). Therefore,

\begin{align*}\lim_{T\downarrow0} \bigg(&\frac{\int_0^T \int_s^T \mathbb{E}[ \sqrt{v_s}\, \mathrm{D}^j_{s} v_y]\mathrm{d}y\mathrm{d}s}{\sqrt{v_0}\,T^{\frac32+H}} - \frac{\int_0^T \int_s^T \mathbb{E}[ \mathrm{D}^j_{s} v_y]\mathrm{d}y\mathrm{d}s}{T^{\frac32+H}} \bigg) \\[3pt] &\qquad\qquad\le C \lim_{T\downarrow0}\left( \sup_{t\le T} \mathbb{E}\left[ \left( \sqrt{\frac{v_t}{v_0}}-1\right)^2\right]^\frac{1}{2} \frac{ \int_0^T \int_s^T (y-s)^{H_{-}}\mathrm{d}y\mathrm{d}s}{T^{\frac32+H}}\right).\end{align*}

Since the fraction is equal to $((H+\frac32)H_+)^{-1}$ and $\limsup_{T\downarrow0} \mathbb{E}\big[(\sqrt{v_t/v_0}-1)^2 \big]$ is null by (iv), we obtain

\[\lim_{T\downarrow0} \frac{\widehat{\mathcal{S}}_T}{T^{H_{-}}}= \sum_{j=1}^N \frac{\rho_j}{2 v_0} \mathbb{E}\left[ \lim_{T\downarrow0} \frac{\int_0^T \int_s^T \mathrm{D}^j_{s} v_y\mathrm{d}y\mathrm{d}s}{T^{\frac32+H}} \right]. \]

6.5.2. Proof of Corollary 1

Since

$$\mathbb{E}[u_s^{-p}] = \mathbb{E}\left[\left(\frac{1}{T-s}\int_s^T v_r\mathrm{d}r \right)^{-\frac{p}{2}}\right],$$

Lemmas 7 and 8 show that the assumptions (i)–(iii) of Proposition 4 hold. Moreover, v has almost-sure continuous paths; hence $\sqrt{\frac{v_t}{v_0}}$ tends to one almost surely and (iv) holds by reverse Fatou’s lemma. For $0\le s\le y$ , (33) implies

$$\mathbb{E}[\mathrm{D}^1_s v_y] = v_0 (y-s)^{H_{-}} \big(\chi \nu +\overline{\chi}\eta\rho\big)\qquad\text{and}\qquad\mathbb{E}[\mathrm{D}^2_s v_y] = v_0 (y-s)^{H_{-}}\overline{\chi}\eta\overline{\rho},$$

and clearly $\mathbb{E}[\mathrm{D}^3_s v_y] =0$ . Therefore, Proposition 4 implies

\[\lim_{T\downarrow0} \frac{\widehat{\mathcal{S}}_T}{T^{H_{-}}} = \frac{\rho_1}{2v_0}\frac{v_0(\chi\nu+\overline{\chi}\eta\rho)}{H_+(H+\frac32)} + \frac{\rho_2}{ 2v_0}\frac{v_0\overline{\chi}\eta\overline{\rho}}{H_+(H+\frac{3}{2})}= \frac{\rho_1 \chi\nu + \eta \overline{\chi} (\rho_1 \rho+\rho_2\overline{\rho})}{(2H_+)(H+\frac32)}.\]