Analysis of variance based instruments for Ornstein–Uhlenbeck type models: swap and price index

Abstract

In this paper a couple of variance dependent instruments in the financial market are studied. Firstly, a number of aspects of the variance swap in connection to the Barndorff-Nielsen and Shephard model are studied. A partial integro-differential equation that describes the dynamics of the arbitrage-free price of the variance swap is formulated. Under appropriate assumptions for the first four cumulants of the driving subordinator, a Večeř-type theorem is proved. The bounds of the arbitrage-free variance swap price are also found. Finally, a price-weighted index modulated by market variance is introduced. The large-basket limit dynamics of the price index and the “error term” are derived. Empirical data driven numerical examples are provided in support of the proposed price index.

Appendices

A Appendix: Proof of Theorem 3.3

Proof

The dynamics of $I_n^{(k)}$ is given by (3.5). Let

$$\begin{aligned} B_n(t) =\int _{0}^{t}I_n^{(k)}(u)\left( k \alpha _1+\frac{k(k-1)}{2}\sigma ^2(u)-\lambda + \beta _k +\frac{\gamma }{\sigma ^2(u)}\right) du. \end{aligned}$$

Clearly, with $X_n = I_n^{(k)}$ we have $M_n = X_n -B_n$ a local martingale, where

$$\begin{aligned} M_n (t)&= \int _{0}^{t}k\sigma (u)\rho _1 I_n^{(k)}(u)dM(u)+ \frac{k\sqrt{1-\rho _1^2}}{n}\sum _{i=1}^{n}\int _{0}^{t}\sigma ^3(u)S_i^k(u)dW_i(u)\\&\quad + \sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^k(u)(e^{k\alpha _2 x}-1)\left( \frac{\sigma ^2(u)}{n}+\frac{x}{n^2}\right) \tilde{N}_i(du, dx)\\&\quad +\frac{1}{n^2}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^k(u)x\tilde{N}_j(du, dx). \end{aligned}$$

Next, we define $A_n(t) = [M_n, M_n](t)$. By construction, clearly $A_n(t)-A_n(s)$ is non-negative definite for $t \ge s \ge 0$. By Doob–Meyer decomposition $M_n^2-A_n$ is a local martingale. Since the jumps occur at distinct times almost surely, we have

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|X_n(t)-X_n(t-)|^2\right]&= {{\mathbb {E}}}\left[ \sup _{1\le i\le n} \sup _{t\le T}\left( \frac{1}{n}(\sigma ^2(t)S_i^k(t)-\sigma ^2(t-)S_i^k(t-))\right) ^2\right] \\&\le \frac{C^2}{n^2}{{\mathbb {E}}}\left[ \sup _{1\le i\le n} \sup _{t\le T}S_i^{2k}(t)\right] . \end{aligned}$$

Since the jump sizes $|A_n(t)-A_n(t-)|$ are essentially same as $|X_n(t)-X_n(t-)|^2$,

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|A_n(t)-A_n(t-)|\right] \le \frac{C^2}{n^2}{{\mathbb {E}}}\left[ \sup _{1\le i\le n} \sup _{t\le T}S_i^{2k}(t)\right] . \end{aligned}$$

Assumptions of this theorem ($E[S_i(0)^{4k}] < \infty$, $\int _{{\mathbb {R}}} e^{4k \alpha _2 x} \nu (dx) < \infty$, and for $t \in [0,T]$, $|\sigma (t)|^2\le C$) imply (see Hambly and Vaicenavicius 2015) that for $t\in [0,T]$, $E[S_i(t)^k]<\infty$. Hence by Hambly and Vaicenavicius (2015) (Lemma 4.3) we obtain

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|X_n(t)-X_n(t-)|^2\right] \rightarrow 0, \end{aligned}$$

and

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|A_n(t)-A_n(t-)|^2\right] \rightarrow 0, \end{aligned}$$

as $n\rightarrow \infty$. Also, since $B_n$ is continuous $\lim _{n\rightarrow \infty }{{\mathbb {E}}}\left[ \sup _{t \le T } |B_n(t)-B_n(t-)|^2\right] = 0$. We observe that

$$\begin{aligned}&B_n(t) -\int _{0}^{t}b(X_n(u))du \\&\quad =\int _{0}^{t}I_n^{(k)}(u)\left( k \alpha _1+\frac{k(k-1)}{2}\sigma ^2(u)-\lambda + \beta _k +\frac{\gamma }{\sigma ^2(u)} + \frac{\mu _k}{n\sigma ^2(u)}\right) du\\&\qquad -\int _{0}^{t}I_n^{(k)}(u)\left( k \alpha _1+\frac{k(k-1)}{2}\sigma ^2(u)-\lambda + \beta _k +\frac{\gamma }{\sigma ^2(u)}\right) dt\\&\quad =\frac{1}{n}\int _{0}^{t}\frac{\mu _k}{\sigma ^2(u)}I_n^{(k)}(u)du \; \rightarrow \; 0, \quad \text {in probability}, \end{aligned}$$

by Lemma 3.1 and the assumption of the theorem. Thus all the conditions except (3.6) are verified for Theorem 3.2. Next, we proceed to verify (3.6). We denote

$$\begin{aligned} G_n(t)&:= \frac{1}{n}\int _{0}^{t}k^2(1-\rho _1^2)\sigma ^4(u)I_n^{(2k)}(u)du \\ H_n^1(t)&:=\frac{1}{n^4}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_p^k(u)S_q^k(u)x^2N_i(du, dx),\\ H_n^2(t)&:=\frac{2}{n^3}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^{2k}(u)\sigma ^2(u)x(e^{k\alpha _2 x}-1)^2N_i(du, dx),\\ H_n^3(t)&:=\frac{1}{n^4}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^{2k}(u)x^2(e^{k\alpha _2 x}-1)^2N_i(du, dx),\\ H_n^4(t)&:= \frac{1}{n^2}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^{2k}(u)\sigma ^4(u)(e^{k\alpha _2 x}-1)^2N_i(du, dx),\\ H_n^5(t)&: = \frac{2}{n^4}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)x(e^{k\alpha _2 x}-1)N_j(du, dx),\\ H_n^6(t)&: = \frac{2}{n^3}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)\sigma ^2(u)x(e^{k\alpha _2 x}-1)N_j(du, dx). \end{aligned}$$

Consider

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}\left| A_n(t)-\int _{0}^{t}a(X_n(t))\right| ^2\right]&\le {{\mathbb {E}}}\left[ \left( \sup _{t\le T}|G_n(t)|+\sum _{i=1}^6 \sup _{t\le T}|H^i_n(t)| \right) ^2\right] \nonumber \\&\le 8{{\mathbb {E}}}\left[ \sup _{t\le T}|G_n(t)|^2\right] + 8 \sum _{i=1}^6 {{\mathbb {E}}}\left[ \sup _{t\le T}|H^i_n(t)|^2\right] . \end{aligned}$$

(5.1)

To verify (3.6), it is sufficient to show that ${{\mathbb {E}}}\left[ \sup _{t\le T}|G_n(t)|^2\right]$ and ${{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^i(t)|^2\right]$ for $1\le i \le 6$ converge to 0 as $n\rightarrow \infty$. Clearly, by Lemma 3.1 and boundedness of $\sigma ^2$, we obtain

$$\begin{aligned}&{{\mathbb {E}}}\left[ \sup _{t\le T}|G_n(t)|^2\right] \\&\quad = \frac{k^4(1-\rho _1^2)^2}{n}{{\mathbb {E}}}\left[ \sup _{t\le T}\left| \int _{0}^{t}\frac{\sigma ^4(u)}{\sqrt{n}}I_n^{(2k)}(u)du\right| ^2\right] \rightarrow 0, \quad \text {as} \quad n \rightarrow \infty . \end{aligned}$$

Also,

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|H^1_n(t)|^2\right]&= {{\mathbb {E}}}\left[ H_n^1(T)^2\right] = \frac{1}{n^6}{{\mathbb {E}}}\left[ \left( \sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{r=1}^{N^1(T)}S_p^k(\tau _r^1)S_q^k(\tau _r^1)({\mathcal {J}}_r^1)^2\right) ^2\right] \\&=\frac{1}{n^4}{{\mathbb {E}}}\left[ \left( \sum _{r=1}^{N^1(T)}S_1^{2k}(\tau _r^1)({\mathcal {J}}_r^1)^2\right) ^2\right] \\&\le \frac{1}{n^4}{{\mathbb {E}}}\left[ N^1(T)\sum _{r=1}^{N^1(T)}S_1^{4k}(\tau _r^1)({\mathcal {J}}_r^1)^4\right] \\&\le \frac{1}{n^4}\sum _{N=1}^{\infty }N^2\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^4]{\mathbb {P}}(N^1(T)=N)\\&\le \frac{1}{n^4}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^4]{{\mathbb {E}}}[N^1(T)^2] \rightarrow 0, \quad \text {as} \quad n \rightarrow \infty .\\ {{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^2(t)|^2\right]&= {{\mathbb {E}}}\left[ H_n^2(T)^2\right] \\&=\frac{4}{n^6}{{\mathbb {E}}}\left[ \left( \sum _{i=1}^{n}\int _{0}^{T}\int _{{\mathbb {R}}}S_i^{2k}(u)\sigma ^2(u)x(e^{k\alpha _2 x}-1)^2N_i(du, dx)\right) ^2\right] \\&\le \frac{4C}{n^5}{{\mathbb {E}}}\left[ \sum _{i=1}^{n}\left( \int _{0}^{T}\int _{{\mathbb {R}}}S_i^{2k}(u)x(e^{k\alpha _2 x}-1)^2N_i(du, dx)\right) ^2\right] \\&\le \frac{4C}{n^4}{{\mathbb {E}}}\left[ N^1(T)\sum _{j=1}^{N^1(T)}S_1^{4k}(\tau _j^1)({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^4\right] \\&\le \frac{4C}{n^4}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^4]{{\mathbb {E}}}[N^1(T)^2] \rightarrow 0,\\&\quad \text {as} \quad n \rightarrow \infty .\\ {{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^3(t)|^2\right]&= {{\mathbb {E}}}\left[ H_n^3(T)^2\right] \\&={{\mathbb {E}}}\left[ \left( \frac{1}{n^4}\sum _{i=1}^{n}\int _{0}^{T}\int _{{\mathbb {R}}}S_i^{2k}(u)x^2(e^{k\alpha _2 x}-1)^2N_i(du, dx)\right) ^2\right] \\&\le \frac{1}{n^7}{{\mathbb {E}}}\left[ \sum _{i=1}^{n}\left( \int _{0}^{T}\int _{{\mathbb {R}}}S_i^{2k}(u)x^2(e^{k\alpha _2 x}-1)^2N_i(du, dx)\right) ^2\right] \\&\le \frac{1}{n^7}{{\mathbb {E}}}\left[ \sum _{i=1}^{n}\left( \sum _{j=1}^{N^i(T)}S_i^{2k}(\tau _j^i)x^2(e^{k\alpha _2 {\mathcal {J}}_j^i}-1)^2\right) ^2\right] \\&\le \frac{1}{n^6}\sum _{N=1}^{\infty }N^2\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^4(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^4]\\&\quad {\mathbb {P}}(N^1(T)=N)\\&\le \frac{1}{n^6}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^4(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^4]{{\mathbb {E}}}[N^1(T)^2] \rightarrow 0, \\&\quad \text {as} \quad n \rightarrow \infty . \end{aligned}$$

In a similar procedure as in the case of $H_n^2(t)$, we can show

$$\begin{aligned}&{{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^4(t)|^2\right] \le \frac{C}{n^3}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^4]{{\mathbb {E}}}[N^1(T)^2] \rightarrow 0, \\&\quad \text {as} \quad n \rightarrow \infty . \end{aligned}$$

Next,

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^5(t)|^2\right]&= {{\mathbb {E}}}\left[ H_n^5(T)^2\right] \\&={{\mathbb {E}}}\left[ \left( \frac{2}{n^4}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{T}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)x(e^{k\alpha _2 x}-1)N_j(du, dx)\right) ^2\right] \\&\le \frac{4}{n^8}{{\mathbb {E}}}\left[ n^2\sum _{i=1}^{n}\sum _{j=1}^{n}\left( \int _{0}^{T}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)x(e^{k\alpha _2 x}-1)N_j(du, dx)\right) ^2\right] \\&\le \frac{4}{n^4}{{\mathbb {E}}}\left[ N^1(T)\sum _{j=1}^{N^1(T)}S_1^{4k}(\tau _r^1)({\mathcal {J}}_r^1)^2(e^{k\alpha _2 {\mathcal {J}}_r^1}-1)^2\right] \\&\le \frac{4}{n^4}\sum _{N=1}^{\infty }N^2\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^2]{\mathbb {P}}(N^1(T)=N)\\&\le \frac{4}{n^4}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^2]{{\mathbb {E}}}[N^1(T)^2] \rightarrow 0,\\&\quad \text {as} \quad n \rightarrow \infty . \end{aligned}$$

Finally,

$$\begin{aligned} {{\mathbb {E}}}\left[ \sup _{t\le T}|H_n^6(t)|^2\right]&= {{\mathbb {E}}}\left[ H_n^6(T)^2\right] \\&={{\mathbb {E}}}\left[ \left( \frac{2}{n^3}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{T}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)\sigma ^2(u)x(e^{k\alpha _2 x}-1)N_j(du, dx)\right) ^2\right] \\&\le \frac{4C}{n^6}{{\mathbb {E}}}\left[ n^2\sum _{i=1}^{n}\sum _{j=1}^{n}\left( \int _{0}^{T}\int _{{\mathbb {R}}}S_i^{k}(u)S_j^{k}(u)x(e^{k\alpha _2 x}-1)N_j(du, dx)\right) ^2\right] \\&\le \frac{4C}{n^2}{{\mathbb {E}}}\left[ N^1(T)\sum _{j=1}^{N^1(T)}S_1^{4k}(\tau _r^1)({\mathcal {J}}_r^1)^2(e^{k\alpha _2 {\mathcal {J}}_r^1}-1)^2\right] \\&\le \frac{4C}{n^2}\sum _{N=1}^{\infty }N^2\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^2]{\mathbb {P}}(N^1(T)=N)\\&\le \frac{4C}{n^2}\left( \sup _{t\le T}{{\mathbb {E}}}[S_1^{4k}(t)]\right) {{\mathbb {E}}}[({\mathcal {J}}_j^1)^2(e^{k\alpha _2 {\mathcal {J}}_j^1}-1)^2]{{\mathbb {E}}}[N^1(T)^2]\rightarrow 0, \\&\quad \text {as} \quad n \rightarrow \infty . \end{aligned}$$

Combining all these results we obtain from (5.1) that ${{\mathbb {E}}}\left[ \sup _{t\le T}\left| A_n(t)-\int _{0}^{t}a(X_n(t))\right| ^2\right] \rightarrow 0$ as $n \rightarrow \infty$. Hence (3.6) is verified and consequently all the assumptions in Theorem 3.2 are verified. Hence the proof is complete. $\square$

B Appendix: Proof of Lemma 3.4

Proof

By (3.7) jumps of $\Pi _n(t)$ are same as jumps of $\sqrt{n}I_n(t)$. Hence,

$$\begin{aligned}&{{\mathbb {E}}}\left[ \sup _{t\le T}|X_n(t)-X_n(t-)|^2\right] = {{\mathbb {E}}}\left[ \sup _{t\le T}\left( \frac{1}{\sqrt{n}}\sum _{i=1}^{n}(\sigma ^2(t)S_i(t)-\sigma ^2(t-)S_i(t-))\right) ^2\right. \\&\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \left. +\left( \frac{\xi ^2}{n}\sum _{i=1}^{n}(\sigma ^2(t)S_i^2(t)-\sigma ^2(t-)S_i^2(t-))\right) ^2 \right] \\&\quad \le C{{\mathbb {E}}}\left[ \sup _{t\le T}\left( \left( \frac{1}{\sqrt{n}}\sum _{i=1}^{n}(S_i(t)-S_i(t-))\right) ^2 +\left( \frac{\xi ^2}{n}\sum _{i=1}^{n}(S_i^2(t)-S_i^2(t-))\right) ^2\right) \right] \\&\quad \le \frac{C}{n}{{\mathbb {E}}}\left[ \sup _{t\le T}\sum _{i=1}^{n}(S_i(t)-S_i(t-))^2\right] + \frac{C\xi ^4}{n^2}{{\mathbb {E}}}\left[ \sup _{t\le T}\sum _{i=1}^{n}(S_i^2(t)-S_i^2(t-))^2\right] \\&\quad \le \frac{C}{n}{{\mathbb {E}}}\left[ \sup _{1\le i\le n}\sup _{t\le T}S_i^2(t)\right] + \frac{C\xi ^4}{n^2}{{\mathbb {E}}}\left[ \sup _{1\le i\le n}\sup _{t\le T}S_i^4(t)\right] , \end{aligned}$$

where we have used repeatedly the fact that no two jumps occur at the same time almost surely. Hence (3.14) is proved. Proof of (3.15) is similar and is skipped here.

We proceed to prove (3.16).

• (i) Case: $i=j=1$. We define

  $$\begin{aligned} U_n(t)&:= A_n^{11} -\int _{0}^{t}a_{11}(X_n(u))du = G_n^{11}+H_n^{11} -\int _{0}^{t}a_{11}(X_n(u))du\\&= -\xi ^2\int _{0}^{t}\sigma ^2(u) I_n^{(2)}(u) du\\&\quad + \frac{1}{n^3}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_p(u)S_q(u)x^2N_i(du, dx)\\&\quad +\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^2(u)(e^{\alpha _2 x}-1)^2\left( \frac{\sigma ^2(u)}{\sqrt{n}}+ \frac{x}{n^{\frac{3}{2}}}\right) ^2N_i(du, dx)\\&\quad + \frac{2}{n^{\frac{3}{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j(u)x(e^{\alpha _2 x}{-}1)\left( \frac{\sigma ^2(u)}{\sqrt{n}}{+} \frac{x}{n^{\frac{3}{2}}}\right) N_j(du, dx). \end{aligned}$$

  After simplification of the above expression, and using (3.11), we obtain the following expression for $U_n(t)$.

  $$\begin{aligned} U_n(t)&= \frac{1}{n^3}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_p(u)S_q(u)x^2N_i(du, dx) \nonumber \\&\quad +\frac{1}{n^3}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^2(u)x^2(e^{\alpha _2 x}-1)^2N_i(du, dx) \nonumber \\&\quad +\frac{1}{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^2(u)\sigma ^4(u)(e^{\alpha _2 x}-1)^2\tilde{N}_i(du, dx) \nonumber \\&\quad +\frac{2}{n^2}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i^2(u)\sigma ^2(u)x(e^{\alpha _2 x}-1)^2N_i(du, dx) \nonumber \\&\quad + \frac{2}{n^4}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j(u)x^2(e^{\alpha _2 x}-1)N_j(du, dx) \nonumber \\&\quad + \frac{2}{n^2}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j(u)\sigma ^2(u)x(e^{\alpha _2 x}-1)N_j(du, dx). \end{aligned}$$

  (6.1)

  A similar proof for Theorem 3.3 (in particular, the analysis related to $\sup _{t \le T} |H_n^i(t)|^2$, for $i=1,2,3,4,5,6$) can be used to show that for $0 \le t \le T$, each of the terms in the right hand side of (6.1) is converging to 0 in probability as $n \rightarrow \infty$. Consequently we conclude that ${{\mathbb {E}}}[\sup _{t \le T} |U_n(t)|^2]\rightarrow 0$ as $n \rightarrow \infty$.

• (ii) Case: $i=1$, $j=2$. In this case clearly $\int _{0}^{t}a_{12}(X_n(u))du =0$.

  $$\begin{aligned}&A_n^{12}(t) - \int _{0}^{t}a_{12}(X_n(u))du = G_n^{12}(t) + H_n^{12}(t)- \int _{0}^{t}a_{12}(X_n(u))du\\&\quad = \frac{\xi ^2}{n^{\frac{7}{2}}}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_p(u)S_q^2(u)x^2N_i(du, dx)\\&\qquad + \frac{\xi ^2}{n^{\frac{3}{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j^2(u)x(e^{2\alpha _2 x}-1)\left( \frac{\sigma ^2(u)}{n}+ \frac{x}{n^2}\right) N_j(du, dx) \\&\qquad +\frac{\xi ^2}{n^2}\sum _{i=1}^{n}\sum _{j=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j^2(u)x(e^{\alpha _2 x}-1)\left( \frac{\sigma ^2(u)}{\sqrt{n}}+ \frac{x}{n^{\frac{3}{2}}}\right) N_j(du, dx) \\&\qquad + \xi ^2\sum _{i=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^3(u)(e^{2\alpha _2 x}-1)(e^{\alpha _2 x}-1)\left( \frac{\sigma ^2(u)}{\sqrt{n}}+ \frac{x}{n^{\frac{3}{2}}}\right) \\&\qquad \times \,\left( \frac{\sigma ^2(u)}{n}+ \frac{x}{n^2}\right) N_i(du, dx), \end{aligned}$$

  and this can be simplified to

  $$\begin{aligned}&A_n^{12}(t) - \int _{0}^{t}a_{12}(X_n(u))du =\frac{\xi ^2}{n^{\frac{7}{2}}}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_p(u)S_q^2(u)x^2N_i(du, dx)\\&\qquad + \frac{2\xi ^2}{n^{\frac{5}{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j^2(u)\sigma ^2(u)x(e^{2\alpha _2 x}-1)N_j(du, dx) \\&\qquad + \frac{2\xi ^2}{n^{\frac{7}{2}}}\sum _{i=1}^{n}\sum _{j=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i(u)S_j^2(u)x^2(e^{2\alpha _2 x}-1)N_j(du, dx)\\&\qquad + \frac{\xi ^2}{n^{\frac{3}{2}}}\sum _{i=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^3(u)\sigma ^4(u)(e^{2\alpha _2 x}-1)(e^{\alpha _2 x}-1)N_i(du, dx)\\&\qquad + \frac{2\xi ^2}{n^{\frac{5}{2}}}\sum _{i=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^3(u)\sigma ^2(u)x(e^{2\alpha _2 x}-1)(e^{\alpha _2 x}-1)N_i(du, dx)\\&\qquad +\frac{\xi ^2}{n^{\frac{7}{2}}}\sum _{i=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^3(u)x^2(e^{2\alpha _2 x}-1)(e^{\alpha _2 x}-1)N_i(du, dx). \end{aligned}$$

  Once again, a similar procedure as in Theorem 3.3 (in particular, the analysis related to $\sup _{t \le T} |H_n^i(t)|^2$, for $i=1,2,3,4,5,6$) can be used to show that for $0 \le t \le T$, each of the terms in the right hand side of the above expression is converging to 0 in probability as $n \rightarrow \infty$.

• (iii) Case: $i=j=2$. We have $\int _{0}^{t}a_{22}(X_n(u))du = \xi ^4\int _{0}^{t}4\sigma ^2(u)(I_n^{(2)}(u))^2du$.

  $$\begin{aligned}&A_n^{22}(t) - \int _{0}^{t}a_{22}(X_n(u))du = G_n^{22}(t) + H_n^{22}(t)- \int _{0}^{t}a_{22}(X_n(u))du\\&\quad =\xi ^4\int _{0}^{t}4\sigma ^2(u)(I_n^{(2)}(u))^2du -\xi ^4\int _{0}^{t}4\sigma ^2(u)(I_n^{(2)}(u))^2du\\&\qquad +\frac{\xi ^4}{n^4}\sum _{p=1}^{n}\sum _{q=1}^{n}\sum _{i=1}^{n}\int _{0}^{t}\int _{{\mathbb {R}}}S_q^2(u)S_q^2(u)x^2N_i(du, dx) \\&\qquad +\xi ^4\sum _{i=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^4(u-)(e^{2\alpha _2 x}-1)^2\left( \frac{\sigma ^2(u)}{n}+ \frac{x}{n^2}\right) ^2N_i(du, dx)\\&\qquad +\frac{2\xi ^4}{n^2}\sum _{i=1}^{n}\sum _{j=1}^{n} \int _{0}^{t}\int _{{\mathbb {R}}}S_i^2(u)S_j^2(u)x(e^{2\alpha _2 x}-1)\left( \frac{\sigma ^2(u)}{n}+ \frac{x}{n^2}\right) N_j(du, dx). \end{aligned}$$

  After expanding the above expression, a similar procedure as in Theorem 3.3 (in particular, the analysis related to $\sup _{t \le T} |H_n^i(t)|^2$, for $i=1,2,3,4,5,6$) can be implemented to show that for $0 \le t \le T$, each of the terms in the right hand side of the above expression is converging to 0 in probability as $n \rightarrow \infty$.

Combining all the above three cases we complete the proof of (3.16). $\square$