Right now I am studying a paper of Sergii Kuchuk-Iatsenko and Yuliya Mishura, Pricing the European Call Option in the Model with Stochastic Volatility Driven by Ornstein-Uhlenbeck Process, Exact Formula. The paper states one possible analytics solution of the option price at $t=0$ is
$$V_0=Se^{rT}\bigg{(}\Phi(k)+\frac{1}{\sqrt{2\pi}}\bigg{(}\int_{k}^{\infty} (\mathbf{Q}(\sigma_0<\sigma_1(s))+\mathbf{Q}(\sigma_0>\sigma_2(s)))e^{-s^2/2}ds-\int_{-\infty}^{-k} (\mathbf{Q}(\sigma_0<\sigma_2(s))-\mathbf{Q}(\sigma_0<\sigma_1(s)))e^{-s^2/2}ds\bigg{)}\bigg{)}-K\bigg{(}\frac{1}{2}+\frac{1}{\sqrt{2\pi}}\bigg{(}\int_{0}^{\infty} (\mathbf{Q}(\sigma_0<\sigma_4(s))-\mathbf{Q}(\sigma_0<\sigma_3(s)))e^{-s^2/2}ds-\int_{-\infty}^{0} (\mathbf{Q}(\sigma_0<\sigma_3(s))+\mathbf{Q}(\sigma_0>\sigma_4(s)))e^{-s^2/2}ds\bigg{)}$$
$\sigma_0$ is a variable random and
$$\sigma_{1,2}(s)=\frac{s}{\sqrt{T}}\mp\frac{\sqrt{s^2T-2T(ln(S/K)+rT)}}{T}$$ $$\sigma_{3,4}(s)=\frac{-s}{\sqrt{T}}\mp\frac{\sqrt{s^2T+2T(ln(S/K)+rT)}}{T}$$
The paper states that under $\mathbf{Q}$, probability measure which is equivalent to objective measure $\mathbf{P}$,
$$\mathbf{Q}(\sigma_0<\sigma_j(s))=\mathbf{Q}(\sigma_0^2<\sigma_j^2(s))=\lim_{\varepsilon\to\infty} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\sigma_j^2(s)} \bigg{(}\int_{-\infty}^{\infty} e^{iyu-\frac{\varepsilon^2u^2}{2}}\phi(u)du\bigg{)}dy$$
where $\phi(u)$ is the characteristic function of $\sigma_0^2$.
I have 2 questions
- The writers states that $\sigma_i$ for $i=1,2,3,4$ is positive over its domain, and the integrals in the $V_0$ formula are positive. Is that the reason why $\mathbf{Q}(\sigma_0<\sigma_j(s))=\mathbf{Q}(\sigma_0^2<\sigma_j^2(s))$?
- Why is $\mathbf{Q}(\sigma_0^2<\sigma_j^2(s))=\lim_{\varepsilon\to\infty} \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\sigma_j^2(s)} \bigg{(}\int_{-\infty}^{\infty} e^{iyu-\frac{\varepsilon^2u^2}{2}}\phi(u)du\bigg{)}dy$? I know, for general, inverse fourier transform formula for PDF of variable random X is $f(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty} e^{-iuX}\phi(u)du$ where $\phi(u)=E[e^{iuX}]$ is a characteristic function of X. So based on the formula, $\mathbf{Q}(\sigma_0^2<\sigma_j^2(s))=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\sigma_j^2(s)} \bigg{(}\int_{-\infty}^{\infty} e^{iuy}\phi(u)du\bigg{)}dy$. Am I missing something here?